12294 lines
302 KiB
Markdown
12294 lines
302 KiB
Markdown
**Exercise Set 5.1**
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Page 296
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Write the first four terms of the sequences defined by the formulas 1-6.
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1. $a_k = \dfrac{k}{10 + k}$, for every integer $k \geq 1$.
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$$ a_1 = \frac{1}{10 + 1} = \frac{1}{11} $$
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$$ a_2 = \frac{2}{10 + 2} = \frac{2}{12} $$
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$$ a_3 = \frac{3}{10 + 3} = \frac{3}{13} $$
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$$ a_4 = \frac{4}{10 + 4} = \frac{4}{14} $$
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2. $b_j = \dfrac{5 - j}{5 + j}$, for every integer $j \geq 1$.
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$$ b_1 = \dfrac{5 - 1}{5 + 1} = \frac{4}{6} $$
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$$ b_2 = \dfrac{5 - 2}{5 + 2} = \frac{3}{7} $$
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$$ b_3 = \dfrac{5 - 3}{5 + 3} = \frac{2}{8} $$
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$$ b_4 = \dfrac{5 - 4}{5 + 4} = \frac{1}{9} $$
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3. $c_i = \dfrac{(-1)^i}{3^i}$, for every integer $i \geq 0$.
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$$ c_0 = \dfrac{(-1)^0}{3^0} = \frac{1}{1} $$
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$$ c_1 = \dfrac{(-1)^1}{3^1} = \frac{-1}{3} $$
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$$ c_2 = \dfrac{(-1)^2}{3^2} = \frac{1}{9} $$
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$$ c_3 = \dfrac{(-1)^3}{3^3} = \frac{-1}{27} $$
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4. $d_m = 1 + \left(\dfrac{1}{2}\right)^m$ for every integer $m \geq 0$.
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$$ d_0 = 1 + \left(\dfrac{1}{2}\right)^0 = 1 $$
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$$ d_1 = 1 + \left(\dfrac{1}{2}\right)^1 = \frac{1}{2} $$
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$$ d_2 = 1 + \left(\dfrac{1}{2}\right)^2 = \frac{1}{4} $$
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$$ d_3 = 1 + \left(\dfrac{1}{2}\right)^3 = \frac{1}{8} $$
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5. $e_n = \left\lfloor \dfrac{n}{2} \right\rfloor \cdot 2$, for every integer
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$n \geq 0$.
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$$ e_0 = \left\lfloor \dfrac{0}{2} \right\rfloor \cdot 2 = 0 $$
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$$ e_1 = \left\lfloor \dfrac{1}{2} \right\rfloor \cdot 2 = 0 $$
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$$ e_2 = \left\lfloor \dfrac{2}{2} \right\rfloor \cdot 2 = 2 $$
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$$ e_3 = \left\lfloor \dfrac{3}{2} \right\rfloor \cdot 2 = 2 $$
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6. $f_n = \left\lfloor \dfrac{n}{4} \right\rfloor \cdot 4$, for every integer
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$n \geq 1$.
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$$ f_1 = \left\lfloor \dfrac{1}{4} \right\rfloor \cdot 4 = 0 $$
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$$ f_2 = \left\lfloor \dfrac{2}{4} \right\rfloor \cdot 4 = 0 $$
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$$ f_3 = \left\lfloor \dfrac{3}{4} \right\rfloor \cdot 4 = 0 $$
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$$ f_4 = \left\lfloor \dfrac{4}{4} \right\rfloor \cdot 4 = 4 $$
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7. Let $a_k = 2k + 1$ and $b_k = (k - 1)^3 + k + 2$ for every integer
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$k \geq 0$. Show that the first three terms of these sequences are identical
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but that their fourth terms differ.
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$$ a_0 = 2(0) + 1 = 1 $$
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$$ a_1 = 2(1) + 1 = 3 $$
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$$ a_2 = 2(2) + 1 = 5 $$
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$$ a_3 = 2(3) + 1 = 7 $$
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$$ b_0 = (0 - 1)^3 + 0 + 2 = 1 $$
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$$ b_1 = (1 - 1)^3 + 1 + 2 = 3 $$
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$$ b_2 = (2 - 1)^3 + 2 + 2 = 5 $$
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$$ b_3 = (3 - 1)^3 + 3 + 2 = 13 $$
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Compute the first fifteen terms of each of the sequences in 8 and 9, and
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describe the general behavior of these sequences in words. (A definition of
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logarithm is given in Section 7.1.)
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8. $g_n = \lfloor \log_{2}n \rfloor$ for every integer $n \geq 1$.
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$$ g_1 = \lfloor \log_{2}(1) \rfloor = 0 $$
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$$ g_2 = \lfloor \log_{2}(2) \rfloor = 1 $$
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$$ g_3 = \lfloor \log_{2}(3) \rfloor = 1 $$
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$$ g_4 = \lfloor \log_{2}(4) \rfloor = 2 $$
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$$ g_5 = \lfloor \log_{2}(5) \rfloor = 2 $$
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$$ g_6 = \lfloor \log_{2}(6) \rfloor = 2 $$
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$$ g_7 = \lfloor \log_{2}(7) \rfloor = 2 $$
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$$ g_8 = \lfloor \log_{2}(8) \rfloor = 3 $$
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$$ g_9 = \lfloor \log_{2}(9) \rfloor = 3 $$
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$$ g_{10} = \lfloor \log_{2}(10) \rfloor = 3 $$
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$$ g_{11} = \lfloor \log_{2}(11) \rfloor = 3 $$
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$$ g_{12} = \lfloor \log_{2}(12) \rfloor = 3 $$
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$$ g_{13} = \lfloor \log_{2}(13) \rfloor = 3 $$
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$$ g_{14} = \lfloor \log_{2}(14) \rfloor = 3 $$
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$$ g_{15} = \lfloor \log_{2}(15) \rfloor = 3 $$
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The general behavior of this sequence is that it increments in binary
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increments, as in it increments every 1, then 2, then 4, then 8 iterations of
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the index $n$.
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9 $h_n = n\lfloor \log_{2}n \rfloor$ for every integer $n \geq 1$.
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$$ h_1 = (1)\lfloor \log_{2}(1) \rfloor = 0 $$
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$$ h_2 = (2)\lfloor \log_{2}(2) \rfloor = 2 $$
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$$ h_3 = (3)\lfloor \log_{2}(3) \rfloor = 3 $$
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$$ h_4 = (4)\lfloor \log_{2}(4) \rfloor = 8 $$
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$$ h_5 = (5)\lfloor \log_{2}(5) \rfloor = 10 $$
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$$ h_6 = (6)\lfloor \log_{2}(6) \rfloor = 12 $$
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$$ h_7 = (7)\lfloor \log_{2}(7) \rfloor = 14 $$
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$$ h_8 = (8)\lfloor \log_{2}(8) \rfloor = 24 $$
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$$ h_9 = (9)\lfloor \log_{2}(9) \rfloor = 27 $$
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$$ h_{10} = (10)\lfloor \log_{2}(10) \rfloor = 30 $$
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$$ h_{11} = (11)\lfloor \log_{2}(11) \rfloor = 33 $$
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$$ h_{12} = (12)\lfloor \log_{2}(12) \rfloor = 36 $$
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$$ h_{13} = (13)\lfloor \log_{2}(13) \rfloor = 39 $$
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$$ h_{14} = (14)\lfloor \log_{2}(14) \rfloor = 42 $$
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$$ h_{15} = (15)\lfloor \log_{2}(15) \rfloor = 45 $$
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The sequence finds the minimal (floor) power of $log_{2}n$ and then multiplies
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it by $n$, which is why there are sudden "jumps" when the floor calculates a
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jump to the next power of $2$. For example, at $n = 7$ to $n = 8$, there is a
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noticeable jump because $\lfloor \log_{2}7 \rfloor$ is $2$, and then
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$\lfloor \log_{2}8 \rfloor$ is $3$.
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Find explicit formulas for sequences of the form $a_1, a_2, a_3, \dots$ with the
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initial terms given in 10-16.
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10. $-1, 1, -1, 1, -1, 1$
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$a_n = (-1)^n$ where $n$ is an integer such that $n \geq 1$.
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11. $0, 1, -2, 3, -4, 5$
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$a_n = (n - 1)(-1)^{n}$ where $n$ is an integer such that $n \geq 1$.
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12. $\dfrac{1}{4}, \dfrac{2}{9}, \dfrac{3}{16}, \dfrac{4}{25}, \dfrac{5}{36}, \dfrac{6}{49}$
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$a_n = \dfrac{n}{(n + 1)^2}$ where $n$ is an integer such that $n \geq 1$.
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13. $1 - \dfrac{1}{2}, \dfrac{1}{2} - \dfrac{1}{3}, \dfrac{1}{3} - \dfrac{1}{4}, \dfrac{1}{4} - \dfrac{1}{5}, \dfrac{1}{5} - \dfrac{1}{6}, \dfrac{1}{6} - \dfrac{1}{7}$
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$a_n = \dfrac{1}{n} - \dfrac{1}{n + 1}$ where $n$ is an integer such that
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$n \geq 1$.
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14. $\dfrac{1}{3}, \dfrac{4}{9}, \dfrac{9}{27}, \dfrac{16}{81}, \dfrac{25}{243}, \dfrac{36}{729}$
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$a_n = \dfrac{n^2}{3^n}$ where $n$ is an integer such that $n \geq 1$.
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15. $0, -\dfrac{1}{2}, \dfrac{2}{3}, -\dfrac{3}{4}, \dfrac{4}{5}, -\dfrac{5}{6}, \dfrac{6}{7}$
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$a_n = \dfrac{(n - 1)(-1)^{n + 1}}{n}$ where $n$ is an integer such that
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$n \geq 1$.
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16. $3, 6, 12, 24, 48, 96$
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$a_n = 3 \cdot 2^{n - 1}$ where $n$ is an integer such that $n \geq 1$.
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17. Consider the sequence defined by $a_n = \dfrac{2n + (-1)^n - 1}{4}$ for
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every integer $n \geq 0$. Find an alternative explicit formula for $a_n$
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that uses the floor notation.
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Omitted.
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18. Let
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$a_0 = 2, a_1 = 3, a_2 = -2, a_3 = 1, a_4 = 0, a_5 = -1, \text{ and } a_6 = -2$.
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Compute each of the summations and products below.
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a. $\sum_{i = 0}^{6}{a_i}$
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$$ \sum_{i = 0}^{6}{a_i} = 2 + 3 + (-2) + 1 + 0 + (-1) + (-2) = 1 $$
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b. $\sum_{i = 0}^{0}{a_i}$
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$$ \sum_{i = 0}^{0}{a_i} = 2 $$
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c. $\sum_{j = 1}^{3}{a_{2j}}$
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$$ \sum_{j = 1}^{3}{a_{2j}} = (-2) + 0 + (-2) = -4 $$
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d. $\prod_{k = 0}^{6}{a_k}$
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$$ \prod_{k = 0}^{6}{a_k} = 2 \cdot 3 \cdot (-2) \cdot 1 \cdot 0 \cdot (-1) \cdot (-2) = 0 $$
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e. $\prod_{k = 2}^{2}{a_k}$
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$$ \prod_{k = 2}^{2}{a_k} = -2 $$
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Compute the summations and products in 19-28.
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19. $\sum_{k = 1}^{5}{(k + 1)}$
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$$ \sum_{k = 1}^{5}{(k + 1)} = (1 + 1) + (2 + 1) + (3 + 1) + (4 + 1) + (5 + 1) = 20 $$
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20. $\prod_{k = 2}^{4}{k^2}$
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$$ \prod_{k = 2}^{4}{k^2} = (2)^2 \cdot (3)^2 \cdot (4)^2 = 576 $$
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21. $\sum_{k = 1}^{3}{(k^2 + 1)}$
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$$ \sum_{k = 1}^{3}{(k^2 + 1)} = ((1)^2 + 1) + ((2)^2 + 1) + ((3)^2 + 1) = 17 $$
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22. $\prod_{j = 0}^{4}{(-1)^j}$
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$$ \prod_{j = 0}^{4}{(-1)^j} = (-1)^{(0)} \cdot (-1)^{(1)} \cdot (-1)^{(2)} \cdot (-1)^{(3)} \cdot (-1)^{(4)} = 1 $$
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23. $\sum_{i = 1}^{1}{i(i + 1)}$
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$$ \sum_{i = 1}^{1}{i(i + 1)} = (1)((1) + 1) = 2 $$
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24. $\sum_{j = 0}^{0}{(j + 2) \cdot 2^j}$
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$$ \sum_{j = 0}^{0}{(j + 2) \cdot 2^j} = (0 + 2) \cdot 2^0 = 2 $$
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25. $\prod_{k = 2}^{2}{\left(1 - \dfrac{1}{k}\right)}$
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$$ \prod_{k = 2}^{2}{\left(1 - \dfrac{1}{k}\right)} = \left(1 - \dfrac{1}{2}\right) = \frac{1}{2} $$
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26. $\sum_{k = -1}^{1}{(k^2 + 3)}$
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$$ \sum_{k = -1}^{1}{(k^2 + 3)} = ((-1)^2 + 3) + ((0)^2 + 3) + ((1)^2 + 3) = 11 $$
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27. $\sum_{n = 1}^{6}{\left(\dfrac{1}{n} - \dfrac{1}{n + 1}\right)}$
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$$ \sum_{n = 1}^{6}{\left(\dfrac{1}{n} - \dfrac{1}{n + 1}\right)} = \left(\dfrac{1}{(1)} - \dfrac{1}{(1) + 1}\right) + \left(\dfrac{1}{(2)} - \dfrac{1}{(2) + 1}\right) + \left(\dfrac{1}{(3)} - \dfrac{1}{(3) + 1}\right) + \left(\dfrac{1}{(4)} - \dfrac{1}{(4) + 1}\right) + \left(\dfrac{1}{(5)} - \dfrac{1}{(5) + 1}\right) + \left(\dfrac{1}{(6)} - \dfrac{1}{(6) + 1}\right) = \frac{6}{7} $$
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28. $\prod_{i = 2}^{5}{\dfrac{i(i + 2)}{(i - 1) \cdot (i + 1)}}$
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$$ \prod_{i = 2}^{5}{\dfrac{i(i + 2)}{(i - 1) \cdot (i + 1)}} = \dfrac{(2)((2) + 2)}{((2) - 1) \cdot ((2) + 1)} + \dfrac{(3)((3) + 2)}{((3) - 1) \cdot ((3) + 1)} + \dfrac{(4)((4) + 2)}{((4) - 1) \cdot ((4) + 1)} + \dfrac{(5)((5) + 2)}{((5) - 1) \cdot ((5) + 1)} = \frac{35}{3} $$
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Write the summations in 29-32 in expanded form.
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29. $\sum_{i = 1}^{n}{(-2)^i}$
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$$ \sum_{i = 1}^{n}{(-2)^i} = (-2)^1 + (-2)^2 + (-2)^3 + \dots + (-2)^{n} $$
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30. $\sum_{j = 1}^{n}{j(j + 1)}$
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$$ \sum_{j = 1}^{n}{j(j + 1)} = ((1)((1) + 1)) + ((2)((2) + 1)) + ((3)((3) + 1)) + \dots + ((n)((n) + 1)) = 2 + 6 + 12 + \dots n(n + 1) $$
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31. $\sum_{k = 0}^{n + 1}{\dfrac{1}{k!}}$
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$$ \sum_{k = 0}^{n + 1}{\dfrac{1}{k!}} = \dfrac{1}{0!} + \dfrac{1}{1!} + \dfrac{1}{2!} + \dots + \dfrac{1}{(n + 1)!} $$
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32. $\sum_{i = 1}^{k + 1}{i(i!)}$
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$$ \sum_{i = 1}^{k + 1}{i(i!)} = 1(1!) + 2(2!) + 3(3!) + \dots + (k + 1)((k + 1)!) $$
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Evaluate the summations and products in 33-36 for the indicated values of the
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variable.
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33. $\dfrac{1}{1^2} + \dfrac{1}{2^2} + \dfrac{1}{3^2} + \dots + \dfrac{1}{n^2}; n = 1$
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$$ \frac{1}{1^2} = 1 $$
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34. $1(1!) + 2(2!) + 3(3!) + \dots + m(m!); m = 2$
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$$ 1(1!) + 2(2!) = 5 $$
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35. $\left(\dfrac{1}{1 + 1}\right)\left(\dfrac{2}{2 + 1}\right)\left(\dfrac{3}{3 + 1}\right) \dots \left(\dfrac{k}{k + 1}\right); k = 3$
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$$ \left(\dfrac{1}{1 + 1}\right)\left(\dfrac{2}{2 + 1}\right)\left(\dfrac{3}{3 + 1}\right) = \frac{1}{4} $$
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36. $\left(\dfrac{1 \cdot 2}{3 \cdot 4}\right)\left(\dfrac{4 \cdot 5}{6 \cdot 7}\right)\left(\dfrac{6 \cdot 7}{8 \cdot 9}\right) \dots \left(\dfrac{m \cdot (m + 1)}{(m + 2) \cdot (m + 3)}\right); m = 1$
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$$ \left(\frac{1 \cdot 2}{3 \cdot 4}\right) = \frac{2}{12} = \frac{1}{6} $$
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Write each of 37-39 as a single summation.
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37. $\sum_{i = 1}^{k}{i^3 + (k + 1)^3}$
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$$ \sum_{i = 1}^{k}{i^3 + (k + 1)^3} = \sum_{i = 1}^{k + 1}{i^3} $$
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38. $\sum_{k = 1}^{m}{\dfrac{k}{k + 1} + \dfrac{m + 1}{m + 2}}$
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$$ \sum_{k = 1}^{m}{\dfrac{k}{k + 1} + \dfrac{m + 1}{m + 2}} = \sum_{k = 1}^{m + 1}{\dfrac{k}{k + 1}} $$
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39. $\sum_{m = 0}^{n}{(m + 1)2^m + (n + 2)2^{n + 1}}$
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$$ \sum_{m = 0}^{n}{(m + 1)2^m + (n + 2)2^{n + 1}} = \sum_{m = 0}^{n + 1}{(m + 1)2^m} $$
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Rewrite 40-42 by separating off the final term.
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40. $\sum_{i = 1}^{k + 1}{i(i!)}$
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$$ \sum_{i = 1}^{k + 1}{i(i!)} = \sum_{i = 1}^{k}{i(i!) + (k + 1)(k + 1)!} $$
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41. $\sum_{k = 1}^{m + 1}{k^2}$
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$$ \sum_{k = 1}^{m + 1}{k^2} = \sum_{k = 1}^{m}{k^2 + (m + 1)^2} $$
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42. $\sum_{m = 1}^{n + 1}{m(m + 1)}$
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$$ \sum_{m = 1}^{n + 1}{m(m + 1)} = \sum_{m = 1}^{n}{m(m + 1) + (n + 1)(n + 2)} $$
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Write each of 43-52 using summation or product notation.
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43. $1^2 - 2^2 + 3^2 - 4^2 + 5^2 - 6^2 + 7^2$
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$$ 1^2 - 2^2 + 3^2 - 4^2 + 5^2 - 6^2 + 7^2 $$
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$$ \sum_{k = 1}^{7}{(-1)^{k + 1}k^2} $$
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44. $(1^3 - 1) - (2^3 - 1) + (3^3 - 1) - (4^3 - 1) + (5^3 - 1)$
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$$ \sum_{k = 1}^{5}{(-1)^{k + 1}(k^3 - 1)} $$
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45. $(2^2 - 1) \cdot (3^2 - 1) \cdot (4^2 - 1)$
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$$ \prod_{k = 2}^{4}{(k^2 - 1)} $$
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46. $\dfrac{2}{3 \cdot 4} - \dfrac{3}{4 \cdot 5} + \dfrac{4}{5 \cdot 6} - \dfrac{5}{6 \cdot 7} + \dfrac{6}{7 \cdot 8}$
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$$ \sum_{k=2}^{6}{\dfrac{(-1)^k \cdot k}{(k + 1) \cdot (k + 2)}} $$
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47. $1 - r + r^2 - r^3 + r^4 - r^5$
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$$ \sum_{k = 0}^{5}{(-1)^kr^{k + 1}} $$
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48. $(1 - t) \cdot (1 - t^2) \cdot (1 - t^3) \cdot (1 - t^4)$
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$$ \prod_{k = 1}^{4}{(1 - t^k)} $$
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49. $1^3 + 2^3 + 3^3 + \dots + n^3$
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$$ \sum_{k}^{n}{k^3} $$
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50. $\dfrac{1}{2!} + \dfrac{2}{3!} + \dfrac{3}{4!} + \dots + \dfrac{n}{(n + 1)!}$
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$$ \sum_{k = 1}^{n}{\frac{k}{(k + 1)!}} $$
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51. $n + (n - 1) + (n - 2) + \dots + 1$
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$$ \sum_{k = 0}^{n - 1}{(n - k)} $$
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52. $n + \dfrac{n - 1}{2!} + \dfrac{n - 2}{3!} + \dfrac{n - 3}{4!} + \dots + \dfrac{1}{n!}$
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$$ \sum_{k = 0}^{n - 1}{\frac{n - k}{(k + 1)!}} $$
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Transform each of 53 and 54 by making the change of variable $i = k + 1$.
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$$ i = k + 1 $$
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$$ i - 1 = k $$
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53. $\sum_{k = 0}^{5}{k(k - 1)}$
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$$ \sum_{k = 0}^{5}{k(k - 1)} = \sum_{i = 1}^{6}{(i - 1)(i - 2)} $$
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54. $\prod_{k = 1}^{n}{\dfrac{k}{k^2 + 4}}$
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$$ \prod_{k = 1}^{n}{\dfrac{k}{k^2 + 4}} = \prod_{i = 2}^{n + 1}{\frac{i - 1}{(i - 1)^2 + 4}} $$
|
|
|
|
Transform each of 55-58 by making the change of variable $j = i - 1$.
|
|
|
|
$$ j = i - 1 $$
|
|
|
|
$$ i = j + 1 $$
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|
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|
55. $\sum_{i = 1}^{n + 1}{\dfrac{(i - 1)^2}{i \cdot n}}$
|
|
|
|
$$ \sum_{i = 1}^{n + 1}{\dfrac{(i - 1)^2}{i \cdot n}} = \sum_{j = 0}^{n}{\frac{j^2}{jn + n}} $$
|
|
|
|
56. $\sum_{i = 3}^{n}{\dfrac{i}{i + n - 1}}$
|
|
|
|
$$ \sum_{i = 3}^{n}{\dfrac{i}{i + n - 1}} = \sum_{j = 2}^{n - 1}{\frac{j + 1}{j + n}} $$
|
|
|
|
57. $\sum_{i = 1}^{n - 1}{\dfrac{i}{(n - i)^2}}$
|
|
|
|
$$ \sum_{i = 1}^{n - 1}{\dfrac{i}{(n - i)^2}} = \sum_{j = 0}^{n - 2}{\frac{j + 1}{(n - j - 1)^2}} $$
|
|
|
|
58. $\prod_{i = n}^{2n}{\dfrac{n - i + 1}{n + i}}$
|
|
|
|
$$ \prod_{i = n}^{2n}{\dfrac{n - i + 1}{n + i}} = \prod_{j = n - 1}^{2n - 1}{\frac{n - j}{n + j + 1}} $$
|
|
|
|
Write each of 59-61 as a single summation or product.
|
|
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|
59. $3 \cdot \sum_{k = 1}^{n}{(2k - 3)} + \sum_{k = 1}^{n}{(4 - 5k)}$
|
|
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|
$$ \sum_{k = 1}^{n}{3(2k - 3) + (4 - 5k)} $$
|
|
|
|
$$ \sum_{k = 1}^{n}{(6k - 9 + 4 - 5k)} $$
|
|
|
|
$$ \sum_{k = 1}^{n}{(k - 5)} $$
|
|
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|
60. $2 \cdot \sum_{k = 1}^{n}{(3k^2 + 4)} + 5 \cdot \sum_{k = 1}^{n}{(2k^2 - 1)}$
|
|
|
|
$$ \sum_{k = 1}^{n}{2(3k^2 + 4) + 5(2k^2 - 1)} $$
|
|
|
|
$$ \sum_{k = 1}^{n}{(6k^2 + 8 + 10k^2 - 5)} $$
|
|
|
|
$$ \sum_{k = 1}^{n}{(16k^2 + 3)} $$
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|
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|
61. $\left(\prod_{k = 1}^{n}{\dfrac{k}{k + 1}}\right) \cdot \left(\prod_{k = 1}^{n}{\dfrac{k + 1}{k + 2}}\right)$
|
|
|
|
$$ \prod_{k = 1}^{n}{\left(\frac{k}{k + 1}\right)\left(\frac{k + 1}{k + 2}\right)} $$
|
|
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|
$$ \prod_{k = 1}^{n}{\left(\frac{k(k + 1)}{(k + 1)(k + 2)}\right)} $$
|
|
|
|
$$ \prod_{k = 1}^{n}{\left(\frac{k\cancel{(k + 1)}}{\cancel{(k + 1)}(k + 2)}\right)} $$
|
|
|
|
$$ \prod_{k = 1}^{n}{\left(\frac{k}{k + 2}\right)} $$
|
|
|
|
Compute each of 62-76. Assume the values of the variables are restricted so that
|
|
the expressions are defined.
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|
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|
62. $\dfrac{4!}{3!}$
|
|
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|
$$ \frac{4!}{3!} = \frac{4 \cdot 3 \cdot 2 \cdot 1}{3 \cdot 2 \cdot 1} = 4 $$
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|
63. $\dfrac{6!}{8!}$
|
|
|
|
$$ \frac{6!}{8!} = \frac{6!}{8 \cdot 7 \cdot 6!} = \frac{1}{56} $$
|
|
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|
64. $\dfrac{4!}{0!}$
|
|
|
|
$$ \frac{4!}{0!} = \frac{4 \cdot 3 \cdot 2 \cdot 1}{1} = 24 $$
|
|
|
|
65. $\dfrac{n!}{(n - 1)!}$
|
|
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|
$$ \frac{n!}{(n - 1)!} = \frac{n(n - 1)!}{(n - 1)!} = n $$
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|
|
|
66. $\dfrac{(n - 1)!}{(n + 1)!}$
|
|
|
|
$$ \frac{(n - 1)!}{(n + 1)!} = \frac{(n - 1)!}{(n + 1)(n)(n - 1)!} = \frac{1}{n(n + 1)} $$
|
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67. $\dfrac{n!}{(n - 2)!}$
|
|
|
|
$$ \dfrac{n!}{(n - 2)!} = \frac{n(n - 1)(n - 2)!}{(n - 2)!} = n(n - 1) $$
|
|
|
|
68. $\dfrac{((n + 1)!)^2}{(n!)^2}$
|
|
|
|
$$ \frac{((n + 1)!)^2}{(n!)^2} = \frac{((n + 1)(n!))^2}{(n!)^2} = (n + 1)^2 $$
|
|
|
|
69. $\dfrac{n!}{(n - k)!}$
|
|
|
|
$$ (n - k)! = (n - k)(n - k - 1)(n - k - 2) \dots (2)(1) $$
|
|
|
|
$$ n! = n(n - 1)(n - 2) \dots (n - k + 1)(n - k)(n - k - 1)(n - k - 2) \dots (2)(1) $$
|
|
|
|
$$ \frac{n!}{(n - k)!} = n(n - 1)(n - 2) \dots (n - k + 1) $$
|
|
|
|
70. $\dfrac{n!}{(n - k + 1)!}$
|
|
|
|
$$ (n - k + 1)! = (n - k + 1)(n - k)(n - k - 1) \dots (2)(1) $$
|
|
|
|
$$ n! = n(n - 1)(n - 2) \dots (n - k + 2)(n - k + 1)(n - k)(n - k - 1) \dots (2)(1)$$
|
|
|
|
$$ \frac{n!}{(n - k + 1)!} = n(n - 1)(n - 2) \dots (n - k + 2) $$
|
|
|
|
71. $\dbinom{5}{3}$
|
|
|
|
$$ \binom{n}{r} = \frac{n!}{r!(n - r)!} $$
|
|
|
|
$$ \binom{5}{3} = \frac{5!}{3!(5 - 3)!} $$
|
|
|
|
$$ = \frac{5 \cdot 4 \cdot 3!}{3!(2)!} $$
|
|
|
|
$$ = \frac{20}{2 \cdot 1} $$
|
|
|
|
$$ = 10 $$
|
|
|
|
72. $\dbinom{7}{4}$
|
|
|
|
$$ \binom{n}{r} = \frac{n!}{r!(n - r)!} $$
|
|
|
|
$$ \binom{7}{4} = \frac{7!}{4!(7 - 4)!} $$
|
|
|
|
$$ = \frac{7 \cdot 6 \cdot 5 \cdot 4!}{4!(3)!} $$
|
|
|
|
$$ = \frac{210}{3!} $$
|
|
|
|
$$ = \frac{210}{6} $$
|
|
|
|
$$ = 35 $$
|
|
|
|
73. $\dbinom{3}{0}$
|
|
|
|
$$ \binom{n}{r} = \frac{n!}{r!(n - r)!} $$
|
|
|
|
$$ \binom{3}{0} = \frac{3!}{0!(3 - 0)!} $$
|
|
|
|
$$ = \frac{3!}{1(3)!} $$
|
|
|
|
$$ = 1 $$
|
|
|
|
74. $\dbinom{5}{5}$
|
|
|
|
$$ \binom{n}{r} = \frac{n!}{r!(n - r)!} $$
|
|
|
|
$$ \binom{5}{5} = \frac{5!}{5!(5 - 5)!} $$
|
|
|
|
$$ = \frac{1}{1(0)!} $$
|
|
|
|
$$ = 1 $$
|
|
|
|
75. $\dbinom{n}{n - 1}$
|
|
|
|
$$ \binom{n}{r} = \frac{n!}{r!(n - r)!} $$
|
|
|
|
$$ \binom{n}{n - 1} = \frac{n!}{(n - 1)!(n - (n - 1))!} $$
|
|
|
|
$$ = \frac{n!}{(n - 1)!(n - n + 1)!} $$
|
|
|
|
$$ = \frac{n!}{(n - 1)!(1)!} $$
|
|
|
|
$$ = \frac{n(n - 1)!}{(n - 1)!(1)!} $$
|
|
|
|
$$ = \frac{n}{1} $$
|
|
|
|
$$ = n $$
|
|
|
|
76. $\dbinom{n + 1}{n - 1}$
|
|
|
|
$$ \binom{n}{r} = \frac{n!}{r!(n - r)!} $$
|
|
|
|
$$ \binom{n + 1}{n - 1} = \frac{(n + 1)!}{(n - 1)!((n + 1) - (n - 1))!} $$
|
|
|
|
$$ = \frac{(n + 1)!}{(n - 1)!(n + 1 - n + 1)!} $$
|
|
|
|
$$ = \frac{(n + 1)!}{(n - 1)!(2)!} $$
|
|
|
|
$$ = \frac{(n + 1)(n)(n - 1)!}{(n - 1)!(2)!} $$
|
|
|
|
$$ = \frac{n(n + 1)}{2} $$
|
|
|
|
77.
|
|
|
|
a. Prove that $n! + 2$ is divisible by $2$, for every integer $n \geq 2$.
|
|
|
|
**Proof:**
|
|
|
|
Suppose that $n$ is any integer such that $n \geq 2$.
|
|
|
|
By the definition of a factorial:
|
|
|
|
$$ n! = n \cdot (n - 1) \dots 3 \cdot 2 \cdot 1 $$
|
|
|
|
Since $n \geq 2$, this can be represented as:
|
|
|
|
$$
|
|
n! =
|
|
\begin{cases}
|
|
2 & \text{if } n = 2 \\
|
|
3 \cdot 2 \cdot 1& \text{if } n = 3 \\
|
|
n \cdot (n - 1) \dots \cdot 2 \cdot 1 & \text{if } n > 3 \\
|
|
\end{cases}
|
|
$$
|
|
|
|
In each case, $n!$ has a factor of $2$. Then:
|
|
|
|
$$ n! + 2 = 2k + 2 $$
|
|
|
|
$$ n! + 2 = 2(k + 1) $$
|
|
|
|
for some integer $k$.
|
|
|
|
Now, $k + 1$ is an integer by the sum of integers.
|
|
|
|
Therefore $n! + 2$ is divisible by $2$.
|
|
|
|
Q.E.D.
|
|
|
|
b. Prove that $n! + k$ is divisible by $k$, for every integer $n \geq 2$ and
|
|
$k = 2, 3, \dots, n$.
|
|
|
|
**Proof:**
|
|
|
|
Suppose $n$ is any integer such that $n \geq 2$, and $k$ is any integer such
|
|
that $2 \leq k \leq n$.
|
|
|
|
Since $2 \leq k \leq n$, it follows that $k$ is one of the factors of $n!$.
|
|
Then:
|
|
|
|
$$ n! = km $$
|
|
|
|
for some integer $m$.
|
|
|
|
By substitution:
|
|
|
|
$$ n! + k = km + k $$
|
|
|
|
$$ = k(m + 1) $$
|
|
|
|
Now, $m + 1$ is an integer by the sum of integers.
|
|
|
|
Therefore $n! + k$ is divisible by $k$.
|
|
|
|
Q.E.D.
|
|
|
|
c. Given any integer $m \geq 2$, is it possible to find a sequence of $m - 1$
|
|
consecutive positive integers none of which is prime? Explain your answer.
|
|
|
|
**Proof:**
|
|
|
|
Suppose $m$ is any integer such that $m \geq 2$.
|
|
|
|
Consider the sequence
|
|
|
|
$$ m! + 2, m! + 3, \dots, m! + m $$
|
|
|
|
This is a sequence of $m - 1$ consecutive positive integers.
|
|
|
|
Let $k$ be any integer such that $2 \leq k \leq m$. The $k - 1$<sup>th</sup>
|
|
term of the sequence is $m! + k$.
|
|
|
|
Since $k \leq m$, it follows that $k \mid m!$ (by part b). Then:
|
|
|
|
$$ m! = kt $$
|
|
|
|
for some integer $t$.
|
|
|
|
Then:
|
|
|
|
$$ m! + k = kt + k = k(t + 1) $$
|
|
|
|
Now, $t + 1$ is an integer by the sum of integers. Thus $k$ divides $m! + k$ and
|
|
since $k \geq 2$ and $(t + 1) > 1$ are both factors greater than or equal to
|
|
$1$, it follows that $m! + k$ is composite.
|
|
|
|
Therefore every term in the sequence is not prime, so there exists a sequence of
|
|
$m - 1$ consecutive positive integers none of which is prime.
|
|
|
|
Q.E.D.
|
|
|
|
78. Prove that for all nonnegative integers $n$ and $r$ with
|
|
|
|
$$ r + 1 \leq n, \binom{n}{r + 1} = \frac{n - r}{r + 1}\binom{n}{r} $$
|
|
|
|
**Proof:**
|
|
|
|
Suppose $n$ and $r$ are any nonnegative integers such that $r + 1 \leq n$.
|
|
|
|
The given equation shown is:
|
|
|
|
$$ \frac{n - r}{r + 1}\binom{n}{r} = \frac{n - r}{r + 1}\left(\frac{n!}{r!(n - r)!}\right) $$
|
|
|
|
$$ = \frac{n!(n - r)}{r!(r + 1)(n - r)!}$$
|
|
|
|
$$ = \frac{n!(n - r)}{r!(r + 1)(n - r)(n - r - 1)!}$$
|
|
|
|
$$ = \frac{n!}{r!(r + 1)(n - r - 1)!}$$
|
|
|
|
$$ = \frac{n!}{r!(r + 1)(n - (r + 1))!}$$
|
|
|
|
Notice that this in the form of a "$n$ choose $r + 1$":
|
|
|
|
$$ \binom{n}{r + 1} $$
|
|
|
|
Therefore, it has been shown that:
|
|
|
|
$$ \binom{n}{r + 1} = \frac{n - r}{r + 1}\binom{n}{r} $$
|
|
|
|
Q.E.D.
|
|
|
|
79. Prove that if $p$ is a prime number and $r$ is an integer with $0 < r < p$,
|
|
then $\dbinom{p}{r}$ is divisible by $p$.
|
|
|
|
**Proof:**
|
|
|
|
Suppose that $p$ is any prime number and $r$ is any integer such that
|
|
$0 < r < p$.
|
|
|
|
_[We need to show that $p \mid \dbinom{p}{r}$.]_
|
|
|
|
Consider:
|
|
|
|
$$ \binom{p}{r} = \frac{p!}{r!(p - r)!} $$
|
|
|
|
Since $0 < r < p$, both $r!$ and $(p - r)!$ are less than $p$. Thus, the
|
|
denominator $r!(p - r)!$ can never have a factor of $p$.
|
|
|
|
The numerator can be expressed as $p! = p(p - 1)!$:
|
|
|
|
$$ \binom{p}{r} = \frac{p(p - 1)!}{r!(p - r)!} $$
|
|
|
|
Factoring $p$ out of the numerator gives:
|
|
|
|
$$ \binom{p}{r} = p \cdot \frac{(p - 1)!}{r!(p - r)!} $$
|
|
|
|
Therefore it has been shown that:
|
|
|
|
$$ p \mid \binom{p}{r} $$
|
|
|
|
Q.E.D.
|
|
|
|
80. Suppose $a[1], a[2], a[3], \dots, a[m]$ is a one-dimensional array and
|
|
consider the following algorithm segment:
|
|
|
|
$\text{sum } := 0\\ \text{\textbf{for }} k := 1 \text{\textbf{ to }} m\\ \ \ \text{sum } := \text{ sum } + a[k]\\ \text{\textbf{next }} k$
|
|
|
|
Fill in the blanks below so that each algorithm segment performs the same job as
|
|
the one shown in the exercise statement.
|
|
|
|
a.
|
|
|
|
$\text{sum } := 0\\ \text{\textbf{for }} i := 0 \text{\textbf{ to \_\_\_\_}}\\ \ \ \text{sum } := \text{\_\_\_\_}\\ \text{\textbf{next }} i$
|
|
|
|
$m - 1$; $\text{sum } + a[i + 1]$
|
|
|
|
b.
|
|
|
|
$\text{sum } := 0\\ \text{\textbf{for }} j := 2 \text{\textbf{ to \_\_\_\_}}\\ \ \ \text{sum } := \text{\_\_\_\_}\\ \text{\textbf{next }} j$
|
|
|
|
$m + 1$; $\text{sum } + a[j - 1]$
|
|
|
|
Use repeated division by $2$ to convert (by hand) the integers in 81-83 from
|
|
base 10 to base 2.
|
|
|
|
81. $90$
|
|
|
|
$$ 90_{10} = 1011010_2 $$
|
|
|
|
82. $98$
|
|
|
|
$$ 98_{10} = 1100010_2 $$
|
|
|
|
83. $205$
|
|
|
|
$$ 205_{10} = 11001101_2 $$
|
|
|
|
Make a trace table to trace the action of Algorithm 5.1.1 on the input in 84-86.
|
|
|
|
84. $23$
|
|
|
|
| | 0 | 1 | 2 | 3 | 4 | 5 |
|
|
| ------ | -- | -- | - | - | - | - |
|
|
| $a$ | 23 | | | | | |
|
|
| $r[i]$ | | 1 | 1 | 1 | 0 | 1 |
|
|
| $q$ | 23 | 11 | 5 | 2 | 1 | 0 |
|
|
| $i$ | 0 | 1 | 2 | 3 | 4 | 5 |
|
|
|
|
Outputs: 10111, which is $23_{10} = 10111_2$.
|
|
|
|
85. $28$
|
|
|
|
| | 0 | 1 | 2 | 3 | 4 | 5 |
|
|
| ------ | -- | -- | - | - | - | - |
|
|
| $a$ | 28 | | | | | |
|
|
| $r[i]$ | | 0 | 0 | 1 | 1 | 1 |
|
|
| $q$ | 28 | 14 | 7 | 3 | 1 | 0 |
|
|
| $i$ | 0 | 1 | 2 | 3 | 4 | 5 |
|
|
|
|
Outputs: 11100, which is $28_{10} = 11100_2$.
|
|
|
|
86. $44$
|
|
|
|
| | 0 | 1 | 2 | 3 | 4 | 5 | 6 |
|
|
| ------ | -- | -- | -- | - | - | - | - |
|
|
| $a$ | 44 | | | | | | |
|
|
| $r[i]$ | | 0 | 0 | 1 | 1 | 0 | 1 |
|
|
| $q$ | 44 | 22 | 11 | 5 | 2 | 1 | 0 |
|
|
| $i$ | 0 | 1 | 2 | 3 | 4 | 5 | 6 |
|
|
|
|
Outputs: 101100, which is $44_{10} = 101100_2$
|
|
|
|
87. Write an informal description of an algorithm (using repeated division
|
|
by 16) to convert a nonnegative integer from decimal notation to hexadecimal
|
|
notation (base 16).
|
|
|
|
**Input:** $a$ _[a nonnegative integer]_
|
|
|
|
**Algorithm Body:**
|
|
|
|
$q := a, i := 0$
|
|
|
|
_[Repeatedly perform the integer division of $q$ by $16$ until $q$ becomes $0$.
|
|
Store successive remainders in a one-dimensional array
|
|
$r[0], r[1], r[2], \dots r[k]$. Even if the initial-value of $q$ equals $0$, the
|
|
loop should execute one time (so that $r[0]$ is computed). Thus the guard
|
|
condition for the **while** loop is $i = 0$ or $q \neq 0$.]_
|
|
|
|
$\text{\textbf{while }}(i = 0 \text{ or } q \neq 0)\\ \ \ r[i] := q \mod 16\\ \ \ q := q \text{ div } 16\\ \ \ \text{[r[i] and q can be obtained by calling the division algorithm.]}\\ \ \ i := i + 1\\ \text{\textbf{end while}}$
|
|
|
|
_[After execution of this step, the values of $r[0], r[1], \dots, r[i - 1]$ are
|
|
all $0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D, E, F$, and
|
|
$a = \left(r[i - 1]r[i - 2] \dots r[2]r[1]r[0]\right)_{16}$.]_
|
|
|
|
**Output:** $r[0], r[1], r[2], \dots, r[i - 1]$ _[a sequence of integers]_
|
|
|
|
Use the algorithm you developed for exercise 87 to convert the integers in 88-90
|
|
to hexadecimal notation.
|
|
|
|
88. $287$
|
|
|
|
$$ 287_{10} = 11F_{16} $$
|
|
|
|
89. $693$
|
|
|
|
$$ 693_{10} = 1BF_{16} $$
|
|
|
|
91. $2,301$
|
|
|
|
$$ 2301_{10} = 8FD_{16} $$
|
|
|
|
91. Write a formal version of the algorithm you developed for exercise 87.
|
|
|
|
Already done.
|
|
|
|
---
|
|
|
|
**Exercise Set 5.2**
|
|
|
|
Page 309
|
|
|
|
1. Use the technique illustrated at the beginning of this section to show that
|
|
the statements in (a) and (b) are true.
|
|
|
|
a. If
|
|
$\left(1 - \dfrac{1}{2}\right)\left(1 - \dfrac{1}{3}\right)\left(1 - \dfrac{1}{4}\right)\left(1 - \dfrac{1}{5}\right) = \dfrac{1}{5}$
|
|
then
|
|
$\left(1 - \dfrac{1}{2}\right)\left(1 - \dfrac{1}{3}\right)\left(1 - \dfrac{1}{4}\right)\left(1 - \dfrac{1}{5}\right)\left(1 - \dfrac{1}{6}\right) = \dfrac{1}{6}$.
|
|
|
|
Since:
|
|
|
|
$$ \left(1 - \dfrac{1}{2}\right)\left(1 - \dfrac{1}{3}\right)\left(1 - \dfrac{1}{4}\right)\left(1 - \dfrac{1}{5}\right) = \dfrac{1}{5} $$
|
|
|
|
then we can say that:
|
|
|
|
$$ \left(1 - \dfrac{1}{2}\right)\left(1 - \dfrac{1}{3}\right)\left(1 - \dfrac{1}{4}\right)\left(1 - \dfrac{1}{5}\right)\left(1 - \dfrac{1}{6}\right) = \dfrac{1}{6} = \frac{1}{5}\left(1 - \dfrac{1}{6}\right) $$
|
|
|
|
Evaluating this right hand side, we find that:
|
|
|
|
$$ \frac{1}{5}\left(1 - \frac{1}{6}\right) $$
|
|
|
|
$$ = \frac{1}{5}\left(\frac{6}{6} - \frac{1}{6}\right) $$
|
|
|
|
$$ = \frac{1}{5}\left(\frac{5}{6}\right) $$
|
|
|
|
$$ = \frac{1}{6} $$
|
|
|
|
Which is equal to the right hand side of the equality to be proved.
|
|
|
|
b. If
|
|
$\left(1 - \dfrac{1}{2}\right)\left(1 - \dfrac{1}{3}\right)\left(1 - \dfrac{1}{4}\right)\left(1 - \dfrac{1}{5}\right)\left(1 - \dfrac{1}{6}\right) = \dfrac{1}{6}$
|
|
then
|
|
$\left(1 - \dfrac{1}{2}\right)\left(1 - \dfrac{1}{3}\right)\left(1 - \dfrac{1}{4}\right)\left(1 - \dfrac{1}{5}\right)\left(1 - \dfrac{1}{6}\right)\left(1 - \dfrac{1}{7}\right) = \dfrac{1}{7}$.
|
|
|
|
Given that:
|
|
|
|
$$ \left(1 - \dfrac{1}{2}\right)\left(1 - \dfrac{1}{3}\right)\left(1 - \dfrac{1}{4}\right)\left(1 - \dfrac{1}{5}\right)\left(1 - \dfrac{1}{6}\right) = \dfrac{1}{6} $$
|
|
|
|
Then, by substitution:
|
|
|
|
$$ \left(1 - \dfrac{1}{2}\right)\left(1 - \dfrac{1}{3}\right)\left(1 - \dfrac{1}{4}\right)\left(1 - \dfrac{1}{5}\right)\left(1 - \dfrac{1}{6}\right)\left(1 - \dfrac{1}{7}\right) = \frac{1}{6}\left(1 - \dfrac{1}{7}\right) $$
|
|
|
|
Evaluating this right hand side, we find:
|
|
|
|
$$ \frac{1}{6}\left(1 - \frac{1}{7}\right) $$
|
|
|
|
$$ = \frac{1}{6}\left(\frac{7}{7} - \frac{1}{7}\right) $$
|
|
|
|
$$ = \frac{1}{6}\left(\frac{6}{7}\right) $$
|
|
|
|
$$ = \frac{1}{7} $$
|
|
|
|
And this is equal to the right hand side of the equality, and therefore shows
|
|
that the statement is true.
|
|
|
|
2. For each positive integer $n$, let $P(n)$ be the formula
|
|
|
|
$$ 1 + 3 + 5 + \dots + (2n - 1) = n^2 $$
|
|
|
|
a. Write $P(1)$. Is $P(1)$ true?
|
|
|
|
$$ P(n) = 1 + 3 + 5 + \dots + (2(n) - 1) = (n)^2 $$
|
|
|
|
By 5.2.1:
|
|
|
|
$$ P(n) = \frac{(2n - 1)((2n - 1) + 1)}{2} $$
|
|
|
|
$$ = \frac{(2n - 1)(2n)}{2} $$
|
|
|
|
$$ = \frac{4n^2 - 2n}{2} $$
|
|
|
|
$$ = 2n^2 - n $$
|
|
|
|
$$ P(1) = 1 + 3 + 5 + \dots + (2(1) - 1) = (1)^2 $$
|
|
|
|
$$ = 2(1)^2 - (1) = (1)^2 $$
|
|
|
|
$$ = 2(1) - (1) = (1) $$
|
|
|
|
$$ = 2 - 1 = 1 $$
|
|
|
|
$$ = 1 = 1 $$
|
|
|
|
$P(1)$ is true.
|
|
|
|
b. Write $P(k)$.
|
|
|
|
$$ P(n) = 1 + 3 + 5 + \dots + (2(n) - 1) = (n)^2 $$
|
|
|
|
$$ P(k) = 1 + 3 + 5 + \dots + (2k - 1) = k^2 $$
|
|
|
|
c. Write $P(k + 1)$.
|
|
|
|
$$ P(n) = 1 + 3 + 5 + \dots + (2(n) - 1) = (n)^2 $$
|
|
|
|
$$ P(k + 1) = 1 + 3 + 5 + \dots + (2(k + 1) - 1) = (k + 1)^2 $$
|
|
|
|
Alternatively:
|
|
|
|
$$ P(k + 1) = 1 + 3 + 5 + \dots + (2k + 2 - 1) = k^2 + 2k + 1 $$
|
|
|
|
$$ P(k + 1) = 1 + 3 + 5 + \dots + (2k + 1) = k^2 + 2k + 1 $$
|
|
|
|
d. In a proof by mathematical induction that the formula holds for every integer
|
|
$n \geq 1$, what must be shown in the inductive step?
|
|
|
|
In a proof by mathematical induction, where $P(n)$ holds for every integer
|
|
$n \geq 1$, the inductive step where for some integer $k$ where it is assumed
|
|
$1 + 3 + 5 + \dots + (2k - 1) = k^2$ is true (inductive hypothesis), then
|
|
$1 + 3 + 5 + \dots + (2(k + 1) - 1) = (k + 1)^2$ must be shown to also be true.
|
|
|
|
3. For each positive integer $n$, let $P(n)$ be the formula
|
|
|
|
$$ 1^2 + 2^2 + \dots + n^2 = \frac{n(n + 1)(2n + 1)}{6} $$
|
|
|
|
a. Write $P(1)$. Is $P(1)$ true?
|
|
|
|
$$ P(n) = 1^2 + 2^2 + \dots + (n)^2 = \frac{(n)((n) + 1)(2(n) + 1)}{6} $$
|
|
|
|
By 5.2.1:
|
|
|
|
$$ P(n) = \frac{(n^2)((n^2) + 1)}{2} $$
|
|
|
|
Then:
|
|
|
|
$$ P(1) = 1^2 + 2^2 + \dots + (1)^2 = \frac{(1)((1) + 1)(2(1) + 1)}{6} $$
|
|
|
|
$$ = 1^2 + 2^2 + \dots + (1)^2 = \frac{(1)((1) + 1)(2(1) + 1)}{6} $$
|
|
|
|
$$ = \frac{((1)^2)(((1)^2) + 1)}{2}= \frac{(1)((1) + 1)(2(1) + 1)}{6} $$
|
|
|
|
$$ = \frac{(1)(1 + 1)}{2}= \frac{(1)(2)(2 + 1)}{6} $$
|
|
|
|
$$ = \frac{(1)(2)}{2}= \frac{(1)(2)(3)}{6} $$
|
|
|
|
$$ = \frac{2}{2} = \frac{6}{6} $$
|
|
|
|
$$ = 1 = 1 $$
|
|
|
|
$P(1)$ is true.
|
|
|
|
b. Write $P(k)$.
|
|
|
|
$$ P(k) = 1^2 + 2^2 + \dots + k^2 = \frac{(k)(k + 1)(2k + 1)}{6} $$
|
|
|
|
c. Write $P(k + 1)$.
|
|
|
|
$$ P(k + 1) = 1^2 + 2^2 + \dots + (k + 1)^2 = \frac{(k + 1)((k + 1) + 1)(2(k + 1) + 1)}{6} $$
|
|
|
|
d. In a proof by mathematical induction that the formula holds for every integer
|
|
$n \geq 1$, what must be shown in the inductive step?
|
|
|
|
In a proof by mathematical induction, where $P(n)$ holds for every integer
|
|
$n \geq 1$, the inductive step where for some integer $k$ where it is assumed
|
|
$1^2 + 2^2 + \dots + k^2 = \frac{(k)(k + 1)(2k + 1)}{6}$ is true (inductive
|
|
hypothesis), then
|
|
$1^2 + 2^2 + \dots + (k + 1)^2 = \frac{(k + 1)((k + 1) + 1)(2(k + 1) + 1)}{6}$
|
|
must be shown to also be true.
|
|
|
|
4. For each integer $n$ with $n \geq 2$, let $P(n)$ be the formula
|
|
|
|
$$ \sum_{i = 1}^{n - 1}{i(i + 1)} = \frac{n(n - 1)(n + 1)}{3} $$
|
|
|
|
a. Write $P(2)$. Is $P(2)$ true?
|
|
|
|
$$ P(n) = \sum_{i = 1}^{(n) - 1}{i(i + 1)} = \frac{(n)((n) - 1)((n) + 1)}{3} $$
|
|
|
|
$$ P(2) = \sum_{i = 1}^{(2) - 1}{i(i + 1)} = \frac{(2)((2) - 1)((2) + 1)}{3} $$
|
|
|
|
Compute left-hand side:
|
|
|
|
$$ \sum_{i = 1}^{(2) - 1}{i(i + 1)} $$
|
|
|
|
$$ \sum_{i = 1}^{1}{i(i + 1)} $$
|
|
|
|
$$ \sum_{i = 1}^{1}{i(i + 1)} = (1)((1) + 1) $$
|
|
|
|
$$ = (1)(2) $$
|
|
|
|
$$ = 2 $$
|
|
|
|
Compute right-hand side:
|
|
|
|
$$ \frac{(2)((2) - 1)((2) + 1)}{3} $$
|
|
|
|
$$ = \frac{(2)(1)(3)}{3} $$
|
|
|
|
$$ = \frac{6}{3} $$
|
|
|
|
$$ = 2 $$
|
|
|
|
Since both the left hand side and the right hand side are equal, $P(2)$ is true.
|
|
|
|
b. Write $P(k)$.
|
|
|
|
$$ P(k) = \sum_{i = 1}^{(k) - 1}{i(i + 1)} = \frac{(k)((k) - 1)((k) + 1)}{3} $$
|
|
|
|
c. Write $P(k + 1)$.
|
|
|
|
$$ P(k + 1) = \sum_{i = 1}^{(k + 1) - 1}{i(i + 1)} = \frac{(k + 1)((k + 1) - 1)((k + 1) + 1)}{3} $$
|
|
|
|
d. In a proof by mathematical induction that the formula holds for every integer
|
|
$n \geq 2$, what must be shown in the inductive step?
|
|
|
|
In a proof by mathematical induction, where $P(n)$ holds for every integer
|
|
$n \geq 2$, the inductive step where for some integer $k$ where it is assumed
|
|
$\sum_{i = 1}^{(k) - 1}{i(i + 1)} = \dfrac{(k)((k) - 1)((k) + 1)}{3}$ is true
|
|
(inductive hypothesis), then
|
|
$\sum_{i = 1}^{(k + 1) - 1}{i(i + 1)} = \dfrac{(k + 1)((k + 1) - 1)((k + 1) + 1)}{3}$
|
|
must be shown to also be true.
|
|
|
|
5. Fill in the missing pieces in the following proof that
|
|
|
|
$$ 1 + 3 + 5 + \dots + (2n - 1) = n^2 $$
|
|
|
|
for every integer $n \geq 1$.
|
|
|
|
**Proof:** Let the property $P(n)$ be the equation
|
|
|
|
$$ 1 + 3 + 5 + \dots + (2n - 1) = n^2 $$
|
|
|
|
_Show that_ $P(1)$ is true:
|
|
|
|
To establish $P(1)$, we must show that when $1$ is substituted in place of $n$,
|
|
the left-hand side equals the right-hand side. But when $n = 1$, the left-hand
|
|
side is the sum of all the odd integers from $1$ to $2 \cdot 1 - 1$, which is
|
|
the sum of the odd integers from $1$ to $1$ and is just $1$. The right-hand side
|
|
is __ (a) __, which also equals $1$. So $P(1)$ is true.
|
|
|
|
_Show that for every integer $k \geq 1$, if $P(k)$ is true then $P(k + 1)$ is
|
|
true:_
|
|
|
|
Let $k$ be any integer with $k \geq 1$.
|
|
|
|
_[Suppose $P(k)$ is true. That is:]_
|
|
|
|
Suppose
|
|
|
|
$1 + 3 + 5 \cdot + (2k - 1) =$ __ (b) __.
|
|
|
|
_[This is the inductive hypothesis.]_
|
|
|
|
_[We must show that $P(k + 1)$ is true. That is:]_
|
|
|
|
We must show that __ \(c\) __ = __ (d) __.
|
|
|
|
Now the left-hand side of $P(k + 1)$ is
|
|
|
|
$$ 1 + 3 + 5 + \dots + (2(k + 1) - 1) $$
|
|
|
|
$$ = 1 + 3 + 5 + \dots + (2k + 1) $$
|
|
|
|
$$ = [1 + 3 + 5 + \dots + (2k - 1)] + (2k + 1) $$
|
|
|
|
the next-to-last term is $2k - 1$ because __ (e) __
|
|
|
|
$$ = k^2 + (2k + 1) $$
|
|
|
|
by __ (f) __
|
|
|
|
$$ = (k + 1)^2 $$
|
|
|
|
which is the right-hand side of $P(k + 1)$ _[as was to be shown]._
|
|
|
|
_[Since we have proved the basis step and the inductive step, we conclude that
|
|
the given statement is true.]_
|
|
|
|
_Note:_ This proof was annotated to help make its logical flow more obvious. In
|
|
standard mathematical writing, such annotation is omitted.
|
|
|
|
a. $(1)^2$
|
|
|
|
b. $k^2$
|
|
|
|
c. $1 + 3 + 5 + \dots + (2(k + 1) - 1)$
|
|
|
|
d. $(k + 1)^2$
|
|
|
|
e. the odd integer just before $2k + 1$ is $2k - 1$
|
|
|
|
f. inductive hypothesis
|
|
|
|
Prove each statement in 6-9 using mathematical induction. Do not derive them
|
|
from Theorem 5.2.1 or Theorem 5.2.2.
|
|
|
|
6. For every integer $n \geq 1$,
|
|
|
|
$$ 2 + 4 + 6 + \dots + 2n = n^2 + n $$
|
|
|
|
**Proof (by mathematical induction):**
|
|
|
|
Let $P(n)$ be the equation
|
|
|
|
$$ 2 + 4 + 6 + \dots + 2n = n^2 + n $$
|
|
|
|
_Basis Step: Show that $P(1)$ is true:_
|
|
|
|
To establish $P(1)$, we must show that when $1$ is substituted in place of $n$,
|
|
the left-hand side equals the right-hand side.
|
|
|
|
When $n = 1$, the left-hand side is the sum of all even integers from $2$ to
|
|
$2(1)$, which is the sum of the even integers from $2$ to $2$ and is just $2$.
|
|
|
|
The right-hand side is $1^2 + 1$, which also equals $2$.
|
|
|
|
Therefore $P(1)$ is true.
|
|
|
|
_Inductive Step:_
|
|
|
|
_Show that for every integer $k \geq 1$, if $P(k)$ is true then $P(k + 1)$ is
|
|
true:_
|
|
|
|
Let $k$ be any integer with $k \geq 1$.
|
|
|
|
Suppose $P(k)$ is true. That is, suppose:
|
|
|
|
$$ 2 + 4 + 6 + \dots + 2k = k^2 + k $$
|
|
|
|
This is the inductive hypothesis.
|
|
|
|
We must show that $P(k + 1)$ is true. That is we must show that:
|
|
|
|
$$ 2 + 4 + 6 + \dots + 2(k + 1) = (k + 1)^2 + (k + 1) $$
|
|
|
|
Now the left-hand side of $P(k + 1)$ is
|
|
|
|
$$ 2 + 4 + 6 + \dots + 2(k + 1) $$
|
|
|
|
$$ = [2 + 4 + 6 + \dots + 2k] + (2(k + 1)) $$
|
|
|
|
Where $2k$ is the next-to-last even term before $2k + 1$. Then, by inductive
|
|
hypothesis:
|
|
|
|
$$ = (k^2 + k) + (2(k + 1)) $$
|
|
|
|
Then, by algebra:
|
|
|
|
$$ = k^2 + 3k + 2 $$
|
|
|
|
Now, the right-hand side is:
|
|
|
|
$$ (k + 1)^2 + (k + 1) $$
|
|
|
|
$$ (k + 1)(k + 1) + (k + 1) $$
|
|
|
|
$$ (k^2 + 2k + 1) + (k + 1) $$
|
|
|
|
$$ k^2 + 3k + 2 $$
|
|
|
|
Thus, the left-hand and right-hand sides of $P(k + 1)$ are equal. Hence
|
|
$P(k + 1)$ is true.
|
|
|
|
Since we have proved the basis step and the inductive step, we conclude that
|
|
$P(n)$ is true for every integer $n \geq 1$.
|
|
|
|
Q.E.D.
|
|
|
|
7. For every integer $n \geq 1$,
|
|
|
|
$$ 1 + 6 + 11 + 16 + \dots + (5n - 4) = \frac{n(5n - 3)}{2} $$
|
|
|
|
**Proof (by mathematical induction):**
|
|
|
|
Let $P(n)$ be the equation
|
|
|
|
$$ 1 + 6 + 11 + 16 + \dots + (5n - 4) = \frac{n(5n - 3)}{2} $$
|
|
|
|
_Basis Step:_
|
|
|
|
We must prove $P(1)$:
|
|
|
|
$$ 1 + 6 + 11 + 16 + \dots + (5(1) - 4) = \frac{(1)(5(1) - 3)}{2} $$
|
|
|
|
When $n = 1$, the left-hand side is the sum of every fifth integer from $1$ to
|
|
$5(1) - 4$, which is $1$.
|
|
|
|
The right-hand side is:
|
|
|
|
$$ \frac{(1)(5(1) - 3)}{2} $$
|
|
|
|
$$ = \frac{1(5 - 3)}{2} $$
|
|
|
|
$$ = \frac{1(2)}{2} $$
|
|
|
|
$$ = 1 $$
|
|
|
|
Both sides of the equality of $P(1)$ are $1$. So $P(1)$ is true.
|
|
|
|
_Inductive Step:_
|
|
|
|
Let $k$ be any integer with $k \geq 1$.
|
|
|
|
Suppose that $P(k)$ is true. That is:
|
|
|
|
$$ 1 + 6 + 11 + 16 + \dots + (5k - 4) = \frac{k(5k - 3)}{2} $$
|
|
|
|
We must show that $P(k + 1)$ is true. That is:
|
|
|
|
$$ 1 + 6 + 11 + 16 + \dots + (5(k + 1) - 4) = \frac{(k + 1)(5(k + 1) - 3)}{2} $$
|
|
|
|
Evaluating the left-hand side:
|
|
|
|
$$ 1 + 6 + 11 + 16 + \dots + (5(k + 1) - 4) $$
|
|
|
|
$$ = [1 + 6 + 11 + 16 + \dots + (5k - 4)] + (5(k + 1) - 4) $$
|
|
|
|
Then, by inductive hypothesis:
|
|
|
|
$$ = \frac{k(5k - 3)}{2} + (5(k + 1) - 4) $$
|
|
|
|
Then by algebra:
|
|
|
|
$$ = \frac{5k^2 - 3k}{2} + (5k + 5 - 4) $$
|
|
|
|
$$ = \frac{5k^2 - 3k}{2} + \frac{2(5k + 5 - 4)}{2} $$
|
|
|
|
$$ = \frac{5k^2 - 3k + 2(5k + 5 - 4)}{2} $$
|
|
|
|
$$ = \frac{5k^2 - 3k + 10k + 10 - 8}{2} $$
|
|
|
|
$$ = \frac{5k^2 + 7k + 2}{2} $$
|
|
|
|
Now, the right-hand side:
|
|
|
|
$$ \frac{(k + 1)(5(k + 1) - 3)}{2} $$
|
|
|
|
$$ = \frac{(k + 1)(5k + 5 - 3)}{2} $$
|
|
|
|
$$ = \frac{5k^2 + 5k + 5k + 5 - 3k - 3}{2} $$
|
|
|
|
$$ = \frac{5k^2 + 10k + 5 - 3k - 3}{2} $$
|
|
|
|
$$ = \frac{5k^2 + 7k + 5 - 3}{2} $$
|
|
|
|
$$ = \frac{5k^2 + 7k + 2}{2} $$
|
|
|
|
which is the left-hand side of $P(k + 1)$. Therefore $P(k + 1)$ is true.
|
|
|
|
Q.E.D.
|
|
|
|
8. For every integer $n \geq 0$,
|
|
|
|
$$ 1 + 2 + 2^2 + \dots + 2^n = 2^{n + 1} - 1 $$
|
|
|
|
**Proof (by mathematical induction):**
|
|
|
|
Let $P(n)$ be the equation:
|
|
|
|
$$ 1 + 2 + 2^2 + \dots + 2^n = 2^{n + 1} - 1 $$
|
|
|
|
_Basis Step:_
|
|
|
|
Prove $P(0)$ is true.
|
|
|
|
$$ P(0) = 1 + 2 + 2^2 + \dots + 2^(0) = 2^{(0) + 1} - 1 $$
|
|
|
|
Evaluate the left-hand side when $n = 0$:
|
|
|
|
$$ 1 + 2 + 2^2 + \dots + 2^(0) = 2^0 = 1 \quad \text{ when } n = 0 $$
|
|
|
|
Evaluate the right-hand side when $n = 0$:
|
|
|
|
$$ 2^{(0) + 1} - 1 $$
|
|
|
|
$$ 2^1 - 1 $$
|
|
|
|
$$ 1 $$
|
|
|
|
Both the left-hand and right-hand sides of $P(0)$ are equal. $P(0)$ is true.
|
|
|
|
_Inductive Step:_
|
|
|
|
Let $k$ be any integer with $k \geq 0$.
|
|
|
|
Suppose $P(k)$ is true. That is:
|
|
|
|
$$ P(k) = 1 + 2 + 2^2 + \dots + 2^k = 2^{k + 1} + 1 $$
|
|
|
|
Prove that $P(k + 1)$ is true:
|
|
|
|
$$ P(k + 1) = 1 + 2 + 2^2 + \dots + 2^(k + 1) = 2^{(k + 1) + 1} + 1 $$
|
|
|
|
$$ 1 + 2 + 2^2 + \dots + 2^(k + 1) = 2^{(k + 1) + 1} + 1 $$
|
|
|
|
Evaluate the left-hand side:
|
|
|
|
$$ 1 + 2 + 2^2 + \dots + 2^(k + 1) $$
|
|
|
|
$$ [1 + 2 + 2^2 + \dots + 2^k] + 2^(k + 1) $$
|
|
|
|
By inductive hypothesis:
|
|
|
|
$$ (2^{k + 1} + 1) + 2^(k + 1) $$
|
|
|
|
$$ 2(2^{k + 1}) + 1 $$
|
|
|
|
$$ 2^{k + 2} + 1 $$
|
|
|
|
Evaluate the right-hand side:
|
|
|
|
$$ 2^{(k + 1) + 1} + 1 $$
|
|
|
|
$$ = 2^{k + 2} + 1 $$
|
|
|
|
Therefore $P(k + 1)$ is true.
|
|
|
|
Q.E.D.
|
|
|
|
9. For every integer $n \geq 3$,
|
|
|
|
$$ 4^3 + 4^4 + 4^5 + \dots + 4^n = \frac{4(4^n - 16)}{3} $$
|
|
|
|
**Proof by mathematical induction:**
|
|
|
|
Let $P(n)$ be the equation:
|
|
|
|
$$ 4^3 + 4^4 + 4^5 + \dots + 4^n = \frac{4(4^n - 16)}{3} $$
|
|
|
|
_Basis Step:_
|
|
|
|
Prove $P(3)$. That is:
|
|
|
|
$$ 4^3 + 4^4 + 4^5 + \dots + 4^3 = \frac{4(4^3 - 16)}{3} $$
|
|
|
|
Evaluate left-hand side when $n = 3$:
|
|
|
|
$$ 4^3 + 4^4 + 4^5 + \dots + 4^3 = 4^3 = 64 \quad \text{ when } n = 3 $$
|
|
|
|
Evaluate right-hand side when $n = 3$:
|
|
|
|
$$ \frac{4(4^3 - 16)}{3} $$
|
|
|
|
$$ = \frac{4(64 - 16)}{3} $$
|
|
|
|
$$ = \frac{4(48)}{3} $$
|
|
|
|
$$ = \frac{192}{3} $$
|
|
|
|
$$ = 64 $$
|
|
|
|
Therefore $P(3)$ is true.
|
|
|
|
_Inductive Step:_
|
|
|
|
Let $k$ be any integer where $k \geq 3$.
|
|
|
|
Suppose $P(k)$. That is:
|
|
|
|
$$ 4^3 + 4^4 + 4^5 + \dots + 4^k = \frac{4(4^k - 16)}{3} $$
|
|
|
|
This is the inductive hypothesis.
|
|
|
|
Prove $P(k + 1)$. That is:
|
|
|
|
$$ 4^3 + 4^4 + 4^5 + \dots + 4^{k + 1} = \frac{4(4^{k + 1} - 16)}{3} $$
|
|
|
|
Evaluate left-hand side:
|
|
|
|
$$ 4^3 + 4^4 + 4^5 + \dots + 4^{k + 1} $$
|
|
|
|
$$ = [4^3 + 4^4 + 4^5 + \dots + 4^k] + 4^{k + 1} $$
|
|
|
|
By inductive hypothesis:
|
|
|
|
$$ = \frac{4(4^k - 16)}{3} + 4^{k + 1} $$
|
|
|
|
$$ = \frac{4^{k + 1} - 64}{3} + \frac{3(4^{k + 1})}{3} $$
|
|
|
|
$$ = \frac{4^{k + 1} - 64 + (3(4^{k + 1}))}{3} $$
|
|
|
|
$$ = \frac{4^{k + 1} + 3(4^{k + 1}) - 64}{3} $$
|
|
|
|
$$ = \frac{1(4^{k + 1}) + 3(4^{k + 1}) - 64}{3} $$
|
|
|
|
$$ = \frac{4(4^{k + 1}) - 64}{3} $$
|
|
|
|
$$ = \frac{4(4^{k + 1} - 16)}{3} $$
|
|
|
|
Evaluate right-hand side:
|
|
|
|
$$ \frac{4(4^{k + 1} - 16)}{3} $$
|
|
|
|
Both the left-hand and right-hand sides of $P(k + 1)$ are equal. $P(k + 1)$ is
|
|
true.
|
|
|
|
Q.E.D.
|
|
|
|
Prove each of the statements in 10-18 by mathematical induction.
|
|
|
|
10. $1^2 + 2^2 + \dots + n^2 = \dfrac{n(n + 1)(2n + 1)}{6}$, for every integer
|
|
$n \geq 1$.
|
|
|
|
**Proof (by mathematical induction):**
|
|
|
|
Let $P(n)$ be the equation:
|
|
|
|
$$ 1^2 + 2^2 + \dots + n^2 = \dfrac{n(n + 1)(2n + 1)}{6} $$
|
|
|
|
_Basis Step:_
|
|
|
|
Prove $P(1)$. That is:
|
|
|
|
$$ 1^2 + 2^2 + \dots + (1)^2 = \dfrac{(1)(1 + 1)(2(1) + 1)}{6} $$
|
|
|
|
Evaluate left-hand side when $n = 1$:
|
|
|
|
$$ 1^2 + 2^2 + \dots + (1)^2 = 1 $$
|
|
|
|
Evaluate right-hand side when $n = 1$:
|
|
|
|
$$ \dfrac{(1)(1 + 1)(2(1) + 1)}{6} $$
|
|
|
|
$$ = \dfrac{(1)(2)(2 + 1)}{6} $$
|
|
|
|
$$ = \dfrac{(1)(2)(3)}{6} $$
|
|
|
|
$$ = \dfrac{6}{6} $$
|
|
|
|
$$ = 1 $$
|
|
|
|
Both the left-hand and right-hand sides of $P(1)$ are equal. Therefore $P(1)$ is
|
|
true.
|
|
|
|
_Inductive Step:_
|
|
|
|
Let $k$ be any integer where $k \geq 1$.
|
|
|
|
Suppose $P(k)$. That is:
|
|
|
|
$$ 1^2 + 2^2 + \dots + k^2 = \dfrac{k(k + 1)(2k + 1)}{6} $$
|
|
|
|
This is the inductive hypothesis.
|
|
|
|
Prove $P(k + 1)$. That is:
|
|
|
|
$$ 1^2 + 2^2 + \dots + (k + 1)^2 = \dfrac{(k + 1)((k + 1) + 1)(2(k + 1) + 1)}{6} $$
|
|
|
|
$$ 1^2 + 2^2 + \dots + (k + 1)^2 = \dfrac{(k + 1)(k + 2)(2k + 2 + 1)}{6} $$
|
|
|
|
$$ 1^2 + 2^2 + \dots + (k + 1)^2 = \dfrac{(k + 1)(k + 2)(2k + 3)}{6} $$
|
|
|
|
Evaluate left-hand side:
|
|
|
|
$$ 1^2 + 2^2 + \dots + (k + 1)^2 $$
|
|
|
|
$$ = [1^2 + 2^2 + \dots + k^2] + (k + 1)^2 $$
|
|
|
|
By inductive hypothesis:
|
|
|
|
$$ = \dfrac{k(k + 1)(2k + 1)}{6} + (k + 1)^2 $$
|
|
|
|
$$ = \dfrac{k(k + 1)(2k + 1)}{6} + \frac{6(k + 1)^2}{6} $$
|
|
|
|
$$ = \dfrac{k(k + 1)(2k + 1) + 6(k + 1)^2}{6} $$
|
|
|
|
$$ = \dfrac{(k + 1)[k(2k + 1) + 6(k + 1)]}{6} $$
|
|
|
|
$$ = \dfrac{(k + 1)[2k^2 + k + 6k + 6]}{6} $$
|
|
|
|
$$ = \dfrac{(k + 1)[2k^2 + 7k + 6]}{6} $$
|
|
|
|
$$ = \dfrac{(k + 1)(k + 2)(2k + 3)}{6} $$
|
|
|
|
Evaluate right-hand side:
|
|
|
|
$$ = \dfrac{(k + 1)(k + 2)(2k + 3)}{6} $$
|
|
|
|
Both the left-hand and right-hand sides of $P(k + 1)$ are equal. Therefore
|
|
$P(k + 1)$ is true.
|
|
|
|
Q.E.D.
|
|
|
|
11. $1^3 + 2^3 + \dots + n^3 = \left[\dfrac{n(n + 1)}{2}\right]^2$, for every
|
|
integer $n \geq 1$.
|
|
|
|
**Proof (by mathematical induction):**
|
|
|
|
Let $P(n)$ be the equation:
|
|
|
|
$$ 1^3 + 2^3 + \dots + n^3 = \left[\dfrac{n(n + 1)}{2}\right]^2 $$
|
|
|
|
_Basis Step:_
|
|
|
|
Prove $P(1)$. That is:
|
|
|
|
$$ 1^3 + 2^3 + \dots + (1)^3 = \left[\dfrac{(1)((1) + 1)}{2}\right]^2 $$
|
|
|
|
Evaluate left-hand when $n = 1$:
|
|
|
|
$$ 1^3 + 2^3 + \dots + (1)^3 = 1 $$
|
|
|
|
Evaluate right-hand when $n = 1$:
|
|
|
|
$$ \left[\dfrac{(1)((1) + 1)}{2}\right]^2 $$
|
|
|
|
$$ = \left[\dfrac{(1)(2)}{2}\right]^2 $$
|
|
|
|
$$ = \left[\dfrac{2}{2}\right]^2 $$
|
|
|
|
$$ = [1]^2 $$
|
|
|
|
$$ = 1 $$
|
|
|
|
Both the left and right hand sides of $P(1)$ are true. Therefore $P(1)$ is true.
|
|
|
|
_Inductive Step:_
|
|
|
|
Let $k$ be any integer where $k \geq 1$.
|
|
|
|
Suppose $P(k)$. That is:
|
|
|
|
$$ 1^3 + 2^3 + \dots + k^3 = \left[\dfrac{k(k + 1)}{2}\right]^2 $$
|
|
|
|
This is the inductive hypothesis.
|
|
|
|
Prove $P(k + 1)$. That is:
|
|
|
|
$$ 1^3 + 2^3 + \dots + (k + 1)^3 = \left[\dfrac{(k + 1)((k + 1) + 1)}{2}\right]^2 $$
|
|
|
|
Evaluate left-hand:
|
|
|
|
$$ 1^3 + 2^3 + \dots + (k + 1)^3 $$
|
|
|
|
$$ = [1^3 + 2^3 + \dots + k^3] + (k + 1)^3 $$
|
|
|
|
By inductive hypothesis:
|
|
|
|
$$ = \left[\dfrac{k(k + 1)}{2}\right]^2 + (k + 1)^3 $$
|
|
|
|
$$ = \dfrac{k^2(k + 1)^2}{4} + (k + 1)^3 $$
|
|
|
|
$$ = \dfrac{k^2(k + 1)^2}{4} + \frac{4(k + 1)^3}{4} $$
|
|
|
|
$$ = \dfrac{k^2(k + 1)^2 + 4(k + 1)^3}{4} $$
|
|
|
|
$$ = \dfrac{k^2(k + 1)^2 + 4(k + 1)^2(k + 1)}{4} $$
|
|
|
|
$$ = \dfrac{(k + 1)^2[k^2 + 4(k + 1)]}{4} $$
|
|
|
|
$$ = \dfrac{(k + 1)^2[k^2 + 4k + 4]}{4} $$
|
|
|
|
$$ = \dfrac{(k + 1)^2(k + 2)^2}{4} $$
|
|
|
|
$$ = \left[\dfrac{(k + 1)(k + 2)}{2}\right]^2 $$
|
|
|
|
Evaluate right-hand:
|
|
|
|
$$ \left[\dfrac{(k + 1)((k + 1) + 1)}{2}\right]^2 $$
|
|
|
|
$$ = \left[\dfrac{(k + 1)(k + 2)}{2}\right]^2 $$
|
|
|
|
Both the left and right hand sides of $P(k + 1)$ are equal. Therefore $P(k + 1)$
|
|
is true.
|
|
|
|
Q.E.D.
|
|
|
|
12. $\dfrac{1}{1 \cdot 2} + \dfrac{1}{2 \cdot 3} + \dots + \dfrac{1}{n(n + 1)} = \dfrac{n}{n + 1}$,
|
|
for every integer $n \geq 1$.
|
|
|
|
**Proof (by mathematical induction):**
|
|
|
|
Let $P(n)$ be the equation:
|
|
|
|
$$ \dfrac{1}{1 \cdot 2} + \dfrac{1}{2 \cdot 3} + \dots + \dfrac{1}{n(n + 1)} = \dfrac{n}{n + 1} $$
|
|
|
|
_Basis Step:_
|
|
|
|
Prove $P(1)$, that is:
|
|
|
|
$$ \dfrac{1}{1 \cdot 2} + \dfrac{1}{2 \cdot 3} + \dots + \dfrac{1}{(1)((1) + 1)} = \dfrac{(1)}{(1) + 1} $$
|
|
|
|
Evaluate left-hand when $n = 1$:
|
|
|
|
$$ \dfrac{1}{1 \cdot 2} + \dfrac{1}{2 \cdot 3} + \dots + \dfrac{1}{(1)((1) + 1)} = \frac{1}{2} $$
|
|
|
|
Evaluate right-hand when $n = 1$:
|
|
|
|
$$ \dfrac{(1)}{(1) + 1} $$
|
|
|
|
$$ = \dfrac{1}{2} $$
|
|
|
|
The left and right hand sides of $P(1)$ are equal. Therefore $P(1)$ is true.
|
|
|
|
_Inductive Step:_
|
|
|
|
Let $k$ be any integer where $k \geq 1$.
|
|
|
|
Suppose $P(k)$. That is:
|
|
|
|
$$ \dfrac{1}{1 \cdot 2} + \dfrac{1}{2 \cdot 3} + \dots + \dfrac{1}{k(k + 1)} = \dfrac{k}{k + 1} $$
|
|
|
|
This is the inductive hypothesis.
|
|
|
|
Prove $P(k + 1)$. That is:
|
|
|
|
$$ \dfrac{1}{1 \cdot 2} + \dfrac{1}{2 \cdot 3} + \dots + \dfrac{1}{(k + 1)((k + 1) + 1)} = \dfrac{(k + 1)}{(k + 1) + 1} $$
|
|
|
|
Alternatively:
|
|
|
|
$$ \dfrac{1}{1 \cdot 2} + \dfrac{1}{2 \cdot 3} + \dots + \dfrac{1}{(k + 1)(k + 2)} = \dfrac{k + 1}{k + 2} $$
|
|
|
|
Evaluate left-hand:
|
|
|
|
$$ \dfrac{1}{1 \cdot 2} + \dfrac{1}{2 \cdot 3} + \dots + \dfrac{1}{(k + 1)(k + 2)} $$
|
|
|
|
$$ = \left[\dfrac{1}{1 \cdot 2} + \dfrac{1}{2 \cdot 3} + \dots + \frac{1}{k(k + 1)}\right] + \dfrac{1}{(k + 1)(k + 2)} $$
|
|
|
|
By the inductive hypothesis:
|
|
|
|
$$ = \dfrac{k}{k + 1} + \dfrac{1}{(k + 1)(k + 2)} $$
|
|
|
|
$$ = \dfrac{k(k + 2)}{(k + 1)(k + 2)} + \dfrac{1}{(k + 1)(k + 2)} $$
|
|
|
|
$$ = \dfrac{k(k + 2) + 1}{(k + 1)(k + 2)} $$
|
|
|
|
$$ = \dfrac{k^2 + 2k + 1}{(k + 1)(k + 2)} $$
|
|
|
|
$$ = \dfrac{(k + 1)(k + 1)}{(k + 1)(k + 2)} $$
|
|
|
|
$$ = \dfrac{k + 1}{k + 2} $$
|
|
|
|
Evaluate right-hand:
|
|
|
|
$$ \dfrac{k + 1}{k + 2} $$
|
|
|
|
Both the left and right hand sides of $P(k + 1)$ are equal. Therefore $P(k + 1)$
|
|
is true.
|
|
|
|
Q.E.D.
|
|
|
|
13. $\sum_{i = 1}^{n - 1}{i(i + 1)} = \dfrac{n(n - 1)(n + 1)}{3}$, for every
|
|
integer $n \geq 2$.
|
|
|
|
Let $P(n)$ be the equation:
|
|
|
|
$$ \sum_{i = 1}^{n - 1}{i(i + 1)} = \dfrac{n(n - 1)(n + 1)}{3} $$
|
|
|
|
_Basis Step:_
|
|
|
|
Prove $P(2)$. That is:
|
|
|
|
$$ \sum_{i = 1}^{(2) - 1}{i(i + 1)} = \dfrac{(2)((2) - 1)((2) + 1)}{3} $$
|
|
|
|
Alternatively:
|
|
|
|
$$ \sum_{i = 1}^{1}{i(i + 1)} = \dfrac{(2)(1)(3)}{3} $$
|
|
|
|
$$ \sum_{i = 1}^{1}{i(i + 1)} = 2 $$
|
|
|
|
Evaluate left-hand when $n = 2$:
|
|
|
|
$$ \sum_{i = 1}^{1}{i(i + 1)} $$
|
|
|
|
$$ = (1)(1 + 1) = 2 $$
|
|
|
|
The left and right hand sides of $P(2)$ are equal. Therefore $P(2)$ is true.
|
|
|
|
_Inductive Step:_
|
|
|
|
Let $k$ be any integer where $k \geq 2$.
|
|
|
|
Suppose $P(k)$. That is:
|
|
|
|
$$ \sum_{i = 1}^{k - 1}{i(i + 1)} = \dfrac{k(k - 1)(k + 1)}{3} $$
|
|
|
|
This is the inductive hypothesis.
|
|
|
|
Prove $P(k + 1)$. That is:
|
|
|
|
$$ \sum_{i = 1}^{(k + 1) - 1}{i(i + 1)} = \dfrac{(k + 1)((k + 1) - 1)((k + 1) + 1)}{3} $$
|
|
|
|
Alternatively:
|
|
|
|
$$ \sum_{i = 1}^{k}{i(i + 1)} = \dfrac{(k + 1)(k)(k + 2)}{3} $$
|
|
|
|
$$ \sum_{i = 1}^{k}{i(i + 1)} = \dfrac{k(k + 1)(k + 2)}{3} $$
|
|
|
|
Evaluate left-hand:
|
|
|
|
$$ \sum_{i = 1}^{k}{i(i + 1)} $$
|
|
|
|
$$ = \left[\sum_{i = 1}^{k - 1}{i(i + 1)}\right] + k(k + 1) $$
|
|
|
|
By the inductive hypothesis:
|
|
|
|
$$ = \dfrac{k(k - 1)(k + 1)}{3} + k(k + 1) $$
|
|
|
|
$$ = \dfrac{k(k - 1)(k + 1)}{3} + \frac{3k(k + 1)}{3} $$
|
|
|
|
$$ = \dfrac{k(k - 1)(k + 1) + 3k(k + 1)}{3} $$
|
|
|
|
$$ = \dfrac{k(k + 1)((k - 1) + 3)}{3} $$
|
|
|
|
$$ = \dfrac{k(k + 1)(k + 2)}{3} $$
|
|
|
|
Evaluate right-hand:
|
|
|
|
$$ \dfrac{k(k + 1)(k + 2)}{3} $$
|
|
|
|
The left and right hand sides of $P(k + 1)$ are equal. Therefore $P(k + 1)$ is
|
|
true.
|
|
|
|
Q.E.D.
|
|
|
|
14. $\sum_{i = 1}^{n + 1}{i \cdot 2^i} = n \cdot 2^{n + 2} + 2$, for every
|
|
integer $n \geq 0$.
|
|
|
|
**Proof (by mathematical induction):**
|
|
|
|
Let $P(n)$ be the equation:
|
|
|
|
$$ \sum_{i = 1}^{n + 1}{i \cdot 2^i} = n \cdot 2^{n + 2} + 2 $$
|
|
|
|
_Basis Step:_
|
|
|
|
Prove $P(0)$. That is:
|
|
|
|
$$ \sum_{i = 1}^{(0) + 1}{i \cdot 2^i} = (0) \cdot 2^{(0) + 2} + 2 $$
|
|
|
|
Alternatively:
|
|
|
|
$$ \sum_{i = 1}^{1}{i \cdot 2^i} = (0) \cdot 2^{2} + 2 $$
|
|
|
|
$$ \sum_{i = 1}^{1}{i \cdot 2^i} = 0 + 2 $$
|
|
|
|
$$ \sum_{i = 1}^{1}{i \cdot 2^i} = 2 $$
|
|
|
|
Evaluate left-hand when $n = 0$:
|
|
|
|
$$ \sum_{i = 1}^{1}{i \cdot 2^i} $$
|
|
|
|
$$ = (1) \cdot 2^(1) $$
|
|
|
|
$$ = 2 $$
|
|
|
|
Both the left and right hand sides of $P(0)$ are equal. Therefore $P(0)$ is
|
|
true.
|
|
|
|
_Inductive Step:_
|
|
|
|
Let $k$ be any integer where $k \geq 0$.
|
|
|
|
Suppose $P(k)$. That is:
|
|
|
|
$$ \sum_{i = 1}^{k + 1}{i \cdot 2^i} = k \cdot 2^{k + 2} + 2 $$
|
|
|
|
This is the inductive hypothesis.
|
|
|
|
Prove $P(k + 1)$. That is:
|
|
|
|
$$ \sum_{i = 1}^{(k + 1) + 1}{i \cdot 2^i} = (k + 1) \cdot 2^{(k + 1) + 2} + 2 $$
|
|
|
|
Alternatively:
|
|
|
|
$$ \sum_{i = 1}^{k + 2}{i \cdot 2^i} = (k + 1) \cdot 2^{k + 3} + 2 $$
|
|
|
|
Evaluate left-hand:
|
|
|
|
$$ \sum_{i = 1}^{k + 2}{i \cdot 2^i} $$
|
|
|
|
$$ = \left[\sum_{i = 1}^{k + 1}{i \cdot 2^i}\right] + (k + 2) \cdot 2^{k + 2} $$
|
|
|
|
By the inductive hypothesis:
|
|
|
|
$$ = k \cdot 2^{k + 2} + 2 + (k + 2) \cdot 2^{k + 2} $$
|
|
|
|
$$ = k(2^{k + 2}) + 2 + (k + 2)(2^{k + 2}) $$
|
|
|
|
$$ = (2^{k + 2})(k + (k + 2)) + 2 $$
|
|
|
|
$$ = (2^{k + 2})(2k + 2) + 2 $$
|
|
|
|
$$ = 2(2^{k + 2})(k + 1) + 2 $$
|
|
|
|
$$ = (2^{k + 3})(k + 1) + 2 $$
|
|
|
|
$$ = (k + 1) \cdot 2^{k + 3} + 2 $$
|
|
|
|
Evaluate right-hand:
|
|
|
|
$$ (k + 1) \cdot 2^{k + 3} + 2 $$
|
|
|
|
The left and right hand sides of $P(k + 1)$ are equal. Therefore $P(k + 1)$ is
|
|
true.
|
|
|
|
Q.E.D.
|
|
|
|
15. $\sum_{i = 1}^{n}{i(i!)} = (n + 1)! - 1$, for every integer $n \geq 1$.
|
|
|
|
**Proof (by mathematical induction):**
|
|
|
|
Let $P(n)$ be the equation:
|
|
|
|
$$ \sum_{i = 1}^{n}{i(i!)} = (n + 1)! - 1 $$
|
|
|
|
_Basis Step:_
|
|
|
|
Prove $P(1)$. That is:
|
|
|
|
$$ \sum_{i = 1}^{(1)}{i(i!)} = ((1) + 1)! - 1 $$
|
|
|
|
Evaluate left-hand side:
|
|
|
|
$$ \sum_{i = 1}^{1}{i(i!)} $$
|
|
|
|
$$ = 1(1!) = 1 $$
|
|
|
|
Evaluate right-hand side:
|
|
|
|
$$ ((1) + 1)! - 1 $$
|
|
|
|
$$ = (2)! - 1 $$
|
|
|
|
$$ = (2 \cdot 1) - 1 $$
|
|
|
|
$$ = 2 - 1 $$
|
|
|
|
$$ = 1 $$
|
|
|
|
Both sides of $P(1)$ are equal. Therefore $P(1)$ is true.
|
|
|
|
_Inductive Step:_
|
|
|
|
Let $k$ be any integer where $k \geq 1$.
|
|
|
|
Suppose $P(k)$. That is:
|
|
|
|
$$ \sum_{i = 1}^{k}{i(i!)} = (k + 1)! - 1 $$
|
|
|
|
This is the inductive hypothesis.
|
|
|
|
Prove $P(k + 1)$. That is:
|
|
|
|
$$ \sum_{i = 1}^{(k + 1)}{i(i!)} = ((k + 1) + 1)! - 1 $$
|
|
|
|
Alternatively:
|
|
|
|
$$ \sum_{i = 1}^{k + 1}{i(i!)} = (k + 2)! - 1 $$
|
|
|
|
Evaluate left-hand:
|
|
|
|
$$ \sum_{i = 1}^{k + 1}{i(i!)} $$
|
|
|
|
$$ = \left[\sum_{i = 1}^{k}{i(i!)}\right] + (k + 1)(k + 1)! $$
|
|
|
|
By the inductive hypothesis:
|
|
|
|
$$ = (k + 1)! - 1 + (k + 1)(k + 1)! $$
|
|
|
|
$$ = (k + 1)! + (k + 1)(k + 1)! - 1 $$
|
|
|
|
$$ = (k + 1)!(1 + (k + 1)) - 1 $$
|
|
|
|
$$ = (k + 1)!(k + 2) - 1 $$
|
|
|
|
$$ = (k + 2)! - 1 $$
|
|
|
|
Evaluate right-hand:
|
|
|
|
$$ (k + 2)! - 1 $$
|
|
|
|
Both sides of $P(k + 1)$ are equal. Therefore $P(k + 1)$ is true.
|
|
|
|
Q.E.D.
|
|
|
|
16. $\left(1 - \dfrac{1}{2^2}\right)\left(1 - \dfrac{1}{3^2}\right) \dots \left(1 - \dfrac{1}{n^2}\right) = \dfrac{n + 1}{2n}$,
|
|
for every integer $n \geq 2$.
|
|
|
|
**Proof (by mathematical induction):**
|
|
|
|
Let $P(n)$ be the equation:
|
|
|
|
$$ \left(1 - \dfrac{1}{2^2}\right)\left(1 - \dfrac{1}{3^2}\right) \dots \left(1 - \dfrac{1}{n^2}\right) = \dfrac{n + 1}{2n} $$
|
|
|
|
_Basis Step:_
|
|
|
|
Prove $P(2)$. That is:
|
|
|
|
$$ \left(1 - \dfrac{1}{2^2}\right)\left(1 - \dfrac{1}{3^2}\right) \dots \left(1 - \dfrac{1}{(2)^2}\right) = \dfrac{(2) + 1}{2(2)} $$
|
|
|
|
Alternatively:
|
|
|
|
$$ \left(1 - \dfrac{1}{2^2}\right)\left(1 - \dfrac{1}{3^2}\right) \dots \left(1 - \dfrac{1}{4}\right) = \dfrac{3}{4} $$
|
|
|
|
Evaluate left-hand side when $n = 2$:
|
|
|
|
$$ \left(1 - \dfrac{1}{2^2}\right)\left(1 - \dfrac{1}{3^2}\right) \dots \left(1 - \dfrac{1}{4}\right) $$
|
|
|
|
$$ = \frac{3}{4} $$
|
|
|
|
Both sides of $P(2)$ are equal. Therefore $P(2)$ is true.
|
|
|
|
_Inductive Step:_
|
|
|
|
Let $k$ be any integer where $k \geq 2$.
|
|
|
|
Suppose $P(k)$. That is:
|
|
|
|
$$ \left(1 - \dfrac{1}{2^2}\right)\left(1 - \dfrac{1}{3^2}\right) \dots \left(1 - \dfrac{1}{k^2}\right) = \dfrac{k + 1}{2k} $$
|
|
|
|
This is the inductive hypothesis.
|
|
|
|
Prove $P(k + 1)$. That is:
|
|
|
|
$$ \left(1 - \dfrac{1}{2^2}\right)\left(1 - \dfrac{1}{3^2}\right) \dots \left(1 - \dfrac{1}{(k + 1)^2}\right) = \dfrac{(k + 1) + 1}{2(k + 1)} $$
|
|
|
|
Alternatively:
|
|
|
|
$$ \left(1 - \dfrac{1}{2^2}\right)\left(1 - \dfrac{1}{3^2}\right) \dots \left(1 - \dfrac{1}{(k + 1)^2}\right) = \dfrac{k + 2}{2k + 2} $$
|
|
|
|
Evaluate left-hand:
|
|
|
|
$$ \left(1 - \dfrac{1}{2^2}\right)\left(1 - \dfrac{1}{3^2}\right) \dots \left(1 - \dfrac{1}{(k + 1)^2}\right) $$
|
|
|
|
$$ = \left[\left(1 - \dfrac{1}{2^2}\right)\left(1 - \dfrac{1}{3^2}\right) \dots \left(1 - \frac{1}{k^2}\right)\right]\left(1 - \dfrac{1}{(k + 1)^2}\right) $$
|
|
|
|
By the inductive hypothesis:
|
|
|
|
$$ = \left(\frac{k + 1}{2k}\right)\left(1 - \dfrac{1}{(k + 1)^2}\right) $$
|
|
|
|
$$ = \left(\frac{k + 1}{2k}\right)\left(\frac{(k + 1)^2}{(k + 1)^2} - \dfrac{1}{(k + 1)^2}\right) $$
|
|
|
|
$$ = \left(\frac{k + 1}{2k}\right)\left(\dfrac{(k + 1)^2 - 1}{(k + 1)^2}\right) $$
|
|
|
|
$$ = \frac{(k + 1)((k + 1)^2 - 1)}{2k(k + 1)^2} $$
|
|
|
|
$$ = \frac{(k + 1)^3 - (k + 1)}{2k(k + 1)^2} $$
|
|
|
|
$$ = \frac{(k + 1)^2 - 1}{2k(k + 1)} $$
|
|
|
|
$$ = \frac{(k + 1)(k + 1) - 1}{2k^2 + 2k} $$
|
|
|
|
$$ = \frac{k^2 + 2k + 1 - 1}{2k^2 + 2k} $$
|
|
|
|
$$ = \frac{k^2 + 2k}{2k^2 + 2k} $$
|
|
|
|
$$ = \frac{k(k + 2)}{k(2k + 2)} $$
|
|
|
|
$$ = \frac{k + 2}{2k + 2} $$
|
|
|
|
Evaluate right-hand:
|
|
|
|
Both sides of $P(k + 1)$ are equal. Therefore $P(k + 1)$ is true.
|
|
|
|
Q.E.D.
|
|
|
|
$$ \dfrac{k + 2}{2k + 2} $$
|
|
|
|
17. $\prod_{i = 0}^{n}{\left(\dfrac{1}{2i + 1} \cdot \dfrac{1}{2i + 2}\right)} = \dfrac{1}{(2n + 2)!}$,
|
|
for every integer $n \geq 0$.
|
|
|
|
**Proof (by mathematical induction):**
|
|
|
|
Let $P(n)$ be the equation:
|
|
|
|
$$ \prod_{i = 0}^{n}{\left(\dfrac{1}{2i + 1} \cdot \dfrac{1}{2i + 2}\right)} = \dfrac{1}{(2n + 2)!} $$
|
|
|
|
_Basis Step:_
|
|
|
|
Prove $P(0)$. That is:
|
|
|
|
$$ \prod_{i = 0}^{(0)}{\left(\dfrac{1}{2i + 1} \cdot \dfrac{1}{2i + 2}\right)} = \dfrac{1}{(2(0) + 2)!} $$
|
|
|
|
Alternatively:
|
|
|
|
$$ \prod_{i = 0}^{(0)}{\left(\dfrac{1}{2i + 1} \cdot \dfrac{1}{2i + 2}\right)} = \dfrac{1}{2!} $$
|
|
|
|
$$ \prod_{i = 0}^{(0)}{\left(\dfrac{1}{2i + 1} \cdot \dfrac{1}{2i + 2}\right)} = \dfrac{1}{2} $$
|
|
|
|
Evaluate left-hand when $n = 0$:
|
|
|
|
$$ \prod_{i = 0}^{(0)}{\left(\dfrac{1}{2i + 1} \cdot \dfrac{1}{2i + 2}\right)} $$
|
|
|
|
$$ = \frac{1}{2} $$
|
|
|
|
Both sides of $P(0)$ are equal. Therefore $P(0)$ is true.
|
|
|
|
_Inductive Step:_
|
|
|
|
Let $k$ be any integer where $k \geq 0$.
|
|
|
|
Suppose $P(k)$. That is:
|
|
|
|
$$ \prod_{i = 0}^{k}{\left(\dfrac{1}{2i + 1} \cdot \dfrac{1}{2i + 2}\right)} = \dfrac{1}{(2k + 2)!} $$
|
|
|
|
This is the inductive hypothesis.
|
|
|
|
Prove $P(k + 1)$. That is:
|
|
|
|
$$ \prod_{i = 0}^{(k + 1)}{\left(\dfrac{1}{2i + 1} \cdot \dfrac{1}{2i + 2}\right)} = \dfrac{1}{(2(k + 1) + 2)!} $$
|
|
|
|
Alternatively:
|
|
|
|
$$ \prod_{i = 0}^{k + 1}{\left(\dfrac{1}{2i + 1} \cdot \dfrac{1}{2i + 2}\right)} = \dfrac{1}{(2k + 2 + 2)!} $$
|
|
|
|
$$ \prod_{i = 0}^{k + 1}{\left(\dfrac{1}{2i + 1} \cdot \dfrac{1}{2i + 2}\right)} = \dfrac{1}{(2k + 4)!} $$
|
|
|
|
Evaluate left-hand:
|
|
|
|
$$ \prod_{i = 0}^{k + 1}{\left(\dfrac{1}{2i + 1} \cdot \dfrac{1}{2i + 2}\right)} $$
|
|
|
|
$$ = \left[\prod_{i = 0}^{k}{\left(\dfrac{1}{2i + 1} \cdot \dfrac{1}{2i + 2}\right)}\right] \cdot \left(\frac{1}{2(k + 1) + 1} \cdot \frac{1}{2(k + 1) + 2}\right) $$
|
|
|
|
By the inductive hypothesis:
|
|
|
|
$$ = \left(\frac{1}{(2k + 2)!}\right) \cdot \left(\frac{1}{2(k + 1) + 1} \cdot \frac{1}{2(k + 1) + 2}\right) $$
|
|
|
|
$$ = \left(\frac{1}{(2k + 2)!}\right) \cdot \left(\frac{1}{2k + 2 + 1} \cdot \frac{1}{2k + 2 + 2}\right) $$
|
|
|
|
$$ = \left(\frac{1}{(2k + 2)!}\right) \cdot \left(\frac{1}{2k + 3} \cdot \frac{1}{2k + 4}\right) $$
|
|
|
|
$$ = \frac{1}{(2k + 4)!} $$
|
|
|
|
Evaluate right-hand:
|
|
|
|
$$ \dfrac{1}{(2k + 4)!} $$
|
|
|
|
Both sides of $P(k + 1)$ are equal. Therefore $P(k + 1)$ is true.
|
|
|
|
Q.E.D.
|
|
|
|
18. $\prod_{i = 2}^{n}{\left(1 - \dfrac{1}{i}\right)} = \dfrac{1}{n}$ for every
|
|
integer $n \geq 2$.
|
|
|
|
_Hint:_ See the discussion at the beginning of this section.
|
|
|
|
**Proof (by mathematical induction):**
|
|
|
|
Let $P(n)$ be the equation:
|
|
|
|
$$ \prod_{i = 2}^{n}{\left(1 - \dfrac{1}{i}\right)} = \dfrac{1}{n} $$
|
|
|
|
_Basis Step:_
|
|
|
|
Prove $P(2)$. That is:
|
|
|
|
$$ \prod_{i = 2}^{2}{\left(1 - \dfrac{1}{i}\right)} = \dfrac{1}{2} $$
|
|
|
|
Evaluate left-hand side when $n = 2$:
|
|
|
|
$$ \prod_{i = 2}^{2}{\left(1 - \dfrac{1}{i}\right)} $$
|
|
|
|
$$ = 1 - \frac{1}{2} $$
|
|
|
|
$$ = \frac{1}{2} $$
|
|
|
|
Both sides of $P(2)$ are equal. Therefore $P(2)$ is true.
|
|
|
|
_Inductive Step:_
|
|
|
|
Let $k$ be any integer where $k \geq 2$.
|
|
|
|
Suppose $P(k)$. That is:
|
|
|
|
$$ \prod_{i = 2}^{k}{\left(1 - \dfrac{1}{i}\right)} = \dfrac{1}{k} $$
|
|
|
|
This is the inductive hypothesis.
|
|
|
|
Prove $P(k + 1)$. That is:
|
|
|
|
$$ \prod_{i = 2}^{k + 1}{\left(1 - \dfrac{1}{i}\right)} = \dfrac{1}{k + 1} $$
|
|
|
|
Evaluate left-hand side:
|
|
|
|
$$ \prod_{i = 2}^{k + 1}{\left(1 - \dfrac{1}{i}\right)} $$
|
|
|
|
$$ = \left[\prod_{i = 2}^{k}{\left(1 - \dfrac{1}{i}\right)}\right] \cdot \left(1 - \frac{1}{k + 1}\right) $$
|
|
|
|
By the inductive hypothesis:
|
|
|
|
$$ = \dfrac{1}{k} \cdot \left(1 - \frac{1}{k + 1}\right) $$
|
|
|
|
$$ = \dfrac{1}{k} \cdot \left(\frac{k + 1}{k + 1} - \frac{1}{k + 1}\right) $$
|
|
|
|
$$ = \dfrac{1}{k} \cdot \left(\frac{(k + 1) - 1}{k + 1}\right) $$
|
|
|
|
$$ = \dfrac{1}{k} \cdot \left(\frac{k}{k + 1}\right) $$
|
|
|
|
$$ = \frac{1}{k + 1} $$
|
|
|
|
Evaluate right-hand side:
|
|
|
|
$$ \dfrac{1}{k + 1} $$
|
|
|
|
Both sides of $P(k + 1)$ are equal. Therefore $P(k + 1)$ is true.
|
|
|
|
Q.E.D.
|
|
|
|
19. (For students who have studied calculus) Use mathematical induction, the
|
|
product rule from calculus, and the facts that $\dfrac{d(x)}{dx} = 1$ and
|
|
that $x^{k + 1} = x \cdot x^k$ to prove that for every integer $n \geq 1$,
|
|
$\dfrac{d(x^n)}{dx} = nx^{n - 1}$.
|
|
|
|
**Proof (by mathematical induction):**
|
|
|
|
Let $P(n)$ be the equation:
|
|
|
|
$$ \frac{d(x^n)}{dx} = nx^{n - 1} $$
|
|
|
|
_Basis Step:_
|
|
|
|
Prove $P(1)$. That is:
|
|
|
|
$$ \frac{d(x^{(1)})}{dx} = (1)x^{(1) - 1} $$
|
|
|
|
Alternatively:
|
|
|
|
$$ \frac{dx}{dx} = 1x^0 $$
|
|
|
|
Evaluate the left-hand side when $n = 1$:
|
|
|
|
$$ \frac{dx}{dx} $$
|
|
|
|
By the given fact that $\dfrac{dx}{dx} = 1$:
|
|
|
|
$$ = 1 $$
|
|
|
|
Evaluate the right-hand side when $n = 1$:
|
|
|
|
$$ = 1x^0 $$
|
|
|
|
$$ = 1 $$
|
|
|
|
Both the left and right hand sides of $P(1)$ are equal. Therefore $P(1)$ is
|
|
true.
|
|
|
|
_Inductive Step:_
|
|
|
|
Let $k$ be any integer where $k \geq 1$.
|
|
|
|
Suppose $P(k)$. That is:
|
|
|
|
$$ \frac{d(x^k)}{dx} = kx^{k - 1} $$
|
|
|
|
This is the inductive hypothesis.
|
|
|
|
Prove $P(k + 1)$. That is:
|
|
|
|
$$ \frac{d(x^{(k + 1)})}{dx} = (k + 1)x^{(k + 1) - 1} $$
|
|
|
|
Alternatively:
|
|
|
|
$$ \frac{d(x^{(k + 1)})}{dx} = (k + 1)x^k $$
|
|
|
|
Evaluate left-hand side:
|
|
|
|
$$ \frac{d(x^{(k + 1)})}{dx} $$
|
|
|
|
$$ \frac{d(x \cdot x^k)}{dx} $$
|
|
|
|
By the product rule, we can separate this out into:
|
|
|
|
$$ \frac{dx}{dx} \cdot x^k + x \cdot \dfrac{d(x^k)}{dx} $$
|
|
|
|
By the given fact that $\dfrac{dx}{dx} = 1$:
|
|
|
|
$$ 1 \cdot x^k + x \cdot \dfrac{d(x^k)}{dx} $$
|
|
|
|
By the inductive hypothesis:
|
|
|
|
$$ 1 \cdot x^k + x \cdot kx^{k - 1} $$
|
|
|
|
$$ x^k + x \cdot kx^{k - 1} $$
|
|
|
|
$$ x^k + kx^{k - 1 + 1} $$
|
|
|
|
$$ x^k + kx^{k} $$
|
|
|
|
$$ x^k(1 + k) $$
|
|
|
|
$$ (k + 1)x^k $$
|
|
|
|
Evaluate right-hand side:
|
|
|
|
$$ (k + 1)x^k $$
|
|
|
|
Both the left and right sides of $P(k + 1)$ are equal. Therefore $P(k + 1)$ is
|
|
true.
|
|
|
|
Q.E.D.
|
|
|
|
Use the formula for the sum of the first $n$ integers and/or the formula for the
|
|
sum of a geometric sequence to evaluate the sums in 20-29 or to write them in
|
|
closed form.
|
|
|
|
20. $4 + 8 + 12 + 16 + \dots + 200$
|
|
|
|
$$ 4 + 8 + 12 + 16 + \dots + 200 $$
|
|
|
|
$$ = 4(1 + 2 + 3 + 4 + \dots + 50) $$
|
|
|
|
$$ = 4\frac{50(51)}{2} $$
|
|
|
|
$$ = 5100 $$
|
|
|
|
21. $5 + 10 + 15 + 20 + \dots + 300$
|
|
|
|
$$ 5 + 10 + 15 + 20 + \dots + 300 $$
|
|
|
|
$$ = 5(1 + 2 + 3 + 4 + \dots 60) $$
|
|
|
|
$$ = 5\left(\frac{(60)(61)}{2}\right) $$
|
|
|
|
$$ = 9150 $$
|
|
|
|
22.
|
|
|
|
a. $3 + 4 + 5 + 6 + \dots + 1000$
|
|
|
|
$$ 3 + 4 + 5 + 6 + \dots + 1000 $$
|
|
|
|
$$ = (1 + 2 + 3 + 4 + \dots + 1000) - (1 + 2) $$
|
|
|
|
$$ = \left(\frac{(1000)(1001)}{2}\right) - 3 $$
|
|
|
|
$$ = 500497 $$
|
|
|
|
b. $3 + 4 + 5 + 6 + \dots + m$
|
|
|
|
$$ 3 + 4 + 5 + 6 + \dots + m $$
|
|
|
|
$$ = (1 + 2 + 3 + 4+ \dots + m) - (1 + 2) $$
|
|
|
|
$$ = \left(\frac{(m)(m + 1)}{2}\right) - 3 $$
|
|
|
|
$$ = \frac{m^2 + m}{2} - 3 $$
|
|
|
|
$$ = \frac{m^2 + m}{2} - \frac{6}{2} $$
|
|
|
|
$$ = \frac{m^2 + m - 6}{2} $$
|
|
|
|
23.
|
|
|
|
a. $7 + 8 + 9 + 10 + \dots + 600$
|
|
|
|
$$ 7 + 8 + 9 + 10 + \dots + 600 $$
|
|
|
|
$$ = (1 + 2 + 3 + 4 + \dots + 600) - (1 + 2 + 3 + 4 + 5 + 6) $$
|
|
|
|
$$ = \left(\frac{(600)(601)}{2}\right) - 21 $$
|
|
|
|
$$ = 180279 $$
|
|
|
|
b. $7 + 8 + 9 + 10 + \dots + k$
|
|
|
|
$$ 7 + 8 + 9 + 10 + \dots + k $$
|
|
|
|
$$ = (1 + 2 + 3 + 4 + \dots + k) - (1 + 2 + 3 + 4 + 5 + 6) $$
|
|
|
|
$$ = \left(\frac{(k)(k + 1)}{2}\right) - 21 $$
|
|
|
|
$$ = \frac{k^2 + k}{2} - 21 $$
|
|
|
|
$$ = \frac{k^2 + k - 42}{2} $$
|
|
|
|
24. $1 + 2 + 3 + \dots + (k - 1)$, where $k$ is any integer with $k \geq 2$.
|
|
|
|
$$ 1 + 2 + 3 + \dots + (k - 1) $$
|
|
|
|
$$ = \frac{(k - 1)((k - 1) + 1)}{2} $$
|
|
|
|
$$ = \frac{(k - 1)(k)}{2} $$
|
|
|
|
$$ = \frac{k^2 - k}{2} $$
|
|
|
|
25.
|
|
|
|
a. $1 + 2 + 2^2 + \dots + 2^{25}$
|
|
|
|
$$ 1 + 2 + 2^2 + \dots + 2^{25} $$
|
|
|
|
$$ = \frac{2^{25 + 1} - 1}{2^{25} - 1} $$
|
|
|
|
$$ = \frac{2^{26} - 1}{2 - 1} $$
|
|
|
|
$$ = 67108863 $$
|
|
|
|
b. $2 + 2^2 + 2^3 + \dots + 2^{26}$
|
|
|
|
$$ 2 + 2^2 + 2^3 + \dots + 2^{26} $$
|
|
|
|
$$ k 2(1 + 2 + 2^2 + \dots + 2^{25}) $$
|
|
|
|
By part a:
|
|
|
|
$$ = 2(67108863) $$
|
|
|
|
$$ = 134217726 $$
|
|
|
|
c. $2 + 2^2 + 2^3 + \dots + 2^n$
|
|
|
|
$$ 2 + 2^2 + 2^3 + \dots + 2^n $$
|
|
|
|
$$ 2(1 + 2 + 2^2 + \dots + 2^{n - 1}) $$
|
|
|
|
$$ 2\left(\frac{2^{(n - 1) + 1} - 1}{2 - 1}\right) $$
|
|
|
|
$$ 2\left(\frac{2^n - 1}{1}\right) $$
|
|
|
|
$$ 2(2^n - 1) $$
|
|
|
|
$$ 2^{n + 1} - 2 $$
|
|
|
|
26. $3 + 3^2 + 3^3 + \dots + 3^n$, where $n$ is any integer with $n \geq 1$.
|
|
|
|
$$ 3 + 3^2 + 3^3 + \dots + 3^n $$
|
|
|
|
$$ 3(1 + 3 + 3^2 + \dots + 3^{n - 1}) $$
|
|
|
|
$$ 3\left(\frac{3^{(n - 1) + 1} - 1}{3 - 1}\right) $$
|
|
|
|
$$ 3\left(\frac{3^n - 1}{2}\right) $$
|
|
|
|
$$ \frac{3^{n + 1} - 3}{2} $$
|
|
|
|
27. $5^3 + 5^4 + 5^5 + \dots + 5^k$, where $k$ is any integer with $k \geq 3$.
|
|
|
|
$$ 5^3 + 5^4 + 5^5 + \dots + 5^k $$
|
|
|
|
$$ = 5^3(1 + 5 + 5^2 + \dots + 5^{k - 3}) $$
|
|
|
|
$$ = 5^3\left(\frac{5^{(k - 3) + 1} - 1}{5 - 1}\right) $$
|
|
|
|
$$ = 5^3\left(\frac{5^{k - 2} - 1}{4}\right) $$
|
|
|
|
$$ = \frac{5^{k - 2 + 3} - 5^3}{4} $$
|
|
|
|
$$ = \frac{5^k - 5^3}{4} $$
|
|
|
|
28. $1 + \dfrac{1}{2} + \dfrac{1}{2^2} + \dots + \dfrac{1}{2^n}$, where $n$ is
|
|
any positive integer.
|
|
|
|
$$ 1 + \dfrac{1}{2} + \dfrac{1}{2^2} + \dots + \dfrac{1}{2^n} $$
|
|
|
|
$$ = \frac{\left(\dfrac{1}{2}\right)^{n + 1} - 1}{\dfrac{1}{2} - 1} $$
|
|
|
|
$$ = \frac{\left(\dfrac{1}{2}\right)^{n + 1} - 1}{-\dfrac{1}{2}} $$
|
|
|
|
$$ = \left[\left(\dfrac{1}{2}\right)^{n + 1} - 1\right](-2) $$
|
|
|
|
$$ = \left(\dfrac{-2}{2}\right)^{n + 1} + 2 $$
|
|
|
|
$$ = 2 - \left(\dfrac{-2}{2}\right)^{n + 1} $$
|
|
|
|
$$ = 2 + \dfrac{1}{2^n} $$
|
|
|
|
29. $1 - 2 + 2^2 - 2^3 + \dots + (-1)^n2^n$, where $n$ is any positive integer.
|
|
|
|
$$ 1 - 2 + 2^2 - 2^3 + \dots + (-1)^n2^n $$
|
|
|
|
$$ = 1 + (-2) + (-2)^2 + (-2)^3 + \dots + (-2)^n $$
|
|
|
|
$$ = \frac{(-2)^{n + 1} - 1}{(-2) - 1} $$
|
|
|
|
$$ = \frac{(-2)^{n + 1} - 1}{-3} $$
|
|
|
|
30. Observe that
|
|
|
|
$$ \frac{1}{1 \cdot 3} = \frac{1}{3} $$
|
|
|
|
$$ \frac{1}{1 \cdot 3} + \frac{1}{3 \cdot 5} = \frac{2}{5} $$
|
|
|
|
$$ \frac{1}{1 \cdot 3} + \frac{1}{3 \cdot 5} + \frac{1}{5 \cdot 7} = \frac{3}{7} $$
|
|
|
|
$$ \frac{1}{1 \cdot 3} + \frac{1}{3 \cdot 5} + \frac{1}{5 \cdot 7} + \frac{1}{7 \cdot 9} = \frac{4}{9} $$
|
|
|
|
Guess a general formula and prove it by mathematical induction.
|
|
|
|
General formula:
|
|
|
|
$$ \frac{1}{1 \cdot 3} + \frac{1}{3 \cdot 5} + \frac{1}{5 \cdot 7} + \dots + \frac{1}{(2n - 1)(2n + 1)} = \frac{n}{2n + 1} $$
|
|
|
|
for all integers $n \geq 1$.
|
|
|
|
**Proof (by mathematical induction):**
|
|
|
|
Let $P(n)$ be the equation:
|
|
|
|
$$ \frac{1}{1 \cdot 3} + \frac{1}{3 \cdot 5} + \frac{1}{5 \cdot 7} + \dots + \frac{1}{(2n - 1)(2n + 1)} = \frac{n}{2n + 1} $$
|
|
|
|
_Basis Step:_
|
|
|
|
Prove $P(1)$:
|
|
|
|
$$ \frac{1}{1 \cdot 3} + \frac{1}{3 \cdot 5} + \frac{1}{5 \cdot 7} + \dots + \frac{1}{(2(1) - 1)(2(1) + 1)} = \frac{(1)}{2(1) + 1} $$
|
|
|
|
Evaluate left-hand side when $n = 1$:
|
|
|
|
$$ \frac{1}{1 \cdot 3} + \frac{1}{3 \cdot 5} + \frac{1}{5 \cdot 7} + \dots + \frac{1}{(2(1) - 1)(2(1) + 1)} $$
|
|
|
|
$$ = \frac{1}{(2 - 1)(2 + 1)}$$
|
|
|
|
$$ = \frac{1}{(1)(3)}$$
|
|
|
|
$$ = \frac{1}{3} $$
|
|
|
|
Evaluate right-hand side when $n = 1$:
|
|
|
|
$$ \frac{(1)}{2(1) + 1} $$
|
|
|
|
$$ \frac{1}{2 + 1} $$
|
|
|
|
$$ \frac{1}{3} $$
|
|
|
|
The left and right hand sides of $P(1)$ are equal. Therefore $P(1)$ is true.
|
|
|
|
_Inductive Step:_
|
|
|
|
Let $k$ be any integer where $k \geq 1$.
|
|
|
|
Suppose $P(k)$. That is:
|
|
|
|
$$ \frac{1}{1 \cdot 3} + \frac{1}{3 \cdot 5} + \frac{1}{5 \cdot 7} + \dots + \frac{1}{(2k - 1)(2k + 1)} = \frac{k}{2k + 1} $$
|
|
|
|
This is the inductive hypothesis.
|
|
|
|
Prove $P(k + 1)$. That is:
|
|
|
|
$$ \frac{1}{1 \cdot 3} + \frac{1}{3 \cdot 5} + \frac{1}{5 \cdot 7} + \dots + \frac{1}{(2(k + 1) - 1)(2(k + 1) + 1)} = \frac{(k + 1)}{2(k + 1) + 1} $$
|
|
|
|
Alternatively:
|
|
|
|
$$ \frac{1}{1 \cdot 3} + \frac{1}{3 \cdot 5} + \frac{1}{5 \cdot 7} + \dots + \frac{1}{(2k + 2 - 1)(2k + 2 + 1)} = \frac{k + 1}{2k + 2 + 1} $$
|
|
|
|
$$ \frac{1}{1 \cdot 3} + \frac{1}{3 \cdot 5} + \frac{1}{5 \cdot 7} + \dots + \frac{1}{(2k + 1)(2k + 3)} = \frac{k + 1}{2k + 3} $$
|
|
|
|
Evaluate the left-hand side:
|
|
|
|
$$ \frac{1}{1 \cdot 3} + \frac{1}{3 \cdot 5} + \frac{1}{5 \cdot 7} + \dots + \frac{1}{(2k + 1)(2k + 3)} $$
|
|
|
|
$$ = \left[\frac{1}{1 \cdot 3} + \frac{1}{3 \cdot 5} + \frac{1}{5 \cdot 7} + \dots + \frac{1}{(2k - 1)(2k + 1)}\right] + \frac{1}{(2k + 1)(2k + 3)} $$
|
|
|
|
By the inductive hypothesis:
|
|
|
|
$$ = \frac{k}{2k + 1} + \frac{1}{(2k + 1)(2k + 3)} $$
|
|
|
|
$$ = \frac{k(2k + 3)}{(2k + 1)(2k + 3)} + \frac{1}{(2k + 1)(2k + 3)} $$
|
|
|
|
$$ = \frac{k(2k + 3) + 1}{(2k + 1)(2k + 3)} $$
|
|
|
|
$$ = \frac{2k^2 + 3k + 1}{(2k + 1)(2k + 3)} $$
|
|
|
|
$$ = \frac{(2k + 1)(k + 1)}{(2k + 1)(2k + 3)} $$
|
|
|
|
$$ = \frac{k + 1}{2k + 3} $$
|
|
|
|
Evaluate the right-hand side:
|
|
|
|
$$ \frac{k + 1}{2k + 3} $$
|
|
|
|
Both sides of $P(k + 1)$ are equal. Therefore $P(k + 1)$ is true.
|
|
|
|
Q.E.D.
|
|
|
|
31. Compute values of the product
|
|
|
|
$$ \left(1 + \frac{1}{1}\right)\left(1 + \frac{1}{2}\right)\left(1 + \frac{1}{3}\right) \dots \left(1 + \frac{1}{n}\right) $$
|
|
|
|
for small values of $n$ in order to conjecture a general formula for the
|
|
product. Prove your conjecture by mathematical induction.
|
|
|
|
32. Observe that
|
|
|
|
$$ 1 = 1 $$
|
|
|
|
$$ 1 - 4 = -(1 + 2) $$
|
|
|
|
$$ 1 - 4 + 9 = 1 + 2 + 3 $$
|
|
|
|
$$ 1 - 4 + 9 - 16 = -(1 + 2 + 3 + 4) $$
|
|
|
|
$$ 1 - 4 + 9 - 16 + 25 = 1 + 2 + 3 + 4 + 5 $$
|
|
|
|
Guess a general formula and prove it by mathematical induction.
|
|
|
|
$$ 1 - 4 + 9 - 16 + \dots + (-1)^{n - 1}(n^2) = (-1)^{n - 1}(1 + 2 + 3 + 4 + \dots + n) $$
|
|
|
|
**Proof (by mathematical induction):**
|
|
|
|
Let $P(n)$ be the equation:
|
|
|
|
$$ 1 - 4 + 9 - 16 + \dots + (-1)^{n - 1}(n^2) = (-1)^{n - 1}(1 + 2 + 3 + 4 + \dots + n) $$
|
|
|
|
for all integers $n \geq 1$.
|
|
|
|
_Basis Step_:
|
|
|
|
Prove $P(1)$. That is:
|
|
|
|
$$ 1 - 4 + 9 - 16 + \dots + (-1)^{(1) - 1}((1)^2) = (-1)^{(1) - 1}(1 + 2 + 3 + 4 + \dots + (1)) $$
|
|
|
|
Evaluate left-hand side when $n = 1$:
|
|
|
|
$$ 1 - 4 + 9 - 16 + \dots + (-1)^{(1) - 1}((1)^2) $$
|
|
|
|
$$ = (-1)^{0}(1^2) $$
|
|
|
|
$$ = 1(1) $$
|
|
|
|
$$ = 1 $$
|
|
|
|
Evaluate right-hand side when $n = 1$:
|
|
|
|
$$ (-1)^{(1) - 1}(1 + 2 + 3 + 4 + \dots + (1)) $$
|
|
|
|
$$ = 1 $$
|
|
|
|
Both sides of $P(1)$ are equal. Therefore $P(1)$ is true.
|
|
|
|
_Inductive Step_:
|
|
|
|
Let $k$ be any integer where $k \geq 1$.
|
|
|
|
Suppose $P(k)$. That is:
|
|
|
|
$$ 1 - 4 + 9 - 16 + \dots + (-1)^{k - 1}(k^2) = (-1)^{k - 1}(1 + 2 + 3 + 4 + \dots + k) $$
|
|
|
|
This is the inductive hypothesis.
|
|
|
|
Prove $P(k + 1)$. That is:
|
|
|
|
$$ 1 - 4 + 9 - 16 + \dots + (-1)^{(k + 1) - 1}((k + 1)^2) = (-1)^{(k + 1) - 1}(1 + 2 + 3 + 4 + \dots + (k + 1)) $$
|
|
|
|
Alternatively:
|
|
|
|
$$ 1 - 4 + 9 - 16 + \dots + (-1)^{k}((k + 1)^2) = (-1)^{k}(1 + 2 + 3 + 4 + \dots + (k + 1)) $$
|
|
|
|
Evaluate left-hand:
|
|
|
|
$$ 1 - 4 + 9 - 16 + \dots + (-1)^{k}((k + 1)^2) $$
|
|
|
|
$$ = \left[1 - 4 + 9 - 16 + \dots + (-1)^{k - 1}(k^2)\right] + (-1)^{k}((k + 1)^2) $$
|
|
|
|
By the inductive hypothesis:
|
|
|
|
$$ = (-1)^{k - 1}(1 + 2 + 3 + 4 + \dots + k) + (-1)^{k}((k + 1)^2) $$
|
|
|
|
$$ = (-1)^{k - 1}\left[(1 + 2 + 3 + 4 + \dots + k) - (k + 1)^2\right] $$
|
|
|
|
By 5.2.1:
|
|
|
|
$$ = (-1)^{k - 1}\left[\frac{k(k + 1)}{2} - (k + 1)^2\right] $$
|
|
|
|
$$ = (-1)^{k - 1}\left[(k + 1)\left(\frac{k}{2} - (k + 1)\right)\right] $$
|
|
|
|
$$ = (-1)^{k - 1}\left[(k + 1)\left(\frac{k}{2} - \frac{2(k + 1)}{2}\right)\right] $$
|
|
|
|
$$ = (-1)^{k - 1}\left[(k + 1)\left(\frac{k - 2(k + 1)}{2}\right)\right] $$
|
|
|
|
$$ = (-1)^{k - 1}\left[(k + 1)\left(\frac{k - 2k - 2)}{2}\right)\right] $$
|
|
|
|
$$ = (-1)^{k - 1}\left[(k + 1)\left(\frac{-k - 2)}{2}\right)\right] $$
|
|
|
|
$$ = (-1)^{k - 1}\left[(-1)(k + 1)\left(\frac{k + 2}{2}\right)\right] $$
|
|
|
|
$$ = (-1)^{k - 1}\left[(-1)\left(\frac{(k + 1)(k + 2)}{2}\right)\right] $$
|
|
|
|
$$ = (-1)^{k}\left(\frac{(k + 1)(k + 2)}{2}\right) $$
|
|
|
|
By 5.2.1:
|
|
|
|
$$ = (-1)^{k}(1 + 2 + 3 + 4 + \dots + (k + 1)) $$
|
|
|
|
Evaluate right-hand:
|
|
|
|
$$ (-1)^{k}(1 + 2 + 3 + 4 + \dots + (k + 1)) $$
|
|
|
|
Both sides of $P(k + 1)$ are equal. Therefore $P(k + 1)$ is true.
|
|
|
|
Q.E.D.
|
|
|
|
33. Find a formula in $n$, $a$, $m$, and $d$ for the sum
|
|
$(a + md) + (a + (m + 1)d) + (a + (m + 2)d) + \dots + (a + (m + n)d)$, where
|
|
$m$ and $n$ are integers, $n \geq 0$, and $a$ and $d$ are real numbers.
|
|
Justify your answer.
|
|
|
|
$$ a + (a + d) + (a + 2d) + \dots (a + nd) = (n + 1)a + d\left(\frac{n(n + 1)}{2}\right) $$
|
|
|
|
34. Find a formula in $a$, $r$, $m$, and $n$ for the sum
|
|
|
|
$$ ar^m + ar^{m + 1} + ar^{m + 2} + \dots + ar^{m + n} $$
|
|
|
|
where $m$ and $n$ are integers, $n \geq 0$, and $a$ and $r$ are real numbers.
|
|
Justify your answer.
|
|
|
|
$$ ar^m + ar^{m + 1} + ar^{m + 2} + \dots + ar^{m + n} = ar^m\left(\frac{r^{n + 1} - 1}{r - 1}\right) $$
|
|
|
|
By factoring out the $ar^m$, this just becomes a geometric series:
|
|
|
|
$$ ar^m(1 + r + r^2 + r^3 + \dots r^n) $$
|
|
|
|
And by 5.2.2, we can substitute that series out with:
|
|
|
|
$$ ar^m\left(\frac{r^{n + 1} - 1}{r - 1}\right) $$
|
|
|
|
35. You have two parents, four grandparents, eight great-grandparents, and so
|
|
forth.
|
|
|
|
a. If all your ancestors were distinct, what would be the total number of your
|
|
ancestors for the past 40 generations (counting your parents' generation as
|
|
number one)? (_Hint:_ Use the formula for the sum of a geometric sequence.)
|
|
|
|
The geometric sequence for this is:
|
|
|
|
$$ 1 + 2 + 2^2 + 2^3 + \dots + 2^n $$
|
|
|
|
So, by 5.2.2, this is:
|
|
|
|
$$ \frac{2^{n + 1} - 1}{2 - 1} $$
|
|
|
|
Where $n$ is the number of generations. Plugging in 39 (since we count as the
|
|
first generation) returns:
|
|
|
|
$$ \frac{2^{39 + 1} - 1}{2 - 1} $$
|
|
|
|
$$ = \frac{2^{40} - 1}{1} $$
|
|
|
|
$$ = 2^{40} - 1 $$
|
|
|
|
$$ = 1099511627775 $$
|
|
|
|
b. Assuming that each generation represents 25 years, how long is 40
|
|
generations?
|
|
|
|
$$ 25 \cdot 1099511627775 $$
|
|
|
|
$$ \approx 2.748779069 \cdot 10^{13} \text{ years} $$
|
|
|
|
c. The total number of people who have ever lived is approximately 10 billion,
|
|
which equals $10^{10}$ people. Compare this fact with the answer to part (a).
|
|
What can you deduce?
|
|
|
|
When demarcated for easier reading, part a's answer reads as:
|
|
|
|
$$ = 1,099,511,627,775 $$
|
|
|
|
Which is 1 trillion, 99 billion, 511 million, 627 thousand, 775 people. Since
|
|
this exceeds the approximate total number of people who have ever lived. We can
|
|
deduce that some(probably many) of my ancestors must have been related to one
|
|
another.
|
|
|
|
Find the mistakes in the proof fragments in 36-38.
|
|
|
|
36.
|
|
|
|
**Theorem:**
|
|
|
|
For any integer $n \geq 1$,
|
|
|
|
$$ 1^2 + 2^2 + \dots + n^2 = \frac{n(n + 1)(2n + 1)}{6} $$
|
|
|
|
**"Proof (by mathematical induction):**
|
|
|
|
Certainly the theorem is true for $n = 1$ because $1^2 = 1$ and
|
|
$\dfrac{1(1 + 1)(2 \cdot 1 + 1)}{6} = 1$ . So the basis step is true. For the
|
|
inductive step, suppose that $k$ is any integer with $k \geq 1$,
|
|
$k^2 = \dfrac{k(k + 1)(2k + 1)}{6}$. We must show that
|
|
$(k + 1)^2 = \dfrac{(k + 1)((k + 1) + 1)(2(k + 1) + 1)}{6}$."
|
|
|
|
In the inductive step, the inductive hypothesis reads:
|
|
|
|
$$ k^2 = \frac{k(k + 1)(2k + 1)}{6} $$
|
|
|
|
But it should read:
|
|
|
|
$$ 1^2 + 2^2 + \dots + k^2 = \frac{k(k + 1)(2k + 1)}{6} $$
|
|
|
|
This error cascades into their proof, which reads:
|
|
|
|
$$ (k + 1)^2 = \frac{(k + 1)((k + 1) + 1)(2(k + 1) + 1)}{6} $$
|
|
|
|
But instead should read:
|
|
|
|
$$ 1^2 + 2^2 + \dots + (k + 1)^2 = \frac{(k + 1)((k + 1) + 1)(2(k + 1) + 1)}{6} $$
|
|
|
|
37.
|
|
|
|
**Theorem:**
|
|
|
|
For any integer $n \geq 0$,
|
|
|
|
$$ 1 + 2 + 2^2 + \dots + 2^n = 2^{n + 1} - 1 $$
|
|
|
|
**"Proof (by mathematical induction):**
|
|
|
|
Let the property $P(n)$ be
|
|
|
|
$$ 1 + 2 + 2^2 + \dots + 2^n = 2^{n + 1} - 1 $$
|
|
|
|
_Show that $P(0)$ is true:_
|
|
|
|
The left-hand side of $P(0)$ is $1 + 2 + 2^2 + \dots + 2^0 = 1$ and the
|
|
right-hand side is $2^{0 + 1} - 1 = 2 - 1 = 1$ also. So $P(0)$ is true."
|
|
|
|
The left-hand side evaluation should instead read:
|
|
|
|
The left-hand side of $P(0)$ is $2^0 = 1$ since when $n = 0$, only the first
|
|
term is evaluated..
|
|
|
|
38.
|
|
|
|
**Theorem:**
|
|
|
|
For any integer $n \geq 1$,
|
|
|
|
$$ \sum_{i = 1}^{n}{i(i!)} = (n + 1)! - 1 $$
|
|
|
|
**"Proof (by mathematical induction):**
|
|
|
|
Let the property $P(n)$ be
|
|
|
|
$$ \sum_{i = 1}^{n}{i(i!)} = (n + 1)! - 1 $$
|
|
|
|
_Show that $P(1)$ is true:_
|
|
|
|
When $n = 1$,
|
|
|
|
$$ \sum_{i = 1}^{i}{i(i!)} = (1 + 1)! - 1 $$
|
|
|
|
So
|
|
|
|
$$ 1(1!) = 2! - 1$$
|
|
|
|
and
|
|
|
|
$$ 1 = 1 $$
|
|
|
|
Thus $P(1)$ is true."
|
|
|
|
The author of this proof fragment incorrectly rewrites the upper limit as $i$
|
|
instead of $1$. They write:
|
|
|
|
$$ \sum_{i = 1}^{i}{i(i!)} = (1 + 1)! - 1 $$
|
|
|
|
When it should be:
|
|
|
|
$$ \sum_{i = 1}^{1}{i(i!)} = (1 + 1)! - 1 $$
|
|
|
|
Then, they should evaluate each side independently, but instead they simply
|
|
evaluate each together, which is incorrect. Instead the basis step should be
|
|
written as:
|
|
|
|
Evaluate the left-hand side when $n = 1$:
|
|
|
|
$$ \sum_{i = 1}^{1}{i(i!)} $$
|
|
|
|
$$ = 1(1!) $$
|
|
|
|
$$ = 1(1) $$
|
|
|
|
$$ = 1 $$
|
|
|
|
Evaluate the right-hand side when $n = 1$:
|
|
|
|
$$ (1 + 1)! - 1 $$
|
|
|
|
$$ = (2)! - 1 $$
|
|
|
|
$$ = (2 \cdot 1) - 1 $$
|
|
|
|
$$ = 2 - 1 $$
|
|
|
|
$$ = 1 $$
|
|
|
|
Both sides of $P(1)$ are equal. Therefore $P(1)$ is true.
|
|
|
|
39. Use Theorem 5.2.1 to prove that if $m$ and $n$ are any positive integers and
|
|
$m$ is odd, then $\sum_{k = 0}^{m - 1}{(n + k)}$ is divisible by $m$. Does
|
|
the conclusion hold if $m$ is even? Justify your answer.
|
|
|
|
Omitted.
|
|
|
|
40. Use Theorem 5.2.1 and the result of exercise 10 to prove that if $p$ is any
|
|
prime number with $p \geq 5$, then the sum of the squares of any $p$
|
|
consecutive integers is divisible by $p$.
|
|
|
|
Omitted.
|
|
|
|
---
|
|
|
|
**Exercise Set 5.3**
|
|
|
|
Page 320
|
|
|
|
1. Use mathematical induction (and the proof of proposition 5.3.1 as a model) to
|
|
show that any amount of money of at least 14¢ can be made up using 3¢ and 8¢
|
|
coins.
|
|
|
|
**Proof (by mathematical induction):**
|
|
|
|
Let $P(n)$ be the sentence:
|
|
|
|
$n$¢ can be obtained using $3$¢ and $8$¢ coins.
|
|
|
|
_Basis Step:_
|
|
|
|
Prove $P(14)$:
|
|
|
|
$P(14)$ is true because $14$¢ can be obtained using one $8$¢ coin and two $3$¢
|
|
coins.
|
|
|
|
_Inductive Step:_
|
|
|
|
Let $k$ be any integer where $k \geq 14$.
|
|
|
|
Suppose $P(k)$ is true. That is:
|
|
|
|
$k$¢ can be obtained using $3$¢ and $8$¢ coins.
|
|
|
|
Prove $P(k + 1)$. That is:
|
|
|
|
$k + 1$¢ can be obtained using $3$¢ and $8$¢ coins.
|
|
|
|
_Case 1 (there is a $8$¢ coin among those used to make up $k$¢):_
|
|
|
|
In this case, replace the $8$¢ coin with three $3$¢ coins. The result will be
|
|
$k + 1$¢.
|
|
|
|
_Case 2 (there is not a $8$¢ coin among those used to make up $k$¢):_
|
|
|
|
In this case, because $k \geq 14$, at least 5 $3$¢ coins must have been used. So
|
|
remove five $3$¢ coins and replace them with two $8$¢ coins. The result will be
|
|
$k + 1$¢.
|
|
|
|
Therefore in either case $(k + 1)$¢ can be obtained using $3$¢ and $8$¢ coins.
|
|
|
|
Q.E.D.
|
|
|
|
2. Use mathematical induction to show that any postage of at least 12¢ can be
|
|
obtained using 3¢ and 7¢ stamps.
|
|
|
|
**Proof (by mathematical induction):**
|
|
|
|
Let $P(n)$ be the sentence:
|
|
|
|
$n$¢ postage can be obtained using $3$¢ and $7$¢ stamps.
|
|
|
|
_Basis Step:_
|
|
|
|
Prove $P(12)$. That is:
|
|
|
|
$12$¢ postage can be obtained using $3$¢ and $7$¢ stamps.
|
|
|
|
$12$¢ can be obtained using four $3$¢ stamps. Therefore $P(12)$ is true.
|
|
|
|
_Inductive Step:_
|
|
|
|
Let $k$ be any integer where $k \geq 12$.
|
|
|
|
Suppose $P(k)$. That is:
|
|
|
|
$k$¢ postage can be obtained using $3$¢ and $7$¢ stamps.
|
|
|
|
Prove $P(k + 1)$. That is:
|
|
|
|
$(k + 1)$¢ postage can be obtained using $3$¢ and $7$¢ stamps.
|
|
|
|
_Case 1 (at least one 2 $3$¢ stamps are used to make up $k$¢):_
|
|
|
|
Replace the two $3$¢ stamps with a $7$¢ stamp. This results in $(k + 1)$¢.
|
|
|
|
_Case 2 (there are 1 or 0 $3$¢ stamps among those used to make up $k$¢):_
|
|
|
|
Replace two $7$¢ stamps with five $3$¢ stamps. This results in $(k + 1)$¢.
|
|
|
|
Therefore, in both cases $(k + 1)$ postage can be obtained using $3$¢ and $7$¢
|
|
stamps.
|
|
|
|
Q.E.D.
|
|
|
|
3. Stamps are sold in packages containing either 5 stamps or 8 stamps.
|
|
|
|
a. Show that a person can obtain 5, 8, 10, 13, 15, 16, 20, 21, 24, or 25 stamps
|
|
by buying a collection of 5-stamp packages and 8-stamp packages.
|
|
|
|
- 5 stamps can be obtained by purchasing one 5 stamp package.
|
|
|
|
- 8 stamps can be obtained by purchasing one 8 stamp package.
|
|
|
|
- 10 stamps can be obtained by purchasing two 5 stamp packages.
|
|
|
|
- 13 stamps can be obtained by purchasing one 5 stamp package and one 8 stamp
|
|
package.
|
|
|
|
- 15 stamps can be obtained by purchasing three 5 stamp packages.
|
|
|
|
- 16 stamps can be obtained by purchasing two 8 stamp packages.
|
|
|
|
- 20 stamps can be obtained by purchasing four 5 stamp packages.
|
|
|
|
- 21 stamps can be obtained by purchasing two 8 stamp packages and one 5 stamp
|
|
package.
|
|
|
|
- 24 stamps can be obtained by purchasing three 8 stamp packages.
|
|
|
|
- 25 stamps can be obtained by purchasing five 5 stamp packages.
|
|
|
|
b. Use mathematical induction to show that any quantity of at least 28 stamps
|
|
can be obtained by buying a collection of 5-stamp packages and 8-stamp packages.
|
|
|
|
**Proof (by mathematical induction):**
|
|
|
|
Let $P(n)$ be the sentence:
|
|
|
|
$n$ stamps can be obtained by buying a collection of 5-stamp packages and
|
|
8-stamp packages.
|
|
|
|
_Basis Step:_
|
|
|
|
Prove $P(28)$. That is:
|
|
|
|
$28$ stamps can be obtained by buying a collection of 5-stamp packages and
|
|
8-stamp packages.
|
|
|
|
$28$ stamps can be obtained by buying four 5-stamp packages and one 8-stamp
|
|
package.
|
|
|
|
Therefore $P(28)$ is true.
|
|
|
|
_Inductive Step:_
|
|
|
|
Let $k$ be any integer where $k \geq 28$.
|
|
|
|
Suppose $P(k)$. That is:
|
|
|
|
$k$ stamps can be obtained by buying a collection of 5-stamp packages and
|
|
8-stamp packages.
|
|
|
|
Prove $P(k + 1)$. That is:
|
|
|
|
$(k + 1)$ stamps can be obtained by buying a collection of 5-stamp packages and
|
|
8-stamp packages.
|
|
|
|
_Case 1 (at least three 5-stamp packages are used in obtaining $k$ stamps):_
|
|
|
|
Replace three 5-stamp packages with two 8-stamp packages. This results in
|
|
$(k + 1)$ stamps.
|
|
|
|
_Case 2 (at most two 5-stamp packages are used in obtaining $k$ stamps):_
|
|
|
|
If there at most two 5-stamp packages, that means that $28-10=18$ must be made
|
|
up of 8-stamp packages. So at least 3 8-stamp packages must be used to exceed
|
|
the 28 minimum.
|
|
|
|
Replace three 8-stamp packages with 5 5-stamp packages. This results in
|
|
$(k + 1)$ stamps.
|
|
|
|
Therefore in both cases $(k + 1)$ stamps can be obtained by buying a collection
|
|
of 5-stamp packages and 8-stamp packages.
|
|
|
|
Q.E.D.
|
|
|
|
4. For each positive integer $n$, let $P(n)$ be the sentence that describes the
|
|
following divisibility property:
|
|
|
|
$$ 5^n - 1 \text{ is divisible by } 4 $$
|
|
|
|
a. Write $P(0)$. Is $P(0)$ true?
|
|
|
|
$$ 5^0 - 1 = 1 - 1 = 0 $$
|
|
|
|
$P(0)$ is true, as $0 = 0 \cdot 4$.
|
|
|
|
b. Write $P(k)$.
|
|
|
|
$$ P(k) = 5^k - 1 \text{ is divisible by } 4 $$
|
|
|
|
c. Write $P(k + 1)$.
|
|
|
|
$$ P(k + 1) = 5^{k + 1} - 1 \text{ is divisible by } 4 $$
|
|
|
|
d. In a proof by mathematical induction that this divisibility property holds
|
|
for every integer $n \geq 0$, what must be shown in the inductive step?
|
|
|
|
It must be shown that supposing that $5^k - 1$ is divisible by $4$ for some
|
|
integer $k \geq 0$, that therefore $5^{k + 1} - 1$ is divisible by $4$.
|
|
|
|
5. For each positive integer $n$, let $P(n)$ be the inequality
|
|
|
|
$$ 2^n < (n + 1)! $$
|
|
|
|
a. Write $P(2)$. Is $P(2)$ true?
|
|
|
|
$$ P(2) = 2^2 < (2 + 1)! $$
|
|
|
|
$$ P(2) = 4 < (3)! $$
|
|
|
|
$$ P(2) = 4 < (3 \cdot 2 \cdot 1) $$
|
|
|
|
$$ P(2) = 4 < 6 $$
|
|
|
|
Yes, $P(2)$ is true because $4$ is less than $6$.
|
|
|
|
b. Write $P(k)$.
|
|
|
|
$$ P(k) = 2^k < (k + 1)! $$
|
|
|
|
c. Write $P(k + 1)$.
|
|
|
|
$$ P(k + 1) = 2^{k + 1} < ((k + 1) + 1)! $$
|
|
|
|
Alternatively:
|
|
|
|
$$ P(k + 1) = 2^{k + 1} < (k + 2)! $$
|
|
|
|
d. In a proof by mathematical induction that this inequality holds for every
|
|
integer $n \geq 2$, what must be shown in the inductive step?
|
|
|
|
It must be shown that supposing $2^k < (k + 1)!$ is true for any integer
|
|
$k \geq 2$, that therefore $2^{k + 1} < (k + 2)!$ is true.
|
|
|
|
6. For each positive integer $n$, let $P(n)$ be the sentence
|
|
|
|
Any checkerboard with dimensions $2 \times 3n$ can be completely covered with
|
|
L-shaped trominoes.
|
|
|
|
a. Write $P(1)$. Is $P(1)$ true?
|
|
|
|
Any checkerboard with dimensions $2 \times 3(1)$ can be completely covered with
|
|
L-shaped trominoes.
|
|
|
|
Yes, this is true, a $2 \times 3$ dimension checkerboard can be completely
|
|
covered with L-shaped trominoes (2 in fact.)
|
|
|
|
b. Write $P(k)$.
|
|
|
|
Any checkerboard with dimensions $2 \times 3k$ can be completely covered with
|
|
L-shaped trominoes.
|
|
|
|
c. Write $P(k + 1)$.
|
|
|
|
Any checkerboard with dimensions $2 \times 3(k + 1)$ can be completely covered
|
|
with L-shaped trominoes.
|
|
|
|
d. In a proof by mathematical induction that $P(n)$ is true for each integer
|
|
$n \geq 1$, what must be shown in the inductive step?
|
|
|
|
It must be shown that supposing any checkerboard with dimensions $2 \times 3k$
|
|
can be completely covered with L-shaped trominoes for any integer $k \geq 1$,
|
|
that therefore any checkerboard with dimensions $2 \times 3(k + 1)$ can be
|
|
completely covered with L-shaped trominoes.
|
|
|
|
7. For each positive integer $n$, let $P(n)$ be the sentence
|
|
|
|
In any round-robin tournament involving $n$ teams, the teams can be labeled
|
|
$T_1$, $T_2$, $T_3$, \dots, $T_n$, so that $T_i$ beats $T_{i + 1}$ for every
|
|
$i = 1, 2, \dots, n$.
|
|
|
|
a. Write $P(2)$. Is $P(2)$ true?
|
|
|
|
In any round-robin tournament involving $2$ teams, the teams can be labeled
|
|
$T_1$, $T_2$, so that $T_i$ beats $T_{i + 1}$ for every $i = 1, 2$.
|
|
|
|
This is true, in a round-robin tournament involving only $2$ teams, one can
|
|
label the teams such that $T_2$ beats $T_1$.
|
|
|
|
b. Write $P(k)$.
|
|
|
|
In any round-robin tournament involving $k$ teams, the teams can be labeled
|
|
$T_1$, $T_2$, $T_3$, \dots, $T_k$, so that $T_i$ beats $T_{i + 1}$ for every
|
|
$i = 1, 2, \dots, k$.
|
|
|
|
c. Write $P(k + 1)$.
|
|
|
|
In any round-robin tournament involving $(k + 1)$ teams, the teams can be
|
|
labeled $T_1$, $T_2$, $T_3$, \dots, $T_{k + 1}$, so that $T_i$ beats $T_{i + 1}$
|
|
for every $i = 1, 2, \dots, (k + 1)$.
|
|
|
|
d. In a proof by mathematical induction that $P(n)$ is true for each integer
|
|
$n \geq 2$, what must be shown in the inductive step?
|
|
|
|
It must be shown that supposing in any round-robin tournament involving $k$
|
|
teams, the teams can be labeled $T_1, T_2, T_3, \dots T_k$, so that $T_i$ beats
|
|
$T_{i + 1}$ for every $i = 1, 2, \dots k$ for any integer $k \geq 2$, then
|
|
therefore in any round-robin tournament involving $(k + 1)$ teams, the teams can
|
|
be labeled $T_1, T_2, T_3, \dots T_{k + 1}$ so that $T_i$ beats $T_{i + 1}$ for
|
|
every $i = 1, 2, \dots (k + 1)$.
|
|
|
|
Prove each statement in 8-23 by mathematical induction.
|
|
|
|
8. $5^n - 1$ is divisible by $4$, for every integer $n \geq 0$.
|
|
|
|
**Proof (by mathematical induction):**
|
|
|
|
Let $P(n)$ be the sentence:
|
|
|
|
$$ 5^n - 1 \text{ is divisible by } 4 $$
|
|
|
|
_Basis Step:_
|
|
|
|
Prove $P(0)$. That is:
|
|
|
|
$$ 5^0 - 1 \text{ is divisible by } 4 $$
|
|
|
|
$$ 1 - 1 \text{ is divisible by } 4 $$
|
|
|
|
$$ 0 \text{ is divisible by } 4 $$
|
|
|
|
This sentence is true as $0 = 0 \cdot 4$, which shows that $0$ is divisible by
|
|
$4$ by the definition of divisibility.
|
|
|
|
Therefore $P(0)$ is true.
|
|
|
|
_Inductive Step:_
|
|
|
|
Let $k$ be any integer where $k \geq 0$.
|
|
|
|
Suppose $P(k)$. That is:
|
|
|
|
$$ 5^k - 1 \text{ is divisible by } 4 $$
|
|
|
|
This is the inductive hypothesis.
|
|
|
|
Prove $P(k + 1)$. That is:
|
|
|
|
$$ 5^{k + 1} - 1 \text{ is divisible by } 4 $$
|
|
|
|
$$ 5^{k + 1} - 1 $$
|
|
|
|
$$ = 5^k \cdot 5 - 1 $$
|
|
|
|
$$ = 5^k \cdot (4 + 1) - 1 $$
|
|
|
|
$$ = 5^k \cdot 4 + 5^k - 1 $$
|
|
|
|
Since we know by the inductive hypothesis that $5^k - 1$ is divisible by $4$. By
|
|
the definition of divisibility:
|
|
|
|
$$ 5^k - 1 = 4r $$
|
|
|
|
for some integer $r$. Our equation now becomes:
|
|
|
|
$$ = 5^k \cdot 4 + 4r $$
|
|
|
|
$$ = 4(5^k + r) $$
|
|
|
|
Now, we know that $5^k + r$ is an integer by the sum and product of integers.
|
|
Therefore, by the definition of divisibility, $5^{k + 1} - 1$ is divisible by
|
|
$4$.
|
|
|
|
Q.E.D.
|
|
|
|
9. $7^n - 1$ is divisible by $6$, for every integer $n \geq 0$.
|
|
|
|
**Proof (by mathematical induction):**
|
|
|
|
Let $P(n)$ be the sentence:
|
|
|
|
$$ 7^n - 1 \text{ is divisible by } 6 $$
|
|
|
|
_Basis Step:_
|
|
|
|
Prove $P(0)$. That is:
|
|
|
|
$$ 7^0 - 1 \text{ is divisible by } 6 $$
|
|
|
|
$$ 7^0 - 1 $$
|
|
|
|
$$ = 1 - 1 $$
|
|
|
|
$$ = 0 $$
|
|
|
|
$0$ is divisible by $6$ because $0 = 0 \cdot 6$.
|
|
|
|
Therefore $P(0)$ is true.
|
|
|
|
_Inductive Step:_
|
|
|
|
Let $k$ be any integer where $k \geq 0$.
|
|
|
|
Suppose $P(k)$. That is:
|
|
|
|
$$ 7^k - 1 \text{ is divisible by } 6 $$
|
|
|
|
Prove $P(k + 1)$. That is:
|
|
|
|
$$ 7^{k + 1} - 1 \text{ is divisible by } 6 $$
|
|
|
|
$$ 7^{k + 1} - 1 $$
|
|
|
|
$$ = 7^k \cdot 7 - 1 $$
|
|
|
|
$$ = 7^k \cdot (6 + 1) - 1 $$
|
|
|
|
$$ = 7^k \cdot 6 + (7^k - 1) $$
|
|
|
|
By the inductive hypothesis and by the definition of divisibility:
|
|
|
|
$$ = 7^k \cdot 6 + 6r $$
|
|
|
|
for some integer $r$.
|
|
|
|
$$ = 6(7^k + r) $$
|
|
|
|
Now, we know that $7^k + r$ is an integer by the sum and product of integers.
|
|
Therefore, by the definition of divisibility, $7^{k + 1} - 1$ is divisible by
|
|
$6$.
|
|
|
|
Q.E.D.
|
|
|
|
10. $n^3 - 7n + 3$ is divisible by $3$, for each integer $n \geq 0$.
|
|
|
|
**Proof (by mathematical induction):**
|
|
|
|
Let $P(n)$ be the sentence:
|
|
|
|
$$ n^3 - 7n + 3 \text{ is divisible by } 3 $$
|
|
|
|
_Basis Step:_
|
|
|
|
Prove $P(0)$. That is:
|
|
|
|
$$ (0)^3 - 7(0) + 3 \text{ is divisible by } 3 $$
|
|
|
|
$$ (0)^3 - 7(0) + 3 $$
|
|
|
|
$$ = 0 - 0 + 3 $$
|
|
|
|
$$ = 3 $$
|
|
|
|
By the definition of divisibility, $3 \mid 3$, as $3 = 1 \cdot 3$.
|
|
|
|
Therefore, $P(0)$ is true.
|
|
|
|
_Inductive Step:_
|
|
|
|
Let $k$ be any integer where $k \geq 0$.
|
|
|
|
Suppose $P(k)$. That is:
|
|
|
|
$$ k^3 - 7k + 3 \text{ is divisible by } 3 $$
|
|
|
|
Prove $P(k + 1)$. That is:
|
|
|
|
$$ (k + 1)^3 - 7(k + 1) + 3 \text{ is divisible by } 3 $$
|
|
|
|
$$ (k + 1)^3 - 7(k + 1) + 3 $$
|
|
|
|
$$ = (k + 1)(k + 1)(k + 1) - 7k - 7 + 3 $$
|
|
|
|
$$ = (k^2 + 2k + 1)(k + 1) - 7k - 7 + 3 $$
|
|
|
|
$$ = (k^2(k + 1) + 2k(k + 1) + 1(k + 1)) - 7k - 7 + 3 $$
|
|
|
|
$$ = (k^3 + k^2 + 2k^2 + 2k + k + 1) - 7k - 7 + 3 $$
|
|
|
|
$$ = (k^3 + 3k^2 + 3k + 1) - 7k - 7 + 3 $$
|
|
|
|
$$ = (k^3 - 7k + 3) + 3k^2 + 3k + 1 - 7 $$
|
|
|
|
$$ = (k^3 - 7k + 3) + 3k^2 + 3k - 6 $$
|
|
|
|
By the inductive hypothesis and definition of divisibility:
|
|
|
|
$$ = (3r) + 3k^2 + 3k - 6 $$
|
|
|
|
for some integer $r$.
|
|
|
|
$$ = 3(r + k^2 + k - 2) $$
|
|
|
|
Now, we know that $r + k^2 + k - 2$ is an integer by the product and sum of
|
|
integers. Thus, by the definition of divisibility, $(k + 1)^3 - 7(k + 1) + 3$ is
|
|
divisible by $3$.
|
|
|
|
Therefore $P(k + 1)$ is true.
|
|
|
|
Q.E.D.
|
|
|
|
11. $3^{2n} - 1$ is divisible by $8$, for every integer $n \geq 0$.
|
|
|
|
**Proof (by mathematical induction):**
|
|
|
|
Let $P(n)$ be the sentence:
|
|
|
|
$$ 3^{2n} - 1 \text{ is divisible by } 8 $$
|
|
|
|
_Basis Step:_
|
|
|
|
Prove $P(0)$. That is:
|
|
|
|
$$ 3^{2(0)} - 1 \text{ is divisible by } 8 $$
|
|
|
|
$$ 3^{2(0)} - 1 $$
|
|
|
|
$$ = 3^0 - 1 $$
|
|
|
|
$$ = 1 - 1 $$
|
|
|
|
$$ = 0 $$
|
|
|
|
$0$ is divisible by $8$ as $0 = 0 \cdot 8$.
|
|
|
|
Therefore $P(0)$ is true.
|
|
|
|
_Inductive Step:_
|
|
|
|
Let $k$ be any integer where $k \geq 0$.
|
|
|
|
Suppose $P(k)$. That is:
|
|
|
|
$$ 3^{2k} - 1 \text{ is divisible by } 8 $$
|
|
|
|
This is the inductive hypothesis.
|
|
|
|
Prove $P(k + 1)$. That is:
|
|
|
|
$$ 3^{2(k + 1)} - 1 \text{ is divisible by } 8 $$
|
|
|
|
$$ 3^{2(k + 1)} - 1 $$
|
|
|
|
$$ = 3^{2k + 2} - 1 $$
|
|
|
|
$$ = 3^{2k} \cdot 3^2 - 1 $$
|
|
|
|
$$ = 3^{2k} \cdot 9 - 1 $$
|
|
|
|
$$ = 3^{2k} \cdot (8 + 1) - 1 $$
|
|
|
|
$$ = 3^{2k} \cdot 8 + (3^{2k} - 1) $$
|
|
|
|
By the inductive hypothesis and the definition of divisibility:
|
|
|
|
$$ = 3^{2k} \cdot 8 + 8r $$
|
|
|
|
for some integer $r$.
|
|
|
|
$$ = 8(3^{2k} + r) $$
|
|
|
|
Now, $3^{2k} + r$ is an integer by the sum and product of integers. Thus
|
|
$3^{2(k + 1)} - 1$ is divisible by $8$ by the definition of divisibility.
|
|
|
|
Therefore $P(k + 1)$ is true.
|
|
|
|
Q.E.D.
|
|
|
|
12. For any integer $n \geq 0$, $7^n - 2^n$ is divisible by $5$.
|
|
|
|
**Proof (by mathematical induction):**
|
|
|
|
Let $P(n)$ be the sentence:
|
|
|
|
$$ 7^n - 2^n \text{ is divisible by } 5 $$
|
|
|
|
_Basis Step:_
|
|
|
|
Prove $P(0)$. That is:
|
|
|
|
$$ 7^0 - 2^0 \text{ is divisible by } 5 $$
|
|
|
|
$$ 7^0 - 2^0 $$
|
|
|
|
$$ = 1 - 1 $$
|
|
|
|
$$ = 0 $$
|
|
|
|
$0$ is divisible by $5$ as $0 = 0 \cdot 5$.
|
|
|
|
Therefore $P(0)$ is true.
|
|
|
|
_Inductive Step:_
|
|
|
|
Let $k$ be any integer where $k \geq 0$.
|
|
|
|
Suppose $P(k)$. That is:
|
|
|
|
$$ 7^k - 2^k \text{ is divisible by } 5 $$
|
|
|
|
This is the inductive hypothesis.
|
|
|
|
Prove $P(k + 1)$. That is:
|
|
|
|
$$ 7^{k + 1} - 2^{k + 1} \text{ is divisible by } 5 $$
|
|
|
|
$$ 7^{k + 1} - 2^{k + 1} $$
|
|
|
|
$$ = 7^k \cdot 7^1 - 2^k \cdot 2^1 $$
|
|
|
|
$$ = 7^k \cdot (5 + 2) - 2^k \cdot 2^1 $$
|
|
|
|
$$ = 7^k \cdot 5 + (2)7^k - 2^k \cdot 2^1 $$
|
|
|
|
$$ = 7^k \cdot 5 + 2(7^k - 2^k) $$
|
|
|
|
By the inductive hypothesis and the definition of divisibility:
|
|
|
|
$$ = 7^k \cdot 5 + 2(5r) $$
|
|
|
|
For some integer $r$.
|
|
|
|
$$ = 5(7^k + 2r) $$
|
|
|
|
Now, $7^k + 2r$ is an integer by the sum and product of integers. Thus
|
|
$7^{k + 1} - 2^{k + 1}$ is divisible by $5$ by the definition of divisibility.
|
|
|
|
Therefore $P(k + 1)$ is true.
|
|
|
|
Q.E.D.
|
|
|
|
13. For any integer $n \geq 0$, $x^n -y^n$ is divisible by $x - y$, where $x$
|
|
and $y$ are any integers with $x \neq y$.
|
|
|
|
**Proof (by mathematical induction):**
|
|
|
|
Suppose $x$ and $y$ are any integers with $x \neq y$.
|
|
|
|
Let $P(n)$ be the sentence:
|
|
|
|
$$ x^n - y^n \text{ is divisible by } x - y $$
|
|
|
|
_Basis Step:_
|
|
|
|
Prove $P(0)$. That is:
|
|
|
|
$$ x^0 - y^0 \text{ is divisible by } x - y $$
|
|
|
|
$$ x^0 - y^0 $$
|
|
|
|
$$ = 1 - 1 $$
|
|
|
|
$$ = 0 $$
|
|
|
|
$0$ is divisible by $(x - y)$ as $0 = 0 \cdot (x - y)$.
|
|
|
|
Therefore $P(0)$ is true.
|
|
|
|
_Inductive Step:_
|
|
|
|
Let $k$ be any integer where $k \geq 0$.
|
|
|
|
Suppose $P(k)$. That is:
|
|
|
|
$$ x^k - y^k \text{ is divisible by } x - y $$
|
|
|
|
This is the inductive hypothesis.
|
|
|
|
Prove $P(k + 1)$. That is:
|
|
|
|
$$ x^{k + 1} - y^{k + 1} \text{ is divisible by } x - y $$
|
|
|
|
$$ x^{k + 1} - y^{k + 1} $$
|
|
|
|
$$ = x^k(x) - y^k(y) $$
|
|
|
|
$$ = x^k(x) - xy^k + xy^k - y^k(y) $$
|
|
|
|
$$ = x(x^k - y^k) + y^k(x - y) $$
|
|
|
|
By the inductive hypothesis:
|
|
|
|
$$ = x(r(x - y)) + y^k(x - y) $$
|
|
|
|
for some integer $r$.
|
|
|
|
$$ = (x - y)(xr + y^k) $$
|
|
|
|
We know $xr + y^k$ is an integer by the sum and product of integers. By the
|
|
definition of divisibility, $x^{k + 1} - y^{k + 1}$ is divisible by $x - y$.
|
|
|
|
Therefore $P(k + 1)$ is true.
|
|
|
|
Q.E.D.
|
|
|
|
14. $n^3 - n$ is divisible by $6$, for each integer $n \geq 0$.
|
|
|
|
**Proof (by mathematical induction):**
|
|
|
|
Let $P(n)$ be the sentence:
|
|
|
|
$$ n^3 - n \text{ is divisible by } 6 $$
|
|
|
|
_Basis Step:_
|
|
|
|
Prove $P(0)$. That is:
|
|
|
|
$$ 0^3 - 0 \text{ is divisible by } 6 $$
|
|
|
|
$$ 0^3 - 0 $$
|
|
|
|
$$ = 0 - 0 $$
|
|
|
|
$$ = 0 $$
|
|
|
|
$0$ is divisible by $6$ because $0 = 0 \cdot 6$.
|
|
|
|
Therefore $P(0)$ is true.
|
|
|
|
_Inductive Step:_
|
|
|
|
Let $k$ be any integer where $k \geq 0$.
|
|
|
|
Suppose $P(k)$. That is:
|
|
|
|
$$ k^3 - k \text{ is divisible by } 6 $$
|
|
|
|
This is the inductive hypothesis.
|
|
|
|
Prove $P(k + 1)$. That is:
|
|
|
|
$$ (k + 1)^3 - (k + 1) \text{ is divisible by } 6 $$
|
|
|
|
$$ (k + 1)^3 - (k + 1) $$
|
|
|
|
$$ = (k + 1)(k + 1)(k + 1) - (k + 1) $$
|
|
|
|
$$ = (k^2 + 2k + 1)(k + 1) - (k + 1) $$
|
|
|
|
$$ = (k^3 + k^2 + 2k^2 + 2k + k + 1) - (k + 1) $$
|
|
|
|
$$ = (k^3 + 3k^2 + 3k + 1) - (k + 1) $$
|
|
|
|
$$ = k^3 + 3k^2 + 3k + 1 - k - 1 $$
|
|
|
|
$$ = k^3 + 3k^2 + 2k $$
|
|
|
|
$$ = (k^3 - k) + 3k^2 + 3k $$
|
|
|
|
$$ = (k^3 - k) + 3k(k + 1) $$
|
|
|
|
By the inductive hypothesis and definition of divisibility:
|
|
|
|
$$ = 6r + 3k(k + 1) $$
|
|
|
|
for some integer $r$.
|
|
|
|
By Theorem 4.5.2, the product of any two consecutive integers must be even.
|
|
|
|
$$ = 6r + 3(2m) $$
|
|
|
|
for some integer $m$.
|
|
|
|
$$ = 6r + 6m $$
|
|
|
|
$$ = 6(r + m) $$
|
|
|
|
Now, $r + m$ is an integer by the sum of integers.
|
|
|
|
Therefore $(k + 1)^3 - (k + 1)$ is divisible by $6$ by the definition of
|
|
divisibility.
|
|
|
|
Therefore $P(k + 1)$ is true.
|
|
|
|
Q.E.D.
|
|
|
|
15. $n(n^2 + 5)$ is divisible by $6$, for each integer $n \geq 0$.
|
|
|
|
**Proof (by mathematical induction):**
|
|
|
|
Let $P(n)$ be the sentence:
|
|
|
|
$$ n(n^2 + 5) \text{ is divisible by } 6 $$
|
|
|
|
_Basis Step:_
|
|
|
|
Prove $P(0)$. That is:
|
|
|
|
$$ 0(0^2 + 5) \text{ is divisible by } 6 $$
|
|
|
|
$$ 0(0^2 + 5) $$
|
|
|
|
$$ = 0(0 + 5) $$
|
|
|
|
$$ = 0(5) $$
|
|
|
|
$$ = 0 $$
|
|
|
|
$0$ is divisible by $6$ as $0 = 0 \cdot 6$.
|
|
|
|
Therefore $P(0)$ is true.
|
|
|
|
_Inductive Step:_
|
|
|
|
Let $k$ be any integer where $k \geq 0$.
|
|
|
|
Suppose $P(k)$. That is:
|
|
|
|
$$ k(k^2 + 5) \text{ is divisible by } 6 $$
|
|
|
|
This is the inductive hypothesis.
|
|
|
|
Prove $P(k + 1)$. That is:
|
|
|
|
$$ (k + 1)((k + 1)^2 + 5) \text{ is divisible by } 6 $$
|
|
|
|
$$ (k + 1)((k + 1)^2 + 5) $$
|
|
|
|
$$ = (k + 1)((k + 1)(k + 1) + 5) $$
|
|
|
|
$$ = (k + 1)(k^2 + 2k + 6) $$
|
|
|
|
$$ = k^3 + k^2 + 2k^2 + 2k + 6k + 6 $$
|
|
|
|
$$ = k^3 + 3k^2 + 8k + 6 $$
|
|
|
|
$$ = k^3 + 3k^2 + 5k + 3k + 6 $$
|
|
|
|
$$ = (k^3 + 5k) + 3k^2 + 3k + 6 $$
|
|
|
|
$$ = k(k^2 + 5) + 3k^2 + 3k + 6 $$
|
|
|
|
$$ = k(k^2 + 5) + 3(k^2 + k + 2) $$
|
|
|
|
By the inductive hypothesis and definition of divisibility:
|
|
|
|
$$ = 6r + 3(k^2 + k + 2) $$
|
|
|
|
for some integer $r$.
|
|
|
|
$$ = 6r + 3(k(k + 1) + 2) $$
|
|
|
|
By Theorem 4.5.2 $k(k + 1)$ is always even:
|
|
|
|
$$ = 6r + 3(2m + 2) $$
|
|
|
|
for some integer $m$.
|
|
|
|
$$ = 6r + 6m + 6 $$
|
|
|
|
$$ = 6(r + m + 1) $$
|
|
|
|
Now, $r + m + 1$ is an integer by the sum of integers. Thus
|
|
$(k + 1)((k + 1)^2 + 5)$ is divisible by $6$ by the definition of divisibility.
|
|
|
|
Therefore $P(k + 1)$ is true.
|
|
|
|
Q.E.D.
|
|
|
|
16. $2^n < (n + 1)!$, for every integer $n \geq 2$.
|
|
|
|
**Proof (by mathematical induction):**
|
|
|
|
Let $P(n)$ be the sentence:
|
|
|
|
$$ 2^n < (n + 1)! $$
|
|
|
|
_Basis Step:_
|
|
|
|
Prove $P(2)$. That is:
|
|
|
|
$$ 2^(2) < (2 + 1)! $$
|
|
|
|
$$ 4 < (3)! $$
|
|
|
|
$$ 4 < (3 \cdot 2 \cdot 1) $$
|
|
|
|
$$ 4 < 6 $$
|
|
|
|
$4$ is less than $6$.
|
|
|
|
Therefore $P(2)$ is true.
|
|
|
|
_Inductive Step:_
|
|
|
|
Let $k$ be any integer where $k \geq 2$.
|
|
|
|
Suppose $P(k)$. That is:
|
|
|
|
$$ 2^k < (k + 1)! $$
|
|
|
|
This is the inductive hypothesis.
|
|
|
|
Prove $P(k + 1)$. That is:
|
|
|
|
$$ 2^{k + 1} < ((k + 1) + 1)! $$
|
|
|
|
Alternatively:
|
|
|
|
$$ 2^{k + 1} < (k + 2)! $$
|
|
|
|
By the inductive hypothesis and the laws of exponents:
|
|
|
|
$$ = 2^{k} \cdot 2 < 2(k + 1)! $$
|
|
|
|
Since $k \geq 2$, then $2 < k + 2$, and so:
|
|
|
|
$$ 2(k + 1)! < (k + 2)(k + 1)! = (k + 2)! $$
|
|
|
|
Combining these inequalities shows:
|
|
|
|
$$ 2^{k + 1} < (k + 2)! $$
|
|
|
|
As was to be shown.
|
|
|
|
Q.E.D.
|
|
|
|
17. $1 + 3n \leq 4^n$, for every integer $n \geq 0$.
|
|
|
|
**Proof (by mathematical induction):**
|
|
|
|
Let $P(n)$ be the inequality:
|
|
|
|
$$ 1 + 3n \leq 4^n $$
|
|
|
|
_Basis Step:_
|
|
|
|
Prove $P(0)$. That is:
|
|
|
|
$$ 1 + 3(0) \leq 4^0 $$
|
|
|
|
$$ = 1 + 0 \leq 1 $$
|
|
|
|
$$ = 1 \leq 1 $$
|
|
|
|
Since $1 = 1$, $1 \leq 1$ is a true statement.
|
|
|
|
Therefore $P(0)$ is true.
|
|
|
|
_Inductive Step:_
|
|
|
|
Let $k$ be any integer where $k \geq 0$.
|
|
|
|
Suppose $P(k)$. That is:
|
|
|
|
$$ 1 + 3k \leq 4^k $$
|
|
|
|
This is the inductive hypothesis.
|
|
|
|
Prove $P(k + 1)$. That is:
|
|
|
|
$$ 1 + 3(k + 1) \leq 4^{k + 1} $$
|
|
|
|
$$ (1 + 3k) + 3 \leq 4^k \cdot 4 $$
|
|
|
|
By the inductive hypothesis:
|
|
|
|
$$ (1 + 3k) + 3 \leq 4^k + 3 $$
|
|
|
|
Now show:
|
|
|
|
$$ 4^k + 3 \leq 4^{k + 1} $$
|
|
|
|
Since:
|
|
|
|
$$ 4^{k + 1} = 4^k \cdot 4 $$
|
|
|
|
it is enough to show:
|
|
|
|
$$ 3 \leq 3 \cdot 4^k $$
|
|
|
|
which is true for all $k \geq 0$.
|
|
|
|
So:
|
|
|
|
$$ 1 + 3(k + 1) \leq 4^k + 3 \leq 4^{k + 1} $$
|
|
|
|
$$ 1 + 3(k + 1) \leq 4^{k + 1} $$
|
|
|
|
Q.E.D.
|
|
|
|
18. $5^n + 9 < 6^n$, for each integer $n \geq 2$.
|
|
|
|
**Proof (by mathematical induction):**
|
|
|
|
Let $P(n)$ be the inequality:
|
|
|
|
$$ 5^n + 9 < 6^n $$
|
|
|
|
_Basis Step:_
|
|
|
|
Prove $P(2)$. That is:
|
|
|
|
$$ 5^2 + 9 < 6^2 $$
|
|
|
|
$$ 25 + 9 < 36 $$
|
|
|
|
$$ 34 < 36 $$
|
|
|
|
Since $34$ is less than $36$, this inequality is true.
|
|
|
|
Therefore $P(2)$ is true.
|
|
|
|
_Inductive Step:_
|
|
|
|
Let $k$ be any integer where $k \geq 2$.
|
|
|
|
Suppose $P(k)$. That is:
|
|
|
|
$$ 5^k + 9 < 6^k $$
|
|
|
|
This is the inductive hypothesis.
|
|
|
|
Prove $P(k + 1)$. That is:
|
|
|
|
$$ 5^{k + 1} + 9 < 6^{k + 1} $$
|
|
|
|
If we multiply the inductive hypothesis by 5:
|
|
|
|
$$ 5(5^k + 9) < 5(6^k) $$
|
|
|
|
$$ 5^{k + 1} + 45 < 5(6^k) $$
|
|
|
|
$$ 5^{k + 1} + 45 < 5(6^k) < 6^{k + 1} $$
|
|
|
|
$$ 5^{k + 1} + 45 < 5(6^k) < 6^k \cdot 6 = 6^{k + 1} $$
|
|
|
|
Note that:
|
|
|
|
$$ 5^{k + 1} + 9 < 5^{k + 1} + 45 < 5(6^k) < 6^k \cdot 6 = 6^{k + 1} $$
|
|
|
|
Therefore:
|
|
|
|
$$ 5^{k + 1} + 9 < 6^{k + 1} $$
|
|
|
|
As was to be shown.
|
|
|
|
Q.E.D.
|
|
|
|
19. $n^2 < 2^n$, for every integer $n \geq 5$.
|
|
|
|
**Proof (by mathematical induction):**
|
|
|
|
Let $P(n)$ be the inequality:
|
|
|
|
$$ n^2 < 2^n $$
|
|
|
|
_Basis Step:_
|
|
|
|
Prove $P(5)$. That is:
|
|
|
|
$$ 5^2 < 2^5 $$
|
|
|
|
$$ 25 < 32 $$
|
|
|
|
Since $25$ is less than $32$, this is a true statement.
|
|
|
|
Therefore $P(5)$ is true.
|
|
|
|
_Inductive Step:_
|
|
|
|
Let $k$ be any integer where $k \geq 5$.
|
|
|
|
Suppose $P(k)$. That is:
|
|
|
|
$$ k^2 < 2^k $$
|
|
|
|
This is the inductive hypothesis.
|
|
|
|
Prove $P(k + 1)$. That is:
|
|
|
|
$$ (k + 1)^2 < 2^{k + 1} $$
|
|
|
|
Now, expanding out the left-hand side:
|
|
|
|
$$ (k + 1)^2 = k^2 + 2k + 1 $$
|
|
|
|
Consider the inductive hypothesis:
|
|
|
|
$$ k^2 < 2^k $$
|
|
|
|
It follows that:
|
|
|
|
$$ k^2 + 2k + 1 < 2^k + 2k + 1 $$
|
|
|
|
By proposition 5.3.2, $2k + 1 < 2^k$ since $k \geq 5 \geq 3$.
|
|
|
|
Hence:
|
|
|
|
$$ (k + 1)^2 = k^2 + 2k + 1 < 2^k + 2k + 1 < 2^k + 2^k = 2^{k + 1} $$
|
|
|
|
$$ (k + 1)^2 < 2^{k + 1} $$
|
|
|
|
As was to be shown.
|
|
|
|
Q.E.D.
|
|
|
|
20. $2^n < (n + 2)!$, for each integer $n \geq 0$.
|
|
|
|
**Proof (by mathematical induction):**
|
|
|
|
Let $P(n)$ be the inequality:
|
|
|
|
$$ 2^n < (n + 2)! $$
|
|
|
|
_Basis Step:_
|
|
|
|
Prove $P(0)$. That is:
|
|
|
|
$$ 2^0 < (0 + 2)! $$
|
|
|
|
$$ 1 < (2)! $$
|
|
|
|
$$ 1 < 2 $$
|
|
|
|
Since $1$ is less than $2$. This is a true statement.
|
|
|
|
Therefore $P(0)$ is true.
|
|
|
|
_Inductive Step:_
|
|
|
|
Let $k$ be any integer such that $k \geq 0$.
|
|
|
|
Suppose $P(k)$. That is:
|
|
|
|
$$ 2^k < (k + 2)! $$
|
|
|
|
Prove $P(k + 1)$. That is:
|
|
|
|
$$ 2^{k + 1} < ((k + 1) + 2)! $$
|
|
|
|
Alternatively:
|
|
|
|
$$ 2^{k + 1} < (k + 3)! $$
|
|
|
|
Expanding out the left-hand side:
|
|
|
|
$$ 2^{k + 1} = 2^k \cdot 2 $$
|
|
|
|
Consider the inductive hypothesis:
|
|
|
|
$$ 2^k < (k + 2)! $$
|
|
|
|
Multiple both sides by $2$:
|
|
|
|
$$ 2(2^k) < 2(k + 2)! $$
|
|
|
|
$$ 2^{k + 1} < 2(k + 2)! $$
|
|
|
|
Now, expanding out the right-hand side:
|
|
|
|
$$ (k + 3)! = (k + 3)(k + 2)! $$
|
|
|
|
Since $k \geq 0$, it follows that $k + 3 \geq 3 \geq 2$. Putting out
|
|
inequalities together then, we get:
|
|
|
|
$$ 2^{k + 1} < 2(k + 2)! < (k + 3)(k + 2)! = (k + 3)! $$
|
|
|
|
And now simplified:
|
|
|
|
$$ 2^{k + 1} < (k + 3)! $$
|
|
|
|
As was to be shown.
|
|
|
|
Q.E.D.
|
|
|
|
21. $\sqrt{n} < \dfrac{1}{\sqrt{1}} + \dfrac{1}{\sqrt{2}} + \dots + \dfrac{1}{\sqrt{n}}$,
|
|
for every integer $n \geq 2$.
|
|
|
|
**Proof (by mathematical induction):**
|
|
|
|
Let $P(n)$ be the inequality:
|
|
|
|
$$ \sqrt{n} < \dfrac{1}{\sqrt{1}} + \dfrac{1}{\sqrt{2}} + \dots + \dfrac{1}{\sqrt{n}} $$
|
|
|
|
_Basis Step:_
|
|
|
|
Prove $P(2)$. That is:
|
|
|
|
$$ \sqrt{2} < \dfrac{1}{\sqrt{1}} + \dfrac{1}{\sqrt{2}} + \dots + \dfrac{1}{\sqrt{2}} $$
|
|
|
|
$\dots \dfrac{1}{\sqrt{2}}$ just ends at term, $\dfrac{1}{\sqrt{2}}$.
|
|
|
|
$$ \sqrt{2} < \dfrac{1}{\sqrt{1}} + \dfrac{1}{\sqrt{2}} $$
|
|
|
|
$$ \sqrt{2} < 1 + \dfrac{1}{\sqrt{2}} $$
|
|
|
|
$$ \sqrt{2} < \frac{\sqrt{2}}{\sqrt{2}} + \dfrac{1}{\sqrt{2}} $$
|
|
|
|
$$ \sqrt{2} < \dfrac{\sqrt{2} + 1}{\sqrt{2}} $$
|
|
|
|
$$ (\sqrt{2})(\sqrt{2}) < \left(\dfrac{\sqrt{2} + 1}{\sqrt{2}}\right)(\sqrt{2}) $$
|
|
|
|
$$ 2 < \sqrt{2} + 1 \approx 2.414213562 $$
|
|
|
|
This statement is true. Therefore $P(2)$ is true.
|
|
|
|
_Inductive Step:_
|
|
|
|
Let $k$ be any integer where $k \geq 2$.
|
|
|
|
Suppose $P(k)$. That is:
|
|
|
|
$$ \sqrt{k} < \dfrac{1}{\sqrt{1}} + \dfrac{1}{\sqrt{2}} + \dots + \dfrac{1}{\sqrt{k}} $$
|
|
|
|
This is the inductive hypothesis.
|
|
|
|
Prove $P(k + 1)$. That is:
|
|
|
|
$$ \sqrt{k + 1} < \dfrac{1}{\sqrt{1}} + \dfrac{1}{\sqrt{2}} + \dots + \dfrac{1}{\sqrt{k + 1}} $$
|
|
|
|
From the inductive hypothesis:
|
|
|
|
$$ \sqrt{k} < \dfrac{1}{\sqrt{1}} + \dfrac{1}{\sqrt{2}} + \dots + \dfrac{1}{\sqrt{k}} $$
|
|
|
|
Add $\dfrac{1}{\sqrt{k + 1}}$ to both sides:
|
|
|
|
$$ \sqrt{k} + \frac{1}{\sqrt{k + 1}} < \dfrac{1}{\sqrt{1}} + \dfrac{1}{\sqrt{2}} + \dots + \dfrac{1}{\sqrt{k}} + \frac{1}{\sqrt{k + 1}} $$
|
|
|
|
From here, it is enough to show:
|
|
|
|
$$ \sqrt{k + 1} \leq \sqrt{k} + \frac{1}{\sqrt{k + 1}} $$
|
|
|
|
$$ \sqrt{k + 1} - \sqrt{k} \leq \frac{1}{\sqrt{k + 1}} $$
|
|
|
|
$$ \left(\sqrt{k + 1} - \sqrt{k}\right)\left(\frac{\sqrt{k + 1} + \sqrt{k}}{\sqrt{k + 1} + \sqrt{k}}\right) \leq \frac{1}{\sqrt{k + 1}} $$
|
|
|
|
$$ \frac{(k + 1) - k}{\sqrt{k + 1} + \sqrt{k}} \leq \frac{1}{\sqrt{k + 1}} $$
|
|
|
|
$$ \frac{1}{\sqrt{k + 1} + \sqrt{k}} \leq \frac{1}{\sqrt{k + 1}} $$
|
|
|
|
Since $\sqrt{k + 1} + \sqrt{k} > \sqrt{k + 1}$, this inequality holds.
|
|
|
|
Simplified, our inequality becomes:
|
|
|
|
$$ \sqrt{k + 1} < \sqrt{k} + \frac{1}{\sqrt{k + 1}} < \dfrac{1}{\sqrt{1}} + \dfrac{1}{\sqrt{2}} + \dots + \dfrac{1}{\sqrt{k + 1}} $$
|
|
|
|
$$ \sqrt{k + 1} < \dfrac{1}{\sqrt{1}} + \dfrac{1}{\sqrt{2}} + \dots + \dfrac{1}{\sqrt{k + 1}} $$
|
|
|
|
As was to be shown.
|
|
|
|
Q.E.D.
|
|
|
|
22. $1 + nx \leq (1 + x)^n$, for every real number $x > -1$ and every integer
|
|
$n \geq 2$.
|
|
|
|
**Proof (by mathematical induction):**
|
|
|
|
Suppose $x$ is any real number where $x > -1$.
|
|
|
|
Let $P(n)$ be the sentence:
|
|
|
|
$$ 1 + nx \leq (1 + x)^n $$
|
|
|
|
_Basis Step:_
|
|
|
|
Prove $P(2)$. That is:
|
|
|
|
$$ 1 + 2x \leq (1 + x)^2 $$
|
|
|
|
$$ 1 + 2x \leq 1 + 2x + x^2 $$
|
|
|
|
$$ 0 \leq x^2 $$
|
|
|
|
This inequality always holds.
|
|
|
|
Therefore $P(2)$ is true.
|
|
|
|
_Inductive Step:_
|
|
|
|
Let $k$ be any integer where $k \geq 2$.
|
|
|
|
Suppose $P(k)$. That is:
|
|
|
|
$$ 1 + kx \leq (1 + x)^k $$
|
|
|
|
This is the inductive hypothesis.
|
|
|
|
Prove $P(k + 1)$. That is:
|
|
|
|
$$ 1 + (k + 1)x \leq (1 + x)^{k + 1} $$
|
|
|
|
Consider the inductive hypothesis:
|
|
|
|
$$ 1 + kx \leq (1 + x)^k $$
|
|
|
|
Multiply each side by $(1 + x)$:
|
|
|
|
$$ (1 + x)(1 + kx) \leq ((1 + x)^k)(1 + x) $$
|
|
|
|
$$ 1 + x + kx + kx^2 \leq (1 + x)^{k + 1} $$
|
|
|
|
Now it is enough to show that the left hand side of $P(k + 1)$ is less than or
|
|
equal to the left-hand side of $(1 + x)(P(k))$:
|
|
|
|
$$ 1 + (k + 1)x \leq 1 + x + kx + kx^2 $$
|
|
|
|
$$ 1 + kx + x \leq 1 + x + kx + kx^2 $$
|
|
|
|
$$ 1 + x + kx \leq 1 + x + kx + kx^2 $$
|
|
|
|
$$ 0 \leq kx^2 $$
|
|
|
|
Since $k \geq 2$, this inequality will always hold.
|
|
|
|
Simplified, our inequality is:
|
|
|
|
$$ 1 + (k + 1)x \leq 1 + x + kx + kx^2 \leq (1 + x)^{k + 1} $$
|
|
|
|
$$ 1 + (k + 1)x \leq (1 + x)^{k + 1} $$
|
|
|
|
As was to be shown.
|
|
|
|
Q.E.D.
|
|
|
|
23.
|
|
|
|
a. $n^3 > 2n + 1$, for each integer $n \geq 2$.
|
|
|
|
**Proof (by mathematical induction):**
|
|
|
|
Let $P(n)$ be the inequality:
|
|
|
|
$$ n^3 > 2n + 1 $$
|
|
|
|
_Basis Step:_
|
|
|
|
Prove $P(2)$. That is:
|
|
|
|
$$ (2)^3 > 2(2) + 1 $$
|
|
|
|
$$ 8 > 4 + 1 $$
|
|
|
|
$$ 8 > 5 $$
|
|
|
|
Since $8$ is greater than $5$, this statement is true.
|
|
|
|
Therefore $P(2)$ is true.
|
|
|
|
_Inductive Step:_
|
|
|
|
Let $k$ be any integer where $k \geq 2$.
|
|
|
|
Suppose $P(k)$. That is:
|
|
|
|
$$ k^3 > 2k + 1 $$
|
|
|
|
This is the inductive hypothesis.
|
|
|
|
Prove $P(k + 1)$. That is:
|
|
|
|
$$ (k + 1)^3 > 2(k + 1) + 1 $$
|
|
|
|
Alternatively:
|
|
|
|
$$ (k + 1)^3 > 2k + 2 + 1 $$
|
|
|
|
$$ (k + 1)^3 > 2k + 3 $$
|
|
|
|
Consider the inductive hypothesis:
|
|
|
|
$$ k^3 > 2k + 1 $$
|
|
|
|
Add $2$ to both sides:
|
|
|
|
$$ k^3 + 2 > 2k + 1 + 2 $$
|
|
|
|
$$ k^3 + 2 > 2k + 3 $$
|
|
|
|
Now it is enough to prove that the left-hand side of this inequality is less
|
|
than the left-hand side of the $P(k + 1)$ inequality:
|
|
|
|
$$ (k + 1)^3 > k^3 + 2 $$
|
|
|
|
$$ (k + 1)(k + 1)(k + 1) > k^3 + 2 $$
|
|
|
|
$$ (k^2 + 2k + 1)(k + 1) > k^3 + 2 $$
|
|
|
|
$$ k^3 + k^2 + 2k^2 + 2k + k + 1 > k^3 + 2 $$
|
|
|
|
$$ k^3 + 3k^2 + 3k + 1 > k^3 + 2 $$
|
|
|
|
$$ 3k^2 + 3k > 1 $$
|
|
|
|
Since $k \geq 2$, this inequality will always hold.
|
|
|
|
Simplified:
|
|
|
|
$$ (k + 1)^3 > k^3 + 2 > 2k + 3 $$
|
|
|
|
$$ (k + 1)^3 > 2k + 3 $$
|
|
|
|
As was to be shown.
|
|
|
|
Q.E.D.
|
|
|
|
b. $n! > n^2$, for each integer $n \geq 4$.
|
|
|
|
**Proof (by mathematical induction):**
|
|
|
|
Let $P(n)$ be the inequality:
|
|
|
|
$$ n! > n^2 $$
|
|
|
|
_Basis Step:_
|
|
|
|
Prove $P(4)$. That is:
|
|
|
|
$$ 4! > 4^2 $$
|
|
|
|
$$ (4 \cdot 3 \cdot 2 \cdot 1) > 16 $$
|
|
|
|
$$ 24 > 16 $$
|
|
|
|
Since $24$ is greater than $16$, this statement is true.
|
|
|
|
Therefore $P(4)$ is true.
|
|
|
|
_Inductive Step:_
|
|
|
|
Let $k$ be any integer where $k \geq 4$.
|
|
|
|
Suppose $P(k)$. That is:
|
|
|
|
$$ k! > k^2 $$
|
|
|
|
This is the inductive hypothesis.
|
|
|
|
Prove $P(k + 1)$. That is:
|
|
|
|
$$ (k + 1)! > (k + 1)^2 $$
|
|
|
|
Take the inductive hypothesis:
|
|
|
|
$$ k! > k^2 $$
|
|
|
|
And multiply each side by $(k + 1)$:
|
|
|
|
$$ (k + 1)k! > k^2(k + 1) $$
|
|
|
|
$$ (k + 1)! > k^2(k + 1) $$
|
|
|
|
Now it is enough to show:
|
|
|
|
$$ k^2(k + 1) > (k + 1)^2 $$
|
|
|
|
$$ k^2 > k + 1 $$
|
|
|
|
And this inequality holds for all $k \geq 4$.
|
|
|
|
Simplified:
|
|
|
|
$$ (k + 1)! > k^2(k + 1) > (k + 1)^2 $$
|
|
|
|
$$ (k + 1)! > (k + 1)^2 $$
|
|
|
|
As was to be shown.
|
|
|
|
Q.E.D.
|
|
|
|
24. A sequence $a_1, a_2, a_3, \dots$ is defined by letting $a_1 = 3$ and
|
|
$a_k = 7a_{k - 1}$ for each integer $k \geq 2$. Show that
|
|
$a_n = 3 \cdot 7^{n - 1}$ for every integer $n \geq 1$.
|
|
|
|
**Proof (by mathematical induction):**
|
|
|
|
Let $P(n)$ be the statement:
|
|
|
|
$$ a_n = 3 \cdot 7^{n - 1} $$
|
|
|
|
_Basis Step:_
|
|
|
|
Prove $P(1)$. That is:
|
|
|
|
$$ a_1 = 3 \cdot 7^{1 - 1} $$
|
|
|
|
$$ = 3 \cdot 7^{0} $$
|
|
|
|
$$ = 3 \cdot 1 $$
|
|
|
|
$$ = 3 $$
|
|
|
|
Since $a_1 = 3$ is defined in the problem statement, this equality is true.
|
|
|
|
Therefore $P(1)$ is true.
|
|
|
|
_Inductive _Step:_
|
|
|
|
Let $k$ be any integer such that $k \geq 1$.
|
|
|
|
Suppose $P(k)$. That is:
|
|
|
|
$$ a_k = 3 \cdot 7^{k - 1} $$
|
|
|
|
This is the inductive hypothesis.
|
|
|
|
Prove $P(k + 1)$. That is:
|
|
|
|
$$ a_{k + 1} = 3 \cdot 7^{(k + 1) - 1} $$
|
|
|
|
Alternatively:
|
|
|
|
$$ a_{k + 1} = 3 \cdot 7^k $$
|
|
|
|
By the definition of the given sequence:
|
|
|
|
$$ a_{k + 1} = 7a_k $$
|
|
|
|
By the inductive hypothesis:
|
|
|
|
$$ = 7(3 \cdot 7^{k - 1}) $$
|
|
|
|
By the laws of exponents:
|
|
|
|
$$ = 3 \cdot 7^k $$
|
|
|
|
And this is the right-hand side of the equality to be shown.
|
|
|
|
Q.E.D.
|
|
|
|
25. A sequence $b_0, b_1, b_2, \dots$ is defined by letting $b_0 = 5$ and
|
|
$b_k = 4 + b_{k - 1}$ for each integer $k \geq 1$. Show that $b_n > 4n$ for
|
|
every integer $n \geq 0$.
|
|
|
|
**Proof (by mathematical induction):**
|
|
|
|
Let $P(n)$ be the inequality:
|
|
|
|
$$ b_n > 4n $$
|
|
|
|
_Basis Step:_
|
|
|
|
Prove $P(0)$. That is:
|
|
|
|
$$ b_0 > 4(0) $$
|
|
|
|
$$ 5 > 4(0) $$
|
|
|
|
$$ 5 > 0 $$
|
|
|
|
This inequality holds. Therefore $P(0)$ is true.
|
|
|
|
_Inductive Step:_
|
|
|
|
Let $k$ be any integer where $k \geq 0$.
|
|
|
|
Suppose $P(k)$. That is:
|
|
|
|
$$ b_k > 4k $$
|
|
|
|
This is the inductive hypothesis.
|
|
|
|
Prove $P(k + 1)$. That is:
|
|
|
|
$$ b_{k + 1} > 4(k + 1) $$
|
|
|
|
By the definition of the sequence:
|
|
|
|
$$ b_{k + 1} = 4 + b_k $$
|
|
|
|
By the inductive hypothesis:
|
|
|
|
$$ > 4 + 4k $$
|
|
|
|
$$ > 4(1 + k) $$
|
|
|
|
$$ > 4(k + 1) $$
|
|
|
|
Q.E.D.
|
|
|
|
26. A sequence $c_0, c_1, c_2, \dots$ is defined by letting $c_0 = 3$ and
|
|
$c_k = (c_{k - 1})^2$ for every integer $k \geq 1$. Show that $c_n = 3^{2n}$
|
|
for each integer $n \geq 0$.
|
|
|
|
**Proof (by mathematical induction):**
|
|
|
|
Let $P(n)$ be the equality:
|
|
|
|
$$ c_n = 3^{2n} $$
|
|
|
|
_Basis Step:_
|
|
|
|
Prove $P(0)$. That is:
|
|
|
|
$$ c_0 = 3^{2(0)} $$
|
|
|
|
$$ c_0 = 3^{0} $$
|
|
|
|
$$ c_0 = 1 $$
|
|
|
|
Stopping here. It is likely Epp has a typo, she means $c_n = 3^{2^n}$, not
|
|
$c_n = 3^{2n}$.
|
|
|
|
27. A sequence $d_1, d_2, d_3, \dots$ is defined by letting $d_1 = 2$ and
|
|
$d_k = \dfrac{d_{k - 1}}{k}$ for each integer $k \geq 2$. Show that for
|
|
every integer $n \geq 1$, $d_n = \dfrac{2}{n!}$.
|
|
|
|
**Proof (by mathematical induction):**
|
|
|
|
Let $P(n)$ be the equation:
|
|
|
|
$$ d_n = \frac{2}{n!} $$
|
|
|
|
_Basis Step:_
|
|
|
|
Prove $P(1)$. That is:
|
|
|
|
$$ d_1 = \frac{2}{1!} $$
|
|
|
|
$$ d_1 = \frac{2}{1} $$
|
|
|
|
$$ d_1 = 2 $$
|
|
|
|
Since the problem statement states that $d_1 = 2$, this matches our equality.
|
|
|
|
Therefore $P(1)$ is true.
|
|
|
|
_Inductive Step:_
|
|
|
|
Let $k$ be any integer such that $k \geq 1$.
|
|
|
|
Suppose $P(k)$, that is:
|
|
|
|
$$ d_k = \frac{2}{k!} $$
|
|
|
|
This is the inductive hypothesis.
|
|
|
|
Prove $P(k + 1)$, that is:
|
|
|
|
$$ d_{k + 1} = \frac{2}{(k + 1)!} $$
|
|
|
|
By the given sequence:
|
|
|
|
$$ d_{k + 1} = \frac{d_k}{k + 1} $$
|
|
|
|
By the inductive hypothesis:
|
|
|
|
$$ = \frac{2}{(k + 1)k!} $$
|
|
|
|
$$ = \frac{2}{(k + 1)!} $$
|
|
|
|
Q.E.D.
|
|
|
|
28. Prove that for every integer $n \geq 1$,
|
|
|
|
$$ \frac{1}{3} = \frac{1 + 3 + 5 + \dots + (2n - 1)}{(2n + 1) + (2n + 3) + \dots + (2n + (2n - 1))} $$
|
|
|
|
**Proof (by mathematical induction):**
|
|
|
|
Let $P(n)$ be the equality:
|
|
|
|
$$ \frac{1}{3} = \frac{1 + 3 + 5 + \dots + (2n - 1)}{(2n + 1) + (2n + 3) + \dots + (2n + (2n - 1))} $$
|
|
|
|
_Basis Step:_
|
|
|
|
Prove $P(1)$, that is:
|
|
|
|
$$ \frac{1}{3} = \frac{1 + 3 + 5 + \dots + (2(1) - 1)}{(2(1) + 1) + (2(1) + 3) + \dots + (2(1) + (2(1) - 1))} $$
|
|
|
|
$$ \frac{1}{3} = \frac{1 + 3 + 5 + \dots + (2 - 1)}{(2 + 1) + (2 + 3) + \dots + (2 + (2 - 1))} $$
|
|
|
|
$$ \frac{1}{3} = \frac{1 + 3 + 5 + \dots + 1}{(2 + 1) + (2 + 3) + \dots + (2 + 1)} $$
|
|
|
|
$$ \frac{1}{3} = \frac{1}{(2 + 1) + (2 + 3) + \dots + 3} $$
|
|
|
|
$$ \frac{1}{3} = \frac{1}{3} $$
|
|
|
|
Therefore $P(1)$ is true.
|
|
|
|
_Inductive Step:_
|
|
|
|
Let $k$ be any integer where $k \geq 1$.
|
|
|
|
Suppose $P(k)$, that is:
|
|
|
|
$$ \frac{1}{3} = \frac{1 + 3 + 5 + \dots + (2k - 1)}{(2k + 1) + (2k + 3) + \dots + (2k + (2k - 1))} $$
|
|
|
|
This is the inductive hypothesis.
|
|
|
|
Prove $P(k + 1)$, that is:
|
|
|
|
$$ \frac{1}{3} = \frac{1 + 3 + 5 + \dots + (2(k + 1) - 1)}{(2(k + 1) + 1) + (2(k + 1) + 3) + \dots + (2(k + 1) + (2(k + 1) - 1))} $$
|
|
|
|
Starting from the inductive hypothesis:
|
|
|
|
$$ \frac{1}{3} = \frac{1 + 3 + 5 + \dots + (2k - 1)}{(2k + 1) + (2k + 3) + \dots + (2k + (2k - 1))} $$
|
|
|
|
Omitted.
|
|
|
|
Exercises 29 and 30 use the definition of string and string length from page 13
|
|
in Section 1.4. Recursive definitions for these terms are given in section 5.9.
|
|
|
|
29. A set $L$ consists of strings obtained by juxtaposing one or more of _abb_,
|
|
_bab_, and _bba_. Use mathematical induction to prove that for every integer
|
|
$n \geq 1$, if a string $s$ in $L$ has a length $3n$, then $s$ contains an
|
|
even number of _b_'s.
|
|
|
|
**Proof (by mathematical induction):**
|
|
|
|
Suppose a set $L$ consists of strings by juxtaposing one or more of _abb_,
|
|
_bab_, and _bba_.
|
|
|
|
Let $P(n)$ be the sentence:
|
|
|
|
If a string $s$ in $L$ has length $3n$, then $s$ contains an even number of
|
|
_b_'s.
|
|
|
|
_Basis Step:_
|
|
|
|
Prove $P(1)$, that is:
|
|
|
|
If a string $s$ in $L$ has length $3(1)$, then $s$ contains an even number of
|
|
_b_'s.
|
|
|
|
Since:
|
|
|
|
$$ L = \{\text{abb}, \text{bab}, \text{bba}\} $$
|
|
|
|
All three string $s$ in $L$ have a length of $3$ and all of them have an even
|
|
number of _b_'s.
|
|
|
|
Therefore $P(1)$ is true.
|
|
|
|
_Inductive Step:_
|
|
|
|
Let $k$ be any integer where $k \geq 1$.
|
|
|
|
Suppose $P(k)$, that is:
|
|
|
|
If a string $s$ in $L$ has length $3k$, then $s$ contains an even number of
|
|
_b_'s.
|
|
|
|
This is the inductive hypothesis.
|
|
|
|
Prove $P(k + 1)$, that is:
|
|
|
|
If a string $s$ in $L$ has length $3(k + 1)$, then $s$ contains an even number
|
|
of _b_'s.
|
|
|
|
Now $3(k + 1) = 3k + 3$ and the strings in $L$ are obtained by juxtaposing
|
|
strings already in $L$ with one of _abb_, _bab_, or _bba_. Thus, either the
|
|
initial or the final three characters in $s$ are _abb_, _bab_, or _bba_.
|
|
Moreoever, the other $3k$ characters in $s$ are also in $L$ by definition of
|
|
$L$, and so, by the inductive hypothesis, the other $3k$ characters in $s$
|
|
contain an even number, say $m$, of _b_'s. Because each of _abb_, _bab_, and
|
|
_bba_ contains 2 _b_'s, the total number of _b_'s in $s$ is $m + 2$, which is a
|
|
sum of even integers and hence is even.
|
|
|
|
Q.E.D.
|
|
|
|
30. A set $S$ consists of strings obtained by juxtaposing one or more copies of
|
|
1110 and 0111. Use mathematical induction to prove that for every integer
|
|
$n \geq 1$, if a string $s$ in $S$ has a length $4n$, then the number of 1's
|
|
in $s$ is a multiple of 3.
|
|
|
|
**Proof (by mathematical induction):**
|
|
|
|
Suppose a set $S$ contains strings obtained by juxtaposing one or more copies of
|
|
1110 and 0111.
|
|
|
|
Let $P(n)$ be the sentence:
|
|
|
|
If a string $s$ in $S$ has length $4n$, then the number of $1$'s in $s$ is a
|
|
multiple of $3$.
|
|
|
|
_Basis Step:_
|
|
|
|
Prove $P(1)$, that is:
|
|
|
|
If a string $s$ in $S$ has length $4(1)$, then the number of $1$'s in $s$ is a
|
|
multiple of $3$.
|
|
|
|
Since $S$ consists only of strings obtained by juxtaposing 1110 and 0111, then
|
|
at a minimum, the strings in $S$ must have a length of $4$. This means that the
|
|
only two strings in $S$ that have a length of $4$ are 1110 and 0111. The number
|
|
of $1$'s in $s$ is a multiple of $3$ in both cases.
|
|
|
|
Therefore $P(1)$ is true.
|
|
|
|
_Inductive Step:_
|
|
|
|
Let $k$ be any integer where $k \geq 1$.
|
|
|
|
Suppose $P(k)$, that is:
|
|
|
|
If a string $s$ in $S$ has length $4k$, then the number of $1$'s in $s$ is a
|
|
multiple of $3$.
|
|
|
|
This is the inductive hypothesis.
|
|
|
|
Prove $P(k + 1)$, that is:
|
|
|
|
If a string $s$ in $S$ has length $4(k + 1)$, then the number of $1$'s in $s$ is
|
|
a multiple of $3$.
|
|
|
|
Now $4(k + 1) = 4k + 4$ and the strings in $S$ are obtained by juxtaposing
|
|
strings already in $S$ with one of 1110 or 0111. Thus, the number of $1$'s is a
|
|
multiple of $3$ in both cases. Moreover, the other $4k$ digits in $s$ are also
|
|
in $S$ by the definition of $S$, and so, by inductive hypothesis, the other $4k$
|
|
characters in $s$ contain an odd number, say $m$ of $1$'s. Because each of 1110
|
|
and 0111 contain 3 $1$'s, the total number of $1$'s in $s$ is $m + 1$, which is
|
|
the sum of odd integers and hence is odd.
|
|
|
|
Q.E.D.
|
|
|
|
31. Use mathematical induction to give an alternative proof for the statement
|
|
proved in Example 4.9.9:
|
|
|
|
For any positive integer $n$, a complete graph on $n$ vertices has
|
|
$\dfrac{n(n - 1)}{2}$ edges. _Hint:_ Let $P(n)$ be the sentence, "the number of
|
|
edges in a complete graph on $n$ vertices is $\dfrac{n(n - 1)}{2}$."
|
|
|
|
Omitted.
|
|
|
|
32. Some $5 \times 5$ checkerboards with one square removed can be completely
|
|
covered by L-shaped trominoes, whereas other $5 \times 5$ checkerboards
|
|
cannot. Find examples of both kinds of checkerboards. Justify your answers.
|
|
|
|
Omitted.
|
|
|
|
33. Consider a $4 \times 6$ checkerboard. Draw a covering of the board by
|
|
L-shaped trominoes.
|
|
|
|
Omitted.
|
|
|
|
34.
|
|
|
|
a. Use mathematical induction to prove that for each integer $n \geq 1$, any
|
|
checkerboard with dimensions $2 \times 3n$ can be completely covered with
|
|
L-shaped trominoes.
|
|
|
|
Omitted.
|
|
|
|
b. Let $n$ be any integer greater than or equal to $1$. Use the result of part
|
|
(a) to prove by mathematical induction that for every integer $m$, any
|
|
checkerboard with dimensions $2m \times 3n$ can be completely covered with
|
|
L-shaped trominoes.
|
|
|
|
Omitted.
|
|
|
|
35. Let $m$ and $n$ be any integers that are greater than or equal to $1$.
|
|
|
|
a. Prove that a necessary condition for an $m \times n$ checkerboard to be
|
|
completely coverable by L-shaped trominoes is that $mn$ be divisible by $3$.
|
|
|
|
Omitted.
|
|
|
|
b. Prove that having $$ be divisible by $3$ is not a sufficient condition for an
|
|
$m \times n$ checkerboard to be completely covered by L-shaped trominoes.
|
|
|
|
Omitted.
|
|
|
|
36. In a round-robin tournament each team plays every other team exactly once
|
|
with ties not allowed. If the teams are labeled $T_1, T_2, \dots, T_n$, then
|
|
the outcome of such a tournament can be represented by a directed graph, in
|
|
which the teams are represented as dots and an arrow is drawn from one dot
|
|
to another if, and only if, the following team represented by the first dot
|
|
beats the team represented by the second dot. For example, the following
|
|
directed graph shows one outcome of a round-robin tournament involving five
|
|
teams, A, B, C, D, and E.
|
|
|
|
See Page 322 for image.
|
|
|
|
Use mathematical induction to show that in any round-robin tournament involving
|
|
$n$ teams, where $n \geq 2$, it is possible to label the teams
|
|
$T_1, T_2, \dots, T_n$ so that $T_i$ beats $T_{i + 1}$ for all
|
|
$i = 1, 2, \dots n - 1$,. (For instance, one such labeling in the example above
|
|
is $T_1 = 1, T_2 = B, T_3 = C, T_4 = E, T_5 = D$.) (_Hint:_ Given $k + 1$ teams,
|
|
pick one - say $T'$ - and apply the inductive hypothesis to the remaining teams
|
|
to obtain an ordering $T_1, T_2, \dots, T_k$. Consider three cases: $T'$ beats
|
|
$T_1$, $T'$ loses to the first $m$ teams (where $1 \leq m \leq k - 1$) and beats
|
|
the $(m + 1)$st team, and $T'$ loses to all the other teams.)
|
|
|
|
Omitted.
|
|
|
|
37. On the outside rim of a circular disk the integers from $1$ through $30$ are
|
|
painted in random order. Show that no matter what this order is, there must
|
|
be three successive integers whose sum is at least 45.
|
|
|
|
Omitted.
|
|
|
|
38. Suppose that $n$ _a_'s and $n$ _b_'s are distributed around the outside of a
|
|
circle. Use mathematical induction to prove that for any integer $n \geq 1$,
|
|
given any such arrangement, it is possible to find a starting point so that
|
|
if you travel around the circle in a clock-wise direction, the number of
|
|
_a_'s you pass is never less than the number of _b_'s you have passed. For
|
|
example, in the diagram shown below, you could start at the _a_ with an
|
|
asterisk.
|
|
|
|
See Page 322 for image.
|
|
|
|
Omitted.
|
|
|
|
39. For a polygon to be **convex** means that given any two points on or inside
|
|
the polygon, the line joining the points lies entirely inside the polygon.
|
|
Use mathematical induction to prove that for every integer $n \geq 3$, the
|
|
angles of any $n$-sided convex polygon add up to $180(n - 2)$ degrees.
|
|
|
|
Omitted.
|
|
|
|
40.
|
|
|
|
a. Prove that in an $8 \times 8$ checkerboard with alternating black and white
|
|
squares, if the squares in the top right and bottom left corners are removed the
|
|
remaining board cannot be covered with dominoes. (_Hint:_ Mathematical induction
|
|
is not needed for this proof.)
|
|
|
|
Omitted.
|
|
|
|
b. Use mathematical induction to prove that for each positive integer $n$, if a
|
|
$2n \times 2n$ checkerboard with alternating black and white squares has one
|
|
white square and one black square removed anywhere on the board, the remaining
|
|
squares can be covered with dominoes.
|
|
|
|
Omitted.
|
|
|
|
41. A group of people are positioned so that the distance between any two people
|
|
is different from the distance between any other two people. Suppose that
|
|
the group contains an odd number of people and each person sends a message
|
|
to their nearest neighbor. Use mathematical induction to prove that at least
|
|
one person does not receive a message from anyone. [This exercise is
|
|
inspired by the article "Odd Pie Fights" by L. Carmony, _The Mathematics
|
|
Teacher_, **72**(1), 1979, 61-64.]
|
|
|
|
Omitted.
|
|
|
|
42. Show that for any integer $n$, it is possible to find a group of $n$ people
|
|
who are all positioned so that the distance between any two people is
|
|
different from the distance between any other two people, so that each
|
|
person sends a message to their nearest neighbor, and so that every person
|
|
in the group receives a message from another person in the group.
|
|
|
|
Omitted.
|
|
|
|
43. Define a game as follows: You begin with an urn that contains a mixture of
|
|
white and black balls, and during the game you have access to as many
|
|
additional white and black balls as you might need. In each move you remove
|
|
two balls from the urn without looking at their colors. If the balls are the
|
|
same color, you put in one black ball. If the balls are different colors,
|
|
you put the white ball back into the urn and keep the black ball out.
|
|
Because each move reduces the number of balls in the urn by one, the game
|
|
will end with a single ball in the urn. If you know how many white balls and
|
|
how many black balls are initially in the urn, can you predict the color of
|
|
the ball at the end of the game? [This exercise is based on one described in
|
|
"Why correctness must be a mathematical concern" by E.W. Djikstra,
|
|
www.cs.utexas.edu/users/EWD/transcriptions/EWD07xx/EWD720.html.]
|
|
|
|
a. Map out all possibilities for playing the game starting with two balls in the
|
|
urn, then three balls, and then four balls. For each case keep track of the
|
|
number of white and black balls you start with and the color of the ball at the
|
|
end of the game.
|
|
|
|
Omitted.
|
|
|
|
b. Does the number of white balls seem to be predictive? Does the number of
|
|
black balls seem to be predictive? Make a conjecture about the color of the ball
|
|
at the end of the game given the numbers of white and black balls at the
|
|
beginning.
|
|
|
|
Omitted.
|
|
|
|
c. Use mathematical induction to prove the conjecture you made in part (b).
|
|
|
|
Omitted.
|
|
|
|
44. Let $P(n)$ be the following sentence: Given any graph $G$ with $n$ vertices
|
|
satisfying the condition that every vertex of $G$ has degree at most $M$,
|
|
then the vertices of $G$ can be colored with at most $M + 1$ colors in such
|
|
a way that no two adjacent vertices have the same color. Use mathematical
|
|
induction to prove this statement is true for every integer $n \geq 1$.
|
|
|
|
In order for a proof by mathematical induction to be valid, the basis statement
|
|
must be true for $n = a$ and the argument of the inductive step must be correct
|
|
for every integer $k \geq a$. IN 45 and 46 find the mistakes in the "proofs" by
|
|
mathematical induction.
|
|
|
|
Omitted.
|
|
|
|
45.
|
|
|
|
**"Theorem:"** For any integer $n \geq 1$, all the numbers in a set of $n$
|
|
numbers are equal to each other.
|
|
|
|
**"Proof (by mathematical induction):** It is obviously true that all the
|
|
numbers in a set consisting of just one number are equal to each other, so the
|
|
basis step is true. For the inductive step, let
|
|
$A = \{a_1, a_2, \dots, a_k, a_{k + 1}\}$ be any set of $k + 1$ numbers. Form
|
|
two subsets each of size $k$:
|
|
|
|
$$ B = \{a_1, a_2, a_3, \dots, a_k\} \text{ and } $$
|
|
|
|
$$ C = \{a_1, a_3, a_4, \dots, a_{k + 1}} $$
|
|
|
|
($B$ consists of all the numbers in $A$ except $a_{k + 1}$, and $C$ consists of
|
|
all the numbers in $A$ except $a_2$.) By inductive hypothesis, all the numbers
|
|
in $B$ equal $a_1$ and all the numbers in $C$ equal $a_1$ (since both sets have
|
|
only $k$ numbers). But every number in $A$ is in $B$ or $C$, so all the numbers
|
|
in $A$ equal $a_1$; hence all are equal to each other."
|
|
|
|
Omitted.
|
|
|
|
46.
|
|
|
|
**"Theorem:"** For every integer $n \geq 1$, $3^n - 2$ is even.
|
|
|
|
**"Proof (by mathematical induction):** Suppose the theorem is true for an
|
|
integer $k$, where $k \geq 1$. That is, suppose that $3^k - 2$ is even. We must
|
|
show that $3^{k + 1} - 2$ is even. Observe that
|
|
|
|
$$ 3^{k + 1} - 2 = 3^k \cdot 3 - 2 = 3^k(1 + 2) - 2 $$
|
|
|
|
$$ = (3^k - 2) + 3^k \cdot 2 $$
|
|
|
|
Now $3^k - 2$ is even by inductive hypothesis and $3^k \cdot 2$ is even by
|
|
inspection. Hence the sum of the two quantities is even (by Theorem 4.1.1). It
|
|
follows that $3^{k + 1} - 2$ is even, which is what we needed to show."
|
|
|
|
Omitted.
|
|
|
|
---
|
|
|
|
**Exercise Set 5.4**
|
|
|
|
Page 333
|
|
|
|
1. Suppose $a_1, a_2, a_3, \dots$ is a sequence defined as follows:
|
|
|
|
$$ a_1 = 1, a_2 = 3, a_k = a_{k - 2} + 2a_{k - 1} $$
|
|
|
|
for each integer $k \geq 3$.
|
|
|
|
Prove that $a_n$ is odd for every integer $n \geq 1$.
|
|
|
|
**Proof (by strong mathematical induction):**
|
|
|
|
Let the property $P(n)$ be the sentence "$a_n$ is odd."rim
|
|
|
|
_Basis Step:_
|
|
|
|
Prove $P(1)$ and $P(2)$ are true. That is:
|
|
|
|
$$ a_1 \text{ is odd} $$
|
|
|
|
and
|
|
|
|
$$ a_2 \text{ is odd} $$
|
|
|
|
Observe from the given definition of the sequence that $a_1 = 1$, which means
|
|
that $P(1)$ is true since $1$ is odd. Also, observe that $a_2 = 3$, which means
|
|
that $P(2)$ is true since $3$ is odd.
|
|
|
|
_Inductive Step:_
|
|
|
|
Let $k$ be any integer with $k \geq 2$. Suppose $a_i$ is odd for each integer
|
|
$i$ with $1 \leq i \leq k$.
|
|
|
|
This is the inductive hypothesis.
|
|
|
|
Prove $P(k + 1)$ is true.
|
|
|
|
By the definition of the sequence, we know that
|
|
|
|
$$ a_{k + 1} = a_{k - 1} + 2a_k $$
|
|
|
|
By the inductive hypothesis, $a_{k - 1}$ is odd.
|
|
|
|
Also, every term in the sequence is an integer by the sum of products of
|
|
integers, and so $2a_k$ is even by the definition of even. It follows that
|
|
$a_{k + 1}$ is the sum of an odd integer and an even integer. By Theorem 4.1.2,
|
|
the sum of an odd and even integer is odd. Therefore $a_{k + 1}$ is odd, and
|
|
$P(k + 1)$ is true.
|
|
|
|
Q.E.D.
|
|
|
|
2. Suppose $b_1, b_2, b_3, \dots$ is a sequence defined as follows:
|
|
|
|
$$ b_1 = 4, b_2 = 12, b_k = b_{k - 2} + b_{k - 1} $$
|
|
|
|
for each integer $k \geq 3$.
|
|
|
|
Prove that $b_n$ is divisible by $4$ for every integer $n \geq 1$.
|
|
|
|
**Proof (by strong mathematical induction):**
|
|
|
|
Let $P(n)$ be the sentence "$b_n$ is divisible by $4$."
|
|
|
|
_Basis Step:_
|
|
|
|
Prove $P(1)$ and $P(2)$. That is:
|
|
|
|
$$ b_1 \text{ is divisible by } 4 $$
|
|
|
|
and
|
|
|
|
$$ b_2 \text{ is divisible by } 4 $$
|
|
|
|
By the given sequence, we know that $b_1 = 4$, which is divisible by $4$ since
|
|
$4 = 4 \cdot 1$. Also $b_2 = 12$, which is divisible by $4$ since
|
|
$12 = 4 \cdot 3$. Therefore $P(1)$ and $P(2)$ are true.
|
|
|
|
_Inductive Step:_
|
|
|
|
Let $k$ be any integer where $k \geq 2$. Suppose that $b_i$ is divisible by $4$
|
|
for each integer $1 \leq i \leq k$.
|
|
|
|
This is the inductive hypothesis.
|
|
|
|
Prove $P(k + 1)$. That is:
|
|
|
|
$$ b_{k + 1} \text{ is divisible by } 4 $$
|
|
|
|
By the definition of the sequence, we know that
|
|
|
|
$$ b_{k + 1} = b_{k - 1} + b_k $$
|
|
|
|
By the inductive hypothesis, we know that $b_{k - 1}$ and $b_k$ are both
|
|
divisible by $4$. By the definition of divisibility, $b_{k + 1}$ can be
|
|
represented as follows:
|
|
|
|
$$ b_{k + 1} = 4r + 4s $$
|
|
|
|
where $r$ and $s$ are some integers. By algebra then:
|
|
|
|
$$ b_{k + 1} = 4(r + s) $$
|
|
|
|
Now, $r + s$ is an integer by the sum of integers. By the definition of
|
|
divisibility, $b_{k + 1}$ is divisible by $4$. Therefore $P(k + 1)$ is true.
|
|
|
|
Q.E.D.
|
|
|
|
3. Suppose that $c_0, c_1, c_2, \dots$ is a sequence defined as follows:
|
|
|
|
$$ c_0 = 2, c_1 = 2, c_2 = 6, c_k = 3c_{k - 3} $$
|
|
|
|
for every integer $k \geq 3$.
|
|
|
|
Prove that $c_n$ is even for each integer $n \geq 0$.
|
|
|
|
**Proof (by strong mathematical induction):**
|
|
|
|
Let $P(n)$ be the sentence "$c_n$ is even."
|
|
|
|
_Basis Step:_
|
|
|
|
Prove $P(0)$, $P(1)$, and $P(2)$. That is:
|
|
|
|
$$ c_0 \text{ is even} $$
|
|
|
|
and
|
|
|
|
$$ c_1 \text{ is even} $$
|
|
|
|
and
|
|
|
|
$$ c_2 \text{ is even} $$
|
|
|
|
By the given sequence $c_0 = 2$, and $2$ is even by the definition of even.
|
|
Also, $c_1 = 2$, and $2$ is even by the definition of even. Also, $c_2 = 6$, and
|
|
$6$ is even by the definition of even. Therefore $P(0)$, $P(1)$, and $P(2)$ are
|
|
true.
|
|
|
|
_Inductive Step:_
|
|
|
|
Let $k$ be any integer where $k \geq 2$. Suppose $c_i$ is even for each integer
|
|
$i$ with $0 \leq i \leq k$.
|
|
|
|
This is the inductive hypothesis.
|
|
|
|
Prove $P(k + 1)$. That is:
|
|
|
|
$$ c_{k + 1} \text{ is even} $$
|
|
|
|
By the given sequence, we know that:
|
|
|
|
$$ c_{k + 1} = 3c_{k - 2} $$
|
|
|
|
By the inductive hypothesis, we know that $c_{k - 2}$ is even. $c_{k + 1}$ can
|
|
then be represented as:
|
|
|
|
$$ c_{k + 1} = 3(2r) $$
|
|
|
|
for some integer $r$.
|
|
|
|
Then, by algebra:
|
|
|
|
$$ c_{k + 1} = 6r $$
|
|
|
|
$$ c_{k + 1} = 2(3r) $$
|
|
|
|
Now, $3r$ is an integer by the product of integers. It follows that $c_{k + 1}$
|
|
is even by the definition of even. Therefore $P(k + 1)$ is true.
|
|
|
|
Q.E.D.
|
|
|
|
4. Suppose that $d_1, d_2, d_3, \dots$ is sequence defined as follows:
|
|
|
|
$$ d_1 = \frac{9}{10}, d_2 = \frac{10}{11}, d_k = d_{k - 1} \cdot d_{k - 2} $$
|
|
|
|
for every integer $k \geq 3$.
|
|
|
|
Prove that $0 < d_n \leq 1$ for each integer $n \geq 1$.
|
|
|
|
**Proof (by strong mathematical induction):**
|
|
|
|
Let $P(n)$ be the sentence "$0 < d_n \leq 1$."
|
|
|
|
_Basis Step:_
|
|
|
|
Prove $P(1)$ and $P(2)$. That is:
|
|
|
|
$$ 0 < d_1 \leq 1 $$
|
|
|
|
and
|
|
|
|
$$ 0 < d_2 \leq 1 $$
|
|
|
|
By the given sequence we know that $d_1 = \dfrac{9}{10}$, and that
|
|
$0 < \dfrac{9}{10} \leq 1$. Also, we know that $d_2 = \dfrac{10}{11}$, and that
|
|
$0 < \dfrac{10}{11} \leq 1$. Therefore $P(1)$ and $P(2)$ are both true.
|
|
|
|
_Inductive Step:_
|
|
|
|
Let $k$ be any integer where $k \geq 2$. Suppose $0 < d_i \leq 1$ for each
|
|
integer $i$ with $1 \leq i \leq k$.
|
|
|
|
This is the inductive hypothesis.
|
|
|
|
Prove $P(k + 1)$. That is:
|
|
|
|
$$ 0 < d_{k + 1} \leq 1 $$
|
|
|
|
By the given sequence, we know that:
|
|
|
|
$$ d_{k + 1} = d_k \cdot d_{k - 1} $$
|
|
|
|
By the inductive hypothesis, we know that $0 < d_k \leq 1$ and that
|
|
$0 < d_{k - 1} \leq 1$. Consequently, $0 < d_{k + 1} \leq 1$ because the product
|
|
of two positive numbers less than or equal to $1$ is itself less than or equal
|
|
to $1$. Therefore $P(k + 1)$ is true.
|
|
|
|
Q.E.D.
|
|
|
|
5. Suppose that $e_0, e_1, e_2, \dots$ is a sequence defined as follows:
|
|
|
|
$$ e_0 = 12, e_1 = 29, e_k, = 5e_{k - 1} - 6e_{k - 2} $$
|
|
|
|
for each integer $k \geq 2$.
|
|
|
|
Prove that $e_n = 5 \cdot 3^n + 7 \cdot 2^n$ for every integer $n \geq 0$.
|
|
|
|
**Proof (by strong mathematical induction):**
|
|
|
|
Let $P(n)$ be the sentence "$e_n = 5 \cdot 3^n + 7 \cdot 2^n$."
|
|
|
|
_Basis Step:_
|
|
|
|
Prove $P(0)$ and $P(1)$. That is:
|
|
|
|
$$ e_0 = 5 \cdot 3^0 + 7 \cdot 2^0 $$
|
|
|
|
and
|
|
|
|
$$ e_1 = 5 \cdot 3^1 + 7 \cdot 2^1 $$
|
|
|
|
By the given sequence, we know that $e_0 = 12$. By algebra/arithmetic:
|
|
|
|
$$ 12 = 5 \cdot 3^0 + 7 \cdot 2^0 $$
|
|
|
|
$$ 12 = 5 \cdot 1 + 7 \cdot 1 $$
|
|
|
|
$$ 12 = 5 + 7 $$
|
|
|
|
$$ 12 = 12 $$
|
|
|
|
By the given sequence, we know that $e_1 = 29$. By algebra/arithmetic:
|
|
|
|
$$ 29 = 5 \cdot 3^1 + 7 \cdot 2^1 $$
|
|
|
|
$$ 29 = 5 \cdot 3 + 7 \cdot 2 $$
|
|
|
|
$$ 29 = 15 + 14 $$
|
|
|
|
$$ 29 = 29 $$
|
|
|
|
Therefore $P(0)$ and $P(1)$ are both true.
|
|
|
|
_Inductive Step:_
|
|
|
|
Let $k$ be any integer where $k \geq 2$. Suppose
|
|
$e_i = 5 \cdot 3^i + 7 \cdot 2^i$ for each integer $i$ with $0 \leq i \leq k$.
|
|
|
|
This is the inductive hypothesis.
|
|
|
|
Prove $P(k + 1)$. That is:
|
|
|
|
$$ e_{k + 1} = 5 \cdot 3^{k + 1} + 7 \cdot 2^{k + 1} $$
|
|
|
|
By the given sequence, we know that:
|
|
|
|
$$ e_{k + 1} = 5e_{k} - 6e_{k - 1} $$
|
|
|
|
By the inductive hypothesis and substitution, $e_{k + 1}$ can be rewritten as:
|
|
|
|
$$ e_{k + 1} = 5(5 \cdot 3^k + 7 \cdot 2^k) - 6(5 \cdot 3^{k - 1} + 7 \cdot 2^{k - 1}) $$
|
|
|
|
$$ = 25 \cdot 3^k + 35 \cdot 2^k - 30 \cdot 3^{k - 1} - 42 \cdot 2^{k - 1} $$
|
|
|
|
$$ = 25 \cdot 3^k + 35 \cdot 2^k - 10 \cdot 3 \cdot 3^{k - 1} - 21 \cdot 2 \cdot 2^{k - 1} $$
|
|
|
|
$$ = 25 \cdot 3^k + 35 \cdot 2^k - 10 \cdot 3^k - 21 \cdot 2^k $$
|
|
|
|
$$ = (25 - 10) \cdot 3^k + (35 - 21) \cdot 2^k $$
|
|
|
|
$$ = 15 \cdot 3^k + 14 \cdot 2^k $$
|
|
|
|
$$ = 5 \cdot 3 \cdot 3^k + 7 \cdot 2 \cdot 2^k $$
|
|
|
|
$$ = 5 \cdot 3^{k + 1} + 7 \cdot 2^{k + 1} $$
|
|
|
|
Therefore $P(k + 1)$ is true.
|
|
|
|
Q.E.D.
|
|
|
|
6. Suppose that $f_0, f_1, f_2, \dots$ is a sequence defined as follows:
|
|
|
|
$$ f_0 = 5, f_1 = 16, f_k = 7f_{k - 1} - 10f_{k - 2} $$
|
|
|
|
for every integer $k \geq 2$.
|
|
|
|
Prove that $f_n = 3 \cdot 2^n + 2 \cdot 5^n$ for each integer $n \geq 0$.
|
|
|
|
**Proof (by strong mathematical induction):**
|
|
|
|
Let $P(n)$ be the sentence "$f_n = 3 \cdot 2^n + 2 \cdot 5^n$."
|
|
|
|
_Basis Step:_
|
|
|
|
Prove $P(0)$ and $P(1)$. That is:
|
|
|
|
$$ f_0 = 3 \cdot 2^0 + 2 \cdot 5^0 $$
|
|
|
|
$$ f_1 = 3 \cdot 2^1 + 2 \cdot 5^1 $$
|
|
|
|
By the given sequence, we know that $f_0 = 5$. So, by algebra/arithmetic:
|
|
|
|
$$ 5 = 3 \cdot 2^0 + 2 \cdot 5^0 $$
|
|
|
|
$$ 5 = 3 \cdot 1 + 2 \cdot 1 $$
|
|
|
|
$$ 5 = 3 + 2 $$
|
|
|
|
$$ 5 = 5 $$
|
|
|
|
By the given sequence, we know that $f_1 = 16$. So, by algebra/arithmetic:
|
|
|
|
$$ 16 = 3 \cdot 2^1 + 2 \cdot 5^1 $$
|
|
|
|
$$ 16 = 3 \cdot 2 + 2 \cdot 5 $$
|
|
|
|
$$ 16 = 6 + 10 $$
|
|
|
|
$$ 16 = 16 $$
|
|
|
|
Therefore $P(0)$ and $P(1)$ are both true.
|
|
|
|
_Inductive Step:_
|
|
|
|
Let $k$ be any integer where $k \geq 2$. Suppose
|
|
$f_i = 3 \cdot 2^i + 2 \cdot 5^i$ for each integer $i$ with $0 \leq i \leq k$.
|
|
|
|
This is the inductive hypothesis.
|
|
|
|
Prove $P(k + 1)$. That is:
|
|
|
|
$$ f_{k + 1} = 3 \cdot 2^{k + 1} + 2 \cdot 5^{k + 1} $$
|
|
|
|
By the given sequence, we know that:
|
|
|
|
$$ f_{k + 1} = 7f_k - 10f_{k - 1} $$
|
|
|
|
By the inductive hypothesis and substitution, $f_{k + 1}$ can be rewritten as:
|
|
|
|
$$ f_{k + 1} = 7(3 \cdot 2^k + 2 \cdot 5^k) - 10(3 \cdot 2^{k - 1} + 2 \cdot 5^{k - 1}) $$
|
|
|
|
$$ = (21 \cdot 2^k + 14 \cdot 5^k) - (30 \cdot 2^{k - 1} + 20 \cdot 5^{k - 1}) $$
|
|
|
|
$$ = (21 \cdot 2 \cdot 2^{k - 1} + 14 \cdot 5 \cdot 5^{k - 1}) - (30 \cdot 2^{k - 1} + 20 \cdot 5^{k - 1}) $$
|
|
|
|
$$ = 42 \cdot 2^{k - 1} + 70 \cdot 5^{k - 1} - 30 \cdot 2^{k - 1} - 20 \cdot 5^{k - 1} $$
|
|
|
|
$$ = (42 - 30) \cdot 2^{k - 1} + (70 - 20) \cdot 5^{k - 1} $$
|
|
|
|
$$ = 12 \cdot 2^{k - 1} + 50 \cdot 5^{k - 1} $$
|
|
|
|
$$ = 3 \cdot 2 \cdot 2 \cdot 2^{k - 1} + 2 \cdot 5 \cdot 5 \cdot 5^{k - 1} $$
|
|
|
|
$$ = 3 \cdot 2^{k + 1} + 2 \cdot 5^{k + 1} $$
|
|
|
|
Therefore $P(k + 1)$ is true.
|
|
|
|
Q.E.D.
|
|
|
|
7. Suppose that $g_1, g_2, g_3, \dots$ is a sequence defined as follows:
|
|
|
|
$$ g_1 = 3, g_2 = 5, g_k = 3g_{k - 1} - 2g_{k - 2} $$
|
|
|
|
for each integer $k \geq 3$.
|
|
|
|
Prove that $g_n = 2^n + 1$ for every integer $n \geq 1$.
|
|
|
|
**Proof (by strong mathematical induction):**
|
|
|
|
Let $P(n)$ be the sentence "$g_n = 2^n + 1$."
|
|
|
|
_Basis Step:_
|
|
|
|
Prove $P(1)$ and $P(2)$. That is:
|
|
|
|
$$ g_1 = 2^1 + 1 $$
|
|
|
|
and
|
|
|
|
$$ g_2 = 2^2 + 1 $$
|
|
|
|
By the given sequence, we know that $g_1 = 3$. By algebra/arithmetic:
|
|
|
|
$$ 3 = 2^1 + 1 $$
|
|
|
|
$$ 3 = 2 + 1 $$
|
|
|
|
$$ 3 = 3 $$
|
|
|
|
By the given sequence, we know that $g_2 = 5$. By algebra/arithmetic:
|
|
|
|
$$ 5 = 2^2 + 1 $$
|
|
|
|
$$ 5 = 4 + 1 $$
|
|
|
|
$$ 5 = 5 $$
|
|
|
|
Therefore $P(1)$ and $P(2)$ are both true.
|
|
|
|
_Inductive Step:_
|
|
|
|
Let $k$ be any integer where $k \geq 2$. Suppose $g_i = 2^i + 1$ for each
|
|
integer $i$ with $1 \leq i \leq k$.
|
|
|
|
This is the inductive hypothesis.
|
|
|
|
Prove $P(k + 1)$. That is:
|
|
|
|
$$ g_{k + 1} = 2^{k + 1} + 1 $$
|
|
|
|
By the given sequence, we know that:
|
|
|
|
$$ g_{k + 1} = 3g_k - 2g_{k - 1} $$
|
|
|
|
By the inductive hypothesis and substitution, $g_{k + 1}$ can be rewritten as:
|
|
|
|
$$ g_{k + 1} = 3(2^k + 1) - 2(2^{k - 1} + 1) $$
|
|
|
|
$$ = 3 \cdot 2^k + 3 - 2 \cdot 2^{k - 1} - 2 $$
|
|
|
|
$$ = 3 \cdot 2 \cdot 2^{k - 1} + 3 - 2 \cdot 2^{k - 1} - 2 $$
|
|
|
|
$$ = 6 \cdot 2^{k - 1} + 3 - 2 \cdot 2^{k - 1} - 2 $$
|
|
|
|
$$ = (6 - 2) \cdot 2^{k - 1} + 3 - 2 $$
|
|
|
|
$$ = 4 \cdot 2^{k - 1} + 1 $$
|
|
|
|
$$ = 2 \cdot 2 \cdot 2^{k - 1} + 1 $$
|
|
|
|
$$ = 2 \cdot 2^{k} + 1 $$
|
|
|
|
$$ = 2^{k + 1} + 1 $$
|
|
|
|
Therefore $P(k + 1)$ is true.
|
|
|
|
Q.E.D.
|
|
|
|
8. Suppose that $h_0, h_1, h_2, \dots$ is a sequence defined as follows:
|
|
|
|
$$ h_0 = 1, h_1 = 2, h_2 = 3, h_k = h_{k - 1} + h_{k - 2} + h_{k - 3} $$
|
|
|
|
for each integer $k \geq 3$.
|
|
|
|
a. Prove that $h_n \leq 3^n$ for every integer $n \geq 0$.
|
|
|
|
**Proof (by strong mathematical induction):**
|
|
|
|
Let $P(n)$ be the sentence "$h_n \leq 3^n$."
|
|
|
|
_Basis Step:_
|
|
|
|
Prove $P(0)$, $P(1)$, and $P(2)$. That is:
|
|
|
|
$$ h_0 \leq 3^0 $$
|
|
|
|
and
|
|
|
|
$$ h_1 \leq 3^1 $$
|
|
|
|
and
|
|
|
|
$$ h_2 \leq 3^2 $$
|
|
|
|
By the given sequence we know that $h_0 = 1$. By substitution:
|
|
|
|
$$ 1 \leq 3^0 $$
|
|
|
|
$$ 1 \leq 1 $$
|
|
|
|
By the given sequence we know that $h_1 = 2$. By substitution:
|
|
|
|
$$ 2 \leq 3^1 $$
|
|
|
|
$$ 2 \leq 3 $$
|
|
|
|
By the given sequence we know that $h_2 = 3$. By substitution:
|
|
|
|
$$ 3 \leq 3^2 $$
|
|
|
|
$$ 3 \leq 9 $$
|
|
|
|
Therefore $P(0)$, $P(1)$, and $P(2)$ are all true.
|
|
|
|
_Inductive Step:_
|
|
|
|
Let $k$ be any integer where $k \geq 3$. Suppose $h_i \leq 3^i$ for each integer
|
|
$i$ with $0 \leq i \leq k$.
|
|
|
|
This is the inductive hypothesis.
|
|
|
|
Prove $P(k + 1)$. That is:
|
|
|
|
$$ h_{k + 1} \leq 3^{k + 1} $$
|
|
|
|
By the given sequence, we know that:
|
|
|
|
$$ h_{k + 1} = h_k + h_{k - 1} + h_{k - 2} $$
|
|
|
|
$$ = h_k + h_{k - 1} + h_{k - 2} \leq 3^k + 3^{k - 1} + 3^{k - 2} $$
|
|
|
|
$$ = h_k + h_{k - 1} + h_{k - 2} \leq 3^{k - 2}(3^2 + 3^1 + 1) $$
|
|
|
|
$$ = h_k + h_{k - 1} + h_{k - 2} \leq 3^{k - 2}(9 + 3 + 1) $$
|
|
|
|
$$ = h_k + h_{k - 1} + h_{k - 2} \leq 13 \cdot 3^{k - 2} $$
|
|
|
|
Since $3^{k + 1} = 3^3 \cdot 3^{k - 2} = 27 \cdot 3^{k - 2}$, we know then that:
|
|
|
|
$$ = h_k + h_{k - 1} + h_{k - 2} \leq 13 \cdot 3^{k - 2} \leq 27 \cdot 3^{k - 2} = 3^{k + 1} $$
|
|
|
|
Therefore $P(k + 1)$ is true.
|
|
|
|
Q.E.D.
|
|
|
|
b. Suppose that $s$ is any real number such that $s^3 \geq s^2 + s + 1$. (This
|
|
implies that $2 > s > 1.83$.) Prove that $h_n \leq s^n$ for every integer
|
|
$n \geq 2$.
|
|
|
|
Omitted.
|
|
|
|
9. Define a sequence $a_1, a_2, a_3, \dots$ as follows: $a_1 = 1, a_2 = 3$, and
|
|
$a_k = a_{k - 1} + a_{k - 2}$ for every integer $k \geq 3$. (This sequence is
|
|
known as the Lucas sequence.) Use strong mathematical induction to prove that
|
|
$a_n \leq \left(\dfrac{7}{4}\right)^n$ for every integer $n \geq 1$.
|
|
|
|
**Proof (by strong mathematical induction):**
|
|
|
|
Let $P(n)$ be the sentence "$a_n \leq \left(\dfrac{7}{4}\right)^n$."
|
|
|
|
_Basis Step:_
|
|
|
|
Prove $P(1)$ and $P(2)$. That is:
|
|
|
|
$$ a_1 \leq \left(\dfrac{7}{4}\right)^1 $$
|
|
|
|
and
|
|
|
|
$$ a_2 \leq \left(\dfrac{7}{4}\right)^2 $$
|
|
|
|
By the given sequence, we know that $a_1 = 1$. By substitution:
|
|
|
|
$$ 1 \leq \left(\dfrac{7}{4}\right)^1 $$
|
|
|
|
$$ 1 \leq \dfrac{7}{4} = 1.75 $$
|
|
|
|
By the given sequence, we know that $a_2 = 3$. By substitution:
|
|
|
|
$$ 3 \leq \left(\dfrac{7}{4}\right)^2 $$
|
|
|
|
$$ 3 \leq \dfrac{49}{16} = 3.0625 $$
|
|
|
|
Therefore $P(1)$ and $P(2)$ are both true.
|
|
|
|
_Inductive Step:_
|
|
|
|
Let $k$ be any integer where $k \geq 2$. Suppose
|
|
$a_i \leq \left(\dfrac{7}{4}\right)^i$ for each integer $i$ with
|
|
$1 \leq i \leq k$.
|
|
|
|
This is the inductive hypothesis.
|
|
|
|
Prove $P(k + 1)$. That is:
|
|
|
|
$$ a_{k + 1} \leq \left(\dfrac{7}{4}\right)^{k + 1} $$
|
|
|
|
By the given sequence, we know that:
|
|
|
|
$$ a_{k + 1} = a_k + a_{k - 1} $$
|
|
|
|
$$ = a_k + a_{k - 1} \leq \left(\frac{7}{4}\right)^k + \left(\frac{7}{4}\right)^{k - 1} $$
|
|
|
|
$$ \leq \left(\frac{7}{4}\right) \cdot \left(\frac{7}{4}\right)^{k - 1} + \left(\frac{7}{4}\right)^{k - 1} $$
|
|
|
|
$$ \leq \left(\frac{7}{4} + 1\right) \cdot \left(\frac{7}{4}\right)^{k - 1} $$
|
|
|
|
$$ \leq \left(\frac{11}{4}\right) \cdot \left(\frac{7}{4}\right)^{k - 1} $$
|
|
|
|
Since we know that:
|
|
|
|
$$ \left(\dfrac{7}{4}\right)^{k + 1} = \frac{7}{4} \cdot \frac{7}{4} \cdot \left(\frac{7}{4}\right)^{k - 1} $$
|
|
|
|
$$ \left(\dfrac{7}{4}\right)^{k + 1} = \frac{49}{16} \cdot \left(\frac{7}{4}\right)^{k - 1} $$
|
|
|
|
Since $\dfrac{11}{4} < \dfrac{49}{16}$, it follows that:
|
|
|
|
$$ a_{k + 1} = \frac{11}{4} \cdot \left(\frac{7}{4}\right)^{k - 1} \leq \frac{49}{16} \cdot \left(\frac{7}{4}\right)^{k - 1} = \left(\dfrac{7}{4}\right)^{k + 1} $$
|
|
|
|
Therefore $P(k + 1)$ is true.
|
|
|
|
Q.E.D.
|
|
|
|
10. The introductory example solved with ordinary mathematical induction in
|
|
Section 5.3 can also be solved using strong mathematical induction. Let
|
|
$P(n)$ be "any $n$¢ can be obtained using a combination of $3$¢ and $5$¢
|
|
coins." Use strong mathematical induction to prove that $P(n)$ is true for
|
|
every integer $n \geq 8$.
|
|
|
|
**Proof (by strong mathematical induction):**
|
|
|
|
Let $P(n)$ be the sentence "any $n$¢ can be obtained using a combination of $3$¢
|
|
and $5$¢ coins."
|
|
|
|
_Basis Step:_
|
|
|
|
Prove $P(8)$ and $P(9)$.
|
|
|
|
$P(8)$ is true because $8$¢ can be obtained by using one $3$¢ coin and one $5$¢
|
|
coin.
|
|
|
|
$P(9)$ is true because $9$¢ can be obtained by using three $3$¢ coins.
|
|
|
|
Therefore $P(8)$ and $P(9)$ are both true.
|
|
|
|
_Inductive Step:_
|
|
|
|
Let $k$ be any integer where $k \geq 8$. Suppose $P(i)$ is true for every
|
|
integer $i$ where $8 \leq i \leq k$.
|
|
|
|
This is the inductive hypothesis.
|
|
|
|
Prove $P(k + 1)$. That is:
|
|
|
|
"any $(k + 1)$¢ can be obtained using a combination of $3$¢ and $5$¢ coins."
|
|
|
|
_Case 1 (There is a $5$¢ coin among those that used to make up the $k$¢.):_
|
|
|
|
In this case replace the $5$¢ coin by two $3$¢ coins; the result will be
|
|
$(k + 1)$¢.
|
|
|
|
_Case 2 (There is not a $5$¢ coin among those used to make up the $k$¢.):_
|
|
|
|
In this case, because $k \geq 8$, at least three $3$¢ coins must have been used.
|
|
So remove three $3$¢ coins and replace them by two $5$¢ coins; the result will
|
|
be $(k + 1)$¢.
|
|
|
|
Thus in either case $(k + 1)$¢ can be obtained using $3$¢ and $5$¢ coins.
|
|
|
|
Q.E.D.
|
|
|
|
11. You begin solving a jigsaw puzzle by finding two pieces that match and
|
|
fitting them together. Every subsequent step of the solution consists of
|
|
fitting together two blocks, each of which is made up of one or more pieces
|
|
that have previously been assembled. Use strong mathematical induction to
|
|
prove that for every integer $n \geq 1$, the number of steps required to put
|
|
together all $n$ pieces of a jigsaw puzzle is $n - 1$.
|
|
|
|
**Proof (by strong mathematical induction):**
|
|
|
|
Let $P(n)$ be the sentence "For every integer $n \geq 1$, the number of steps
|
|
required to put together all $n$ pieces of a jigsaw puzzle is $n - 1$."
|
|
|
|
_Basis Step:_
|
|
|
|
Prove $P(1)$. That is:
|
|
|
|
"For every integer $1 \geq 1$, the number of steps required to put together all
|
|
$1$ pieces of a jigsaw puzzle is $1 - 1 = 0$."
|
|
|
|
Since there is only $1$ piece of the jigsaw puzzle, it follows that it takes $0$
|
|
steps to complete the jigsaw puzzle.
|
|
|
|
Therefore $P(1)$ is true.
|
|
|
|
_Inductive Step:_
|
|
|
|
Let $k$ be any integer where $k \geq 1$. Suppose that $P(i)$ is true, where
|
|
$1 \leq i \leq k$.
|
|
|
|
This is the inductive hypothesis.
|
|
|
|
Prove $P(k + 1)$. That is:
|
|
|
|
"For every integer $(k + 1) \geq 1$, the number of steps required to put
|
|
together all $(k + 1)$ pieces of a jigsaw puzzle is $(k + 1) - 1 = k$."
|
|
|
|
Consider assembling a jigsaw puzzle consisting of $k + 1$ pieces. The last step
|
|
involves fitting together two blocks. Suppose one of the blocks consists of $r$
|
|
pieces and the other consists of $s$ pieces (where $r$ and $s$ are some
|
|
integers.) Then $r + s = k + 1$ and $1 \leq r \leq k$ and $1 \leq s \leq k$.
|
|
|
|
By the inductive hypothesis, the number of steps required to assemble the blocks
|
|
are $r - 1$ and $s - 1$, respectively.
|
|
|
|
Then, the total number of steps required to assemble the puzzle is
|
|
$(r - 1) + (s - 1) + 1 = (r + s) - 1 = (k + 1) - 1 = k$.
|
|
|
|
Therefore $P(k + 1)$ is true.
|
|
|
|
Q.E.D.
|
|
|
|
12. The sides of a circular track contain a sequence of $n$ cans of gasoline.
|
|
For each integer $n \geq 1$, the total amount in the cans is sufficient to
|
|
enable a certain car to make one complete circuit of the track. In addition,
|
|
all the gasoline could fit into the car's gas tank at one time. Use
|
|
mathematical induction to prove that it is possible to find an initial
|
|
location for placing the car so that it will be able to traverse the entire
|
|
track by using the various amounts of gasoline in the cans that it
|
|
encounters along the way.
|
|
|
|
**Proof (by strong mathematical induction):**
|
|
|
|
Let $P(n)$ be the sentence:
|
|
|
|
For any circular track containing $n$ gasoline cans whose total gasoline is
|
|
enough for one complete circuit (and all gasoline fits in the tank), there
|
|
exists an initial location at which the car can start and successfully traverse
|
|
the entire track.
|
|
|
|
_Basis Step:_
|
|
|
|
Prove $P(1)$. That is:
|
|
|
|
For any circular track containing $1$ gasoline cans whose total gasoline is
|
|
enough for one complete circuit (and all gasoline fits in the tank), there
|
|
exists an initial location at which the car can start and successfully traverse
|
|
the entire track.
|
|
|
|
It follows from the given problem statement that since there is $1$ gasoline can
|
|
whose total gasoline is enough for one complete circuit, that the initial
|
|
location at which the car can start and successfully traverse the entire track
|
|
is the location of this $1$ gasoline can.
|
|
|
|
Therefore $P(1)$ is true.
|
|
|
|
_Inductive Step:_
|
|
|
|
Let $k$ be any integer where $k \geq 1$. Suppose $P(i)$ is true for every
|
|
integer $i$ where $1 \leq i \leq k$.
|
|
|
|
This is the inductive hypothesis.
|
|
|
|
Prove $P(k + 1)$. That is:
|
|
|
|
For any circular track containing $(k + 1)$ gasoline cans whose total gasoline
|
|
is enough for one complete circuit (and all gasoline fits in the tank), there
|
|
exists an initial location at which the car can start and successfully traverse
|
|
the entire track.
|
|
|
|
Consider an arbitrary circular track with $k + 1$ gasoline cans. Since the total
|
|
amount of gasoline in the cans is sufficient to enable the car to make one
|
|
complete circuit of the track, at least one gasoline can must contain enough
|
|
gasoline to enable the car to travel to the next can.
|
|
|
|
Take such a can and transfer its gasoline to the can immediately preceding it in
|
|
the direction of travel. This reduces the number of cans from $k + 1$ to $k$.
|
|
|
|
By the inductive hypothesis, the resulting configuration with $k$ cans can be
|
|
traversed starting from some initial location. This starting location also works
|
|
for the $k + 1$ can configuration, since the redistribution of gasoline does not
|
|
prevent traversal of the track.
|
|
|
|
Q.E.D.
|
|
|
|
13. Use strong mathematical induction to prove the existence part of the unique
|
|
factorization of integers theorem (Theorem 4.4.5). In other words, prove
|
|
that every integer greater than $1$ is either a prime number or a product of
|
|
prime numbers.
|
|
|
|
**Proof (by strong mathematical induction):**
|
|
|
|
Let $P(n)$ be the sentence "$n$ is either a prime number or a product of prime
|
|
numbers."
|
|
|
|
_Basis Step:_
|
|
|
|
Prove $P(2)$. That is:
|
|
|
|
"$2$ is either a prime number or a product of prime numbers."
|
|
|
|
By the definition of prime numbers, $2$ is a prime number.
|
|
|
|
Therefore $P(2)$ is true.
|
|
|
|
_Inductive Step:_
|
|
|
|
Let $k$ be any integer where $k > 1$. Suppose $P(i)$, for every $i$ where
|
|
$2 \leq i \leq k$, that is:
|
|
|
|
"$i$ is either a prime number or a product of prime numbers."
|
|
|
|
Prove $P(k + 1)$. That is:
|
|
|
|
"$(k + 1)$ is either a prime number or a product of prime numbers."
|
|
|
|
_Case where $(k + 1)$ is prime:_
|
|
|
|
Since $(k + 1)$ is prime, $P(k + 1)$ is true.
|
|
|
|
_Case where $(k + 1)$ is composite (not prime):_
|
|
|
|
Since $(k + 1)$ is composite, this means that $k + 1$ can be written as:
|
|
|
|
$$ k + 1 = a \cdot b $$
|
|
|
|
where $a$ and $b$ are some integers such that $2 \leq a \leq k$ and
|
|
$2 \leq b \leq k$.
|
|
|
|
By the inductive hypothesis, this means that both $P(a)$ and $P(b)$ are true. It
|
|
follows then that $a \cdot b$ is a product of primes and that $k + 1$ is a
|
|
product of primes. Therefore $P(k + 1)$ is true.
|
|
|
|
Q.E.D.
|
|
|
|
14. Any product of two or more integers is a result of successive
|
|
multiplications of two integers at a time. For instance, here are a few of
|
|
the ways in which $a_1a_2a_3a_4$ might be computed: $(a_1a_2)(a_3a_4)$ or
|
|
$(((a_1a_2)a_3)a_4)$ or $a_1((a_2a_3)a_4)$. Use strong mathematical
|
|
induction to prove that any product of two or more odd integers is odd.
|
|
|
|
**Proof (by strong mathematical induction):**
|
|
|
|
Let $P(n)$ be the sentence "any product of $n \geq 2$ odd integers is odd."
|
|
|
|
_Basis Step:_
|
|
|
|
Prove $P(2)$. That is:
|
|
|
|
"any product of $2$ odd integers is odd."
|
|
|
|
The following is a proof from 4.2 (exercise 20) that proves this:
|
|
|
|
Suppose $n$ is any odd integer and $m$ is any odd integer.
|
|
|
|
Since $n$ and $m$ are odd integers, $n = 2k + 1$ and $m = 2s + 1$ where $k$ is
|
|
some integer and $s$ is some integer.
|
|
|
|
Then:
|
|
|
|
$$ n \cdot m = (2k + 1)(2s + 1) \quad \text{ by substitution} $$
|
|
|
|
$$ \quad = 4ks + 2s + 2k + 1 $$
|
|
|
|
$$ \quad = 2(2ks + s + k) + 1 \quad \text{ by algebra} $$
|
|
|
|
Let $t = 2ks + s + k$.
|
|
|
|
Then $n \cdot m = 2(2ks + s + k) + 1 = 2t + 1$ where $t$ is an integer because
|
|
the products and sums of integers is an integer.
|
|
|
|
Therefore $n \cdot m$ is odd by the definition of odd integers and $P(2)$ is
|
|
true.
|
|
|
|
_Inductive Step:_
|
|
|
|
Let $k$ be any integer where $k \geq 2$. Suppose $P(i)$ for every integer $i$
|
|
where $2 \leq i \leq k$. That is:
|
|
|
|
"any product of $i$ odd integers is odd."
|
|
|
|
This is the inductive hypothesis.
|
|
|
|
Prove $P(k + 1)$. That is:
|
|
|
|
"any product of $(k + 1)$ odd integers is odd."
|
|
|
|
Consider the product of a series of odd integers up until $k + 1$ integers:
|
|
|
|
$$ [a_1 \cdot a_2 \cdot a_3 \cdot \dots \cdot a_k] \cdot a_{k + 1} $$
|
|
|
|
By the inductive hypothesis we know that
|
|
$[a_1 \cdot a_2 \cdot a_3 \cdot \dots \cdot a_k]$ is odd. Thus, we can rewrite
|
|
this as:
|
|
|
|
$$ (2r + 1) \cdot a_{k + 1} $$
|
|
|
|
where $r$ is some integer.
|
|
|
|
Now $2r + 1$ is an integer by the sum and product of integers and $2r + 1$ is
|
|
odd by the definition of odd. The product of $(2r + 1) \cdot a_{k + 1}$ is odd
|
|
by the proof provided in the basis step. Thus the product of $k + 1$ odd
|
|
integers is odd.
|
|
|
|
Therefore $P(k + 1)$ is true.
|
|
|
|
Q.E.D.
|
|
|
|
15. Define the "sum" of one integer to be that integer, and use strong
|
|
mathematical induction to prove that for every integer $n \geq 1$, any sum
|
|
of $n$ even integers is even.
|
|
|
|
**Proof (by strong mathematical induction):**
|
|
|
|
Let $P(n)$ be the sentence "any sum of $n$ even integers is even."
|
|
|
|
_Basis Step:_
|
|
|
|
Prove $P(1)$. That is:
|
|
|
|
"any sum of $1$ even integers is even."
|
|
|
|
Let $r$ be any even integer. Since $r$ is even, $r = 2s$ for some integer $s$.
|
|
|
|
By the problem statement, the sum of one integer is that integer. Therefore the
|
|
sum of $r$ is $r$, which is even.
|
|
|
|
Therefore $P(1)$ is true.
|
|
|
|
_Inductive Step:_
|
|
|
|
Let $k$ be any integer where $k \geq 1$. Suppose $P(i)$ is true for every
|
|
integer $i$ where $1 \leq i \leq k$. That is:
|
|
|
|
"any sum of $i$ even integers is even."
|
|
|
|
This is the inductive hypothesis.
|
|
|
|
Prove $P(k + 1)$. That is:
|
|
|
|
"any sum of $(k + 1)$ even integers is even."
|
|
|
|
Consider a series of even integers up until $k + 1$ integers:
|
|
|
|
$$ a_1, a_2, a_3, \dots, a_{k + 1} $$
|
|
|
|
Now consider the sum of these even integers:
|
|
|
|
$$ a_1 + a_2 + a_3 + \dots + a_{k + 1} $$
|
|
|
|
This can also be written as:
|
|
|
|
$$ [a_1 + a_2 + a_3 + \dots + a_k] + a_{k + 1} $$
|
|
|
|
By the inductive hypothesis we know that $[a_1 + a_2 + a_3 + \dots + a_k]$ is
|
|
even. Then we can rewrite this sum as:
|
|
|
|
$$ (2q) + a_{k + 1} $$
|
|
|
|
for some integer $q$.
|
|
|
|
Also, since we know that $a_{k + 1}$ is even, we can further rewrite this as:
|
|
|
|
$$ (2q) + (2u) $$
|
|
|
|
for some integer u.
|
|
|
|
Then this becomes, by algebra:
|
|
|
|
$$ 2(q + u) $$
|
|
|
|
Now $q + u$ is an integer by the sum of integers, and $2(q + u)$ is even by the
|
|
definition of even. Thus, the sum of $k + 1$ integers is even.
|
|
|
|
Therefore $P(k + 1)$ is true.
|
|
|
|
Q.E.D.
|
|
|
|
16. Use strong mathematical induction to prove that for every integer
|
|
$n \geq 2$, if $n$ is even, then any sum of $n$ odd integers is even, and if
|
|
$n$ is odd, then any sum of $n$ odd integers is odd.
|
|
|
|
**Proof (by strong mathematical induction):**
|
|
|
|
Let $P(n)$ be the sentence "If $n$ is even, then any sum of $n$ odd integers is
|
|
even, and if $n$ is odd, then any sum of $n$ odd integers is odd."
|
|
|
|
_Basis Step:_
|
|
|
|
Prove $P(2)$ and $P(3)$.
|
|
|
|
For $P(2)$:
|
|
|
|
"If $2$ is even, then any sum of $2$ odd integers is even, and if $2$ is odd,
|
|
then any sum of $2$ odd integers is odd."
|
|
|
|
Since $2$ is even:
|
|
|
|
Let $m$ and $p$ be any $2$ odd integers. Since both $m$ and $p$ are odd,
|
|
$m = 2q + 1$ and $p = 2r + 1$ for some integers $q$ and $r$.
|
|
|
|
Their sum then is:
|
|
|
|
$$ m + p = 2q + 1 + 2r + 1 $$
|
|
|
|
$$ = 2q + 2r + 2 $$
|
|
|
|
$$ = 2(q + r + 1) $$
|
|
|
|
Now $q + r + 1$ is an integer by the sum of integers. Also, $2(q + r + 1)$ is
|
|
even by the definition of even. Thus $P(2)$ is true.
|
|
|
|
and
|
|
|
|
For $P(3)$:
|
|
|
|
"If $3$ is even, then any sum of $3$ odd integers is even, and if $3$ is odd,
|
|
then any sum of $3$ odd integers is odd."
|
|
|
|
Since $3$ is odd:
|
|
|
|
Let $a$, $b$, and $c$ be any $3$ odd integers. Since $a$, $b$, and $c$ are odd,
|
|
then $a = 2z + 1$, $b = 2y + 1$, and $c = 2x + 1$, for some integers $z$, $y$,
|
|
and $x$.
|
|
|
|
Their sum then is:
|
|
|
|
$$ a + b + c = (2z + 1) + (2y + 1) + (2x + 1) $$
|
|
|
|
$$ = 2z + 2y + 2x + 2 + 1 $$
|
|
|
|
$$ = 2(z + y + x + 1) + 1 $$
|
|
|
|
Now, $z + y + x + 1$ is an integer by the sum of integers, and
|
|
$2(z + y + x + 1) + 1$ is odd by the definition of odd. Thus $P(3)$ is true.
|
|
|
|
Therefore $P(2)$ and $P(3)$ are both true.
|
|
|
|
_Inductive Step:_
|
|
|
|
Let $k$ be any integer where $k \geq 2$. Suppose $P(i)$ for every integer $i$
|
|
where $2 \leq i \leq k$. That is:
|
|
|
|
"If $i$ is even, then any sum of $i$ odd integers is even, and if $i$ is odd,
|
|
then any sum of $i$ odd integers is odd."
|
|
|
|
Prove $P(k + 1)$. That is:
|
|
|
|
"If $(k + 1)$ is even, then any sum of $(k + 1)$ odd integers is even, and if
|
|
$(k + 1)$ is odd, then any sum of $(k + 1)$ odd integers is odd."
|
|
|
|
_Case $(k + 1)$ is odd:_
|
|
|
|
Consider a series of odd integers up until $k + 1$ integers:
|
|
|
|
$$ a_1, a_2, a_3, \dots, a_{k + 1} $$
|
|
|
|
Their sum would be:
|
|
|
|
$$ a_1 + a_2 + a_3 + \dots + a_{k + 1} $$
|
|
|
|
Alternatively:
|
|
|
|
$$ [a_1 + a_2 + a_3 + \dots + a_k] + a_{k + 1} $$
|
|
|
|
By the definition of odd, if $k + 1$ is odd, then $k$ is even. By the inductive
|
|
hypothesis then, we know that $[a_1 + a_2 + a_3 + \dots + a_k]$ is even. Thus,
|
|
we can rewrite our summation as:
|
|
|
|
$$ 2r + a_{k + 1} $$
|
|
|
|
for some integer $r$.
|
|
|
|
Since we know that $a_{k + 1}$ is odd, we can further rewrite our summation as:
|
|
|
|
$$ 2r + (2s + 1) $$
|
|
|
|
for some integer $s$.
|
|
|
|
Then, by algebra:
|
|
|
|
$$ 2(r + s) + 1 $$
|
|
|
|
Now, $r + s$ is an integer by the sum of integers, and $2(r + s) + 1$ is odd by
|
|
the definition of odd.
|
|
|
|
Thus $P(k + 1)$ is true in this case.
|
|
|
|
_Case $(k + 1)$ is even:_
|
|
|
|
Consider a series of odd integers up until $k + 1$ integers:
|
|
|
|
$$ a_1, a_2, a_3, \dots, a_{k + 1} $$
|
|
|
|
Their sum would be:
|
|
|
|
$$ a_1 + a_2 + a_3 + \dots + a_{k + 1} $$
|
|
|
|
Alternatively:
|
|
|
|
$$ [a_1 + a_2 + a_3 + \dots + a_k] + a_{k + 1} $$
|
|
|
|
By the definition of even, if $k + 1$ is even, then $k$ is odd. By the inductive
|
|
hypothesis then, we know that $[a_1 + a_2 + a_3 + \dots + a_k]$ is odd. Thus, we
|
|
can rewrite our summation as:
|
|
|
|
$$ (2r + 1) + a_{k + 1} $$
|
|
|
|
for some integer $r$.
|
|
|
|
Since we know that $a_{k + 1}$ is odd, we can further rewrite our summation as:
|
|
|
|
$$ (2r + 1) + (2s + 1) $$
|
|
|
|
for some integer $s$.
|
|
|
|
Then, by algebra:
|
|
|
|
$$ 2r + 2s + 2 $$
|
|
|
|
$$ 2(r + s + 1) $$
|
|
|
|
Now, $r + s + 1$ is an integer by the sum of integers, and $2(r + s + 1)$ is
|
|
even by the definition of even. Thus $P(k + 1)$ is true in this case.
|
|
|
|
Therefore in both cases $P(k + 1)$ is true.
|
|
|
|
Q.E.D.
|
|
|
|
17. Compute $4^1, 4^2, 4^3, 4^4, 4^5, 4^6, 4^7,$ and $4^8$. Make a conjecture
|
|
about the units digit of $4^n$ where $n$ is a positive integer. Use strong
|
|
mathematical induction to prove your conjecture.
|
|
|
|
$$
|
|
4^1 = 4 \\
|
|
4^2 = 16 \\
|
|
4^3 = 64 \\
|
|
4^4 = 256 \\
|
|
4^5 = 1024 \\
|
|
4^6 = 4096 \\
|
|
4^7 = 16384 \\
|
|
4^8 = 65536 \\
|
|
$$
|
|
|
|
**Conjecture:**
|
|
|
|
For some integer $n \geq 1$, if $n$ is odd, then the units digit of $4^n$ is
|
|
$4$, if $n$ is even, then the units digit of $4^n$ is $6$.
|
|
|
|
**Proof (by strong mathematical induction):**
|
|
|
|
Let $P(n)$ be the sentence: "the units digit of $4^n$ is $4$ if $n$ is odd and
|
|
$6$ if $n$ is even."
|
|
|
|
_Basis Step:_
|
|
|
|
Prove $P(1)$ and $P(2)$.
|
|
|
|
For $P(1)$, since $1$ is odd, then the units of digit of $4^1$ should be $4$.
|
|
Evaluating $4^1$:
|
|
|
|
$$ 4^1 = 4 $$
|
|
|
|
The units digit of $4^1$ is $4$, so $P(1)$ is true.
|
|
|
|
For $P(2)$, since $2$, is even, then the units digit of $4^2$ should be $6$.
|
|
Evaluating $4^2$:
|
|
|
|
$$ 4^2 = 16 $$
|
|
|
|
The units digit of $4^2$ is $6$, so $P(2)$ is true.
|
|
|
|
Therefore both $P(1)$ and $P(2)$ are true.
|
|
|
|
_Inductive Step:_
|
|
|
|
Let $k$ be any integer where $k \geq 1$. Suppose $P(i)$ for every integer $i$
|
|
where $1 \leq i \leq k$. That is:
|
|
|
|
"the units digit of $4^i$ is $4$ if $i$ is odd and $6$ if $i$ is even."
|
|
|
|
Prove $P(k + 1)$. That is:
|
|
|
|
"the units digit of $4^{k + 1}$ is $4$ if $(k + 1)$ is odd and $6$ if $(k + 1)$
|
|
is even."
|
|
|
|
_Case $(k + 1)$ is even:_
|
|
|
|
Consider the following:
|
|
|
|
$$ 4^{k + 1} = 4 \cdot 4^k $$
|
|
|
|
By the definition of even, if $k + 1$ is even, then $k$ is odd. Thus $4^k$ is
|
|
$4$ to the power of an odd integer. By the inductive hypothesis, we know that
|
|
this means that:
|
|
|
|
$$ 4^{k + 1} = 4 \cdot (10m + 4) $$
|
|
|
|
for some integer $m$.
|
|
|
|
By algebra:
|
|
|
|
$$ = 40m + 16 $$
|
|
|
|
$$ = 10(4m + 1) + 6 $$
|
|
|
|
Where $4m + 1$ is an integer by the sum and product of integers. Thus the units
|
|
digit of $4^{k + 1}$ is $6$.
|
|
|
|
Therefore $P(k + 1)$ is true in this case.
|
|
|
|
_Case $(k + 1)$ is odd:_
|
|
|
|
Consider the following:
|
|
|
|
$$ 4^{k + 1} = 4 \cdot 4^k $$
|
|
|
|
By the definition of odd, if $k + 1$ is odd, then $k$ is even. Thus $4^k$ is $4$
|
|
to the power of an even integer. By the inductive hypothesis, we know that this
|
|
means that:
|
|
|
|
$$ 4^{k + 1} = 4 \cdot (10m + 6) $$
|
|
|
|
for some integer $m$.
|
|
|
|
By algebra:
|
|
|
|
$$ = 40m + 24 $$
|
|
|
|
$$ = 10(4m + 2) + 4 $$
|
|
|
|
Where $4m + 2$ is an integer by the sum and product of integers. Thus the units
|
|
digit of $4^{k + 1}$ is $4$.
|
|
|
|
Therefore $P(k + 1)$ is true in this case.
|
|
|
|
Therefore, in both cases $P(k + 1)$ is true.
|
|
|
|
Q.E.D.
|
|
|
|
18. Compute $9^0, 9^1, 9^2, 9^3, 9^4,$ and $9^5$. Make a conjecture about the
|
|
units digit of $9^n$ where $n$ is a positive integer. Use strong
|
|
mathematical induction to prove your conjecture.
|
|
|
|
$$
|
|
9^0 = 1 \\
|
|
9^1 = 9 \\
|
|
9^2 = 81 \\
|
|
9^3 = 729 \\
|
|
9^4 = 6561 \\
|
|
9^5 = 59049 \\
|
|
$$
|
|
|
|
**Conjecture:**
|
|
|
|
For any integer $n \geq 0$, the units digit of $9^n$ is $1$ if $n$ is even, and
|
|
$9$ if $n$ is odd.
|
|
|
|
**Proof (by strong induction):**
|
|
|
|
Let $P(n)$ be the sentence: "the units digit of $9^n$ is $1$ if $n$ is even, and
|
|
$9$ if $n$ is odd."
|
|
|
|
_Basis Step:_
|
|
|
|
Prove $P(0)$ and $P(1)$.
|
|
|
|
For $P(0)$:
|
|
|
|
Since $0$ is even, the units digit of $9^0$ is claimed to be $1$. Evaluate
|
|
$9^0$:
|
|
|
|
$$ 9^0 = 1 $$
|
|
|
|
Thus $P(0)$ is true.
|
|
|
|
For $P(1)$:
|
|
|
|
Since $1$ is odd, the units digit of $9^1$ is claimed to be $9$. Evaluate $9^1$;
|
|
|
|
$$ 9^1 = 9 $$
|
|
|
|
Thus $P(1)$ is true.
|
|
|
|
Therefore both $P(0)$ and $P(1)$ are true.
|
|
|
|
_Inductive Step:_
|
|
|
|
Let $k$ be any integer where $k \geq 0$. Suppose $P(i)$ for every integer $i$
|
|
where $0 \leq i \leq k$. That is:
|
|
|
|
"the units digit of $9^i$ is $1$ if $i$ is even, and $9$ if $i$ is odd."
|
|
|
|
Prove $P(k + 1)$. That is:
|
|
|
|
"the units digit of $9^{k + 1}$ is $1$ if $(k + 1)$ is even, and $9$ if
|
|
$(k + 1)$ is odd."
|
|
|
|
_Case where $(k + 1)$ is even:_
|
|
|
|
Consider:
|
|
|
|
$$ 9^{k + 1} = 9 \cdot 9^k $$
|
|
|
|
By the definition of even, if $k + 1$ is even, then $k$ is odd.
|
|
|
|
By the inductive hypothesis, we know that the units digit of $9^k$ is $9$ if $k$
|
|
is odd. We can then rewrite $9^{k + 1}$ as:
|
|
|
|
$$ 9^{k + 1} = 9 \cdot (10m + 9) $$
|
|
|
|
for some integer $m$.
|
|
|
|
Then, by algebra:
|
|
|
|
$$ 9^{k + 1} = 90m + 81 $$
|
|
|
|
$$ = 10(9m + 8) + 1 $$
|
|
|
|
Where $9m + 8$ is an integer by the sum and product of integers. Thus the units
|
|
digit of $9^{k + 1}$ is $1$.
|
|
|
|
Therefore $P(k + 1)$ is true in this case.
|
|
|
|
_Case where $(k + 1)$ is odd:_
|
|
|
|
Consider:
|
|
|
|
$$ 9^{k + 1} = 9 \cdot 9^k $$
|
|
|
|
By the definition of odd, if $k + 1$ is odd, then $k$ is even.
|
|
|
|
By the inductive hypothesis, we know that the units digit of $9^k$ is $1$ if $k$
|
|
is even. We can then rewrite $9^{k + 1}$ as:
|
|
|
|
$$ 9^{k + 1} = 9 \cdot (10m + 1) $$
|
|
|
|
for some integer $m$.
|
|
|
|
Then, by algebra:
|
|
|
|
$$ 9^{k + 1} = 90m + 9 $$
|
|
|
|
$$ = 10(9m) + 9 $$
|
|
|
|
Where $9m$ is an integer by the product of integers. Thus the units digit of
|
|
$9^{k + 1}$ is $9$.
|
|
|
|
Therefore $P(k + 1)$ is true in this case.
|
|
|
|
Therefore $P(k + 1)$ is true in all cases.
|
|
|
|
Q.E.D.
|
|
|
|
19. Suppose that $a_1, a_2, a_3, \dots$ is a sequence defined as follows:
|
|
|
|
$a_1 = 1$ $a_k = 2 \cdot a_{\lfloor \frac{k}{2} \rfloor}$
|
|
|
|
for every integer $k \geq 2$.
|
|
|
|
Prove that $a_n \leq n$ for each integer $n \geq 1$.
|
|
|
|
**Proof (by strong mathematical induction):**
|
|
|
|
Let $a_1, a_2, a_3 \dots$ be a sequence that satisfies the recurrence relation
|
|
$a_k = 2 \cdot a_{\lfloor \frac{k}{2} \rfloor}$ for every integer $k \geq 2$,
|
|
with the initial condition $a_1 = 1$.
|
|
|
|
Let $P(n)$ be the inequality $a_n \leq n$.
|
|
|
|
_Basis Step:_
|
|
|
|
Prove $P(1)$. By the given sequence, we know that $a_1 = 1$. Then:
|
|
|
|
$$ 1 \leq 1 $$
|
|
|
|
This is a true statement, therefore $P(1)$ is true.
|
|
|
|
_Inductive Step:_
|
|
|
|
Let $k$ be any integer where $k \geq 1$. Suppose $P(i)$ for every integer $i$
|
|
where $1 \leq i \leq k$. That is:
|
|
|
|
$$ a_k \leq k $$
|
|
|
|
Prove $P(k + 1)$. That is:
|
|
|
|
$$ a_{k + 1} \leq (k + 1) $$
|
|
|
|
By the given sequence, we know that:
|
|
|
|
$$ a_{k + 1} = 2 \cdot a_{\lfloor \frac{k + 1}{2} \rfloor} $$
|
|
|
|
$$ \leq 2 \cdot \lfloor \frac{k + 1}{2} \rfloor $$
|
|
|
|
By the inductive hypothesis:
|
|
|
|
$$
|
|
\leq
|
|
\begin{cases}
|
|
2 \cdot \left(\frac{k + 1}{2}\right) & \text{if } k \text{ is odd} \\
|
|
2 \cdot \left(\frac{k}{2}\right) & \text{if } k \text{ is even}
|
|
\end{cases}
|
|
$$
|
|
|
|
$$
|
|
\leq
|
|
\begin{cases}
|
|
k + 1 & \text{if } k \text{ is odd} \\
|
|
k & \text{if } k \text{ is even}
|
|
\end{cases}
|
|
$$
|
|
|
|
$$ \leq k + 1 $$
|
|
|
|
In both cases $a_{k + 1} \leq (k + 1)$. Therefore $P(k + 1)$ is true.
|
|
|
|
Q.E.D.
|
|
|
|
20. Suppose that $b_1, b_2, b_3, \dots$ is a sequence defined as follows:
|
|
|
|
$b_1 = 0, b_2 = 3, b_k = 5 \cdot b_{\frac{k}{2}} + 6$
|
|
|
|
for every integer $k \geq 3$.
|
|
|
|
Prove that $b_n$ is divisible by $3$ for each integer $n \geq 1$.
|
|
|
|
**Proof (by strong mathematical induction):**
|
|
|
|
Let the sequence, $b_1, b_2, b_3, \dots$ be the sequence that satisfies the
|
|
recurrence relation $b_k = 5 \cdot b_{\lfloor \frac{k}{2} \rfloor} + 6$ for
|
|
every integer $k \geq 3$, with the initial conditions $b_1 = 0$ and $b_2 = 3$.
|
|
|
|
Let $P(n)$ be the sentence "$b_n$ is divisible by $3$" where $n \geq 1$.
|
|
|
|
_Basis Step:_
|
|
|
|
Prove $P(1)$ and $P(2)$.
|
|
|
|
For $P(1)$:
|
|
|
|
By the given sequence $b_1 = 0$, and $0$ is divisible by $3$ since
|
|
$0 = 0 \cdot 3$.
|
|
|
|
For $P(2)$:
|
|
|
|
By the given sequence $b_2 = 3$, and $3$ is divisible by $3$ since
|
|
$3 = 1 \cdot 3$.
|
|
|
|
Therefore both $P(1)$ and $P(2)$ are true.
|
|
|
|
_Inductive Step:_
|
|
|
|
Let $k$ be any integer where $k \geq 1$. Suppose $P(i)$ for every integer $i$
|
|
such that $1 \leq i \leq k$. That is:
|
|
|
|
"$b_i$ is divisible by $3$"
|
|
|
|
Prove $P(k + 1)$. That is:
|
|
|
|
"$b_{k + 1}$ is divisible by $3$"
|
|
|
|
By the given sequence, we know that:
|
|
|
|
$$ b_{k + 1} = 5 \cdot b_{\lfloor \frac{k + 1}{2} \rfloor} + 6 $$
|
|
|
|
Since $1 \leq \lfloor \dfrac{k + 1}{2} \rfloor \leq k$, then, by the inductive
|
|
hypothesis, $b_{\lfloor \frac{k + 1}{2} \rfloor}$ is divisible by $3$.
|
|
|
|
By the definition of divisibility, we can then rewrite $b_{k + 1}$ as:
|
|
|
|
$$ b_{k + 1} = 5 \cdot 3m + 6 $$
|
|
|
|
for some integer $m$.
|
|
|
|
Then, by algebra:
|
|
|
|
$$ = 15m + 6 $$
|
|
|
|
$$ = 3(5m + 2) $$
|
|
|
|
Now, $5m + 2$ is an integer by the sum and product of integers. Thus
|
|
$3 \mid b_{k + 1}$.
|
|
|
|
Therefore $P(k + 1)$ is true.
|
|
|
|
Q.E.D.
|
|
|
|
21. Suppose that $c_1, c_2, c_3, \dots$ is a sequence defined as follows:
|
|
|
|
$$ c_0 = 1, c_1 = 1, c_k = c_{\lfloor \frac{k}{2} \rfloor} + c_{\lfloor \frac{k}{2} \rfloor} $$
|
|
|
|
for every integer $k \geq 2$.
|
|
|
|
Prove that $c_n = n$ for each integer $n \geq 1$.
|
|
|
|
**Proof (by strong mathematical induction):**
|
|
|
|
Let the sequence, $c_0, c_1, c_2, \dots$ be the sequence that satisfies the
|
|
recurrence relation
|
|
$c_k = c_{\lfloor \frac{k}{2} \rfloor} + c_{\lfloor \frac{k}{2} \rfloor}$ for
|
|
every integer $k \geq 2$, with the initial conditions $c_0 = 1$ and $c_1 = 1$.
|
|
|
|
Let $P(n)$ be the equality $c_n = n$ for each integer $n \geq 1$.
|
|
|
|
_Basis Step:_
|
|
|
|
Prove $P(1)$ and $P(2)$.
|
|
|
|
For $P(1)$:
|
|
|
|
Based on the given sequence, we know that $c_1 = 1$. Thus $1 = 1$ is a true
|
|
statement.
|
|
|
|
For $P(2)$:
|
|
|
|
Based on the given recurrence relation:
|
|
|
|
$$ c_2 = c_{\lfloor \frac{2}{2} \rfloor} + c_{\lfloor \frac{2}{2} \rfloor} $$
|
|
|
|
$$ c_2 = c_1 + c_1 $$
|
|
|
|
Based on the given sequence, we know that $c_1 = 1$. By substitution:
|
|
|
|
$$ c_2 = 1 + 1 $$
|
|
|
|
$$ c_2 = 2 $$
|
|
|
|
$2 = 2$ is a true statement.
|
|
|
|
Therefore $P(1)$ and $P(2)$ are both true.
|
|
|
|
_Inductive Step:_
|
|
|
|
Let $k$ be any integer where $k \geq 1$. Suppose $P(i)$ for every integer $i$
|
|
such that $1 \leq i \leq k$. That is:
|
|
|
|
$$ c_i = i $$
|
|
|
|
This is the inductive hypothesis.
|
|
|
|
Prove $P(k + 1)$. That is:
|
|
|
|
$$ c_{k + 1} = k + 1 $$
|
|
|
|
By the given sequence, we know that:
|
|
|
|
$$ c_{k + 1} = c_{\lfloor \frac{k + 1}{2} \rfloor} + c_{\lfloor \frac{k + 1}{2} \rfloor} $$
|
|
|
|
Since we know that $1 \leq \lfloor \dfrac{k + 1}{2} \rfloor \leq k$, by the
|
|
inductive hypothesis, we then know that
|
|
$c_{\lfloor \frac{k + 1}{2} \rfloor} = \dfrac{k + 1}{2}$. By substitution:
|
|
|
|
$$ c_{k + 1} = \frac{k + 1}{2} + \frac{k + 1}{2} $$
|
|
|
|
$$ = 2\left(\frac{k + 1}{2}\right) $$
|
|
|
|
$$ = k + 1 $$
|
|
|
|
Therefore $P(k + 1)$ is true.
|
|
|
|
Q.E.D.
|
|
|
|
22. One version of the game NIM starts with two piles of objects such as coins,
|
|
stones, or matchsticks. In each turn a player is required to remove from one
|
|
to three objects from one of the piles. The two players take turns doing
|
|
this until both piles are empty. The loser is the first player who can't
|
|
make a move. Use strong mathematical induction to show that if both piles
|
|
contain the same number of objects at the start of the game, the player who
|
|
goes second can always win.
|
|
|
|
Omitted.
|
|
|
|
23. Define a game $G$ as follows: Begin with a pile of $n$ stones and $0$
|
|
points. In the first move split the pile into two possibly unequal
|
|
sub-piles, multiply the number of stones in one sub-pile times the number of
|
|
stones in the other sub-pile, and add the product to your score. In the
|
|
second move, split each of the newly created piles into a pair of possibly
|
|
unequal sub-piles, multiply the number of stones in each sub-pile times the
|
|
number of stones in the paired sub-pile, and add the new products to your
|
|
score. Continue by successively splitting each newly created pile of stones
|
|
that has at least two stones into a pair of sub-piles, multiplying the
|
|
number of stones in each sub-pile times the number of stones in the paired
|
|
sub-pile, and adding the new products to your score. The game $G$ ends when
|
|
no pile contains more than one stone.
|
|
|
|
a. Play $G$ starting with $10$ stones and using the following initial moves. In
|
|
move $1$ split the pile of $10$ stones into two sub-piles with $3$ and $7$
|
|
stones respectively, compute $3 \cdot 7 = 21$, and find that your score is $21$.
|
|
In move $2$ split the pile of $3$ stones into two sub-piles, with $1$ and $2$
|
|
stones respectively, and split the pile of $7$ stones into two sub-piles, with
|
|
$4$ and $3$ stones respectively, compute $1 \cdot 2 = 2$ and $4 \cdot 3 = 12$,
|
|
and find that your score is $21 + 2 + 12 = 35$. In move $3$ split the pile of
|
|
$4$ stones into two sub-piles, each with $2$ stones, and split the pile of $3$
|
|
<F2>tones into two sub-piles, with $1$ and $2$ stones respectively, and find
|
|
your new score. Continue splitting piles and computing your score until no pile
|
|
has more than one stone. Show your final score along with a record of the
|
|
numbers of stones in the piles you created with your moves.
|
|
|
|
Omitted.
|
|
|
|
b. Play $G$ again starting with $10$ stones, but use a different initial move
|
|
from the one in part (a). Show your final score along with a record of the
|
|
numbers of stones in the piles you created with your moves.
|
|
|
|
Omitted.
|
|
|
|
c. Show that you can use strong mathematical induction to prove that for every
|
|
integer $n \geq 1$, given the set-up of game $G$, no matter how you split the
|
|
piles in the various moves, your final score is $\dfrac{n(n - 1)}{2}$. The basis
|
|
step may look a little strange because a pile consisting of one stone cannot be
|
|
split into any sub-piles. Another way to say this is that it can only be split
|
|
into zero piles, and that gives an answer that agrees with the general formula
|
|
for the final score.
|
|
|
|
Omitted.
|
|
|
|
24. Imagine a situation in which eight people, numbered consecutively 1-8, are
|
|
arranged in a circle. Starting from person #1, every second person in the
|
|
circle is eliminated. The elimination process continues until only one
|
|
person remains. In the first round the people numbered $2$, $4$, $6$, and
|
|
$8$ are eliminated, in the second round the people numbered $3$ and $7$ are
|
|
eliminated, and in the third round person #5 is eliminated, so after the
|
|
third round only person #1 remains, as shown on the next page.
|
|
|
|
See page 336 for image.
|
|
|
|
a. Given a set of sixteen people arranged in a circle and numbered,
|
|
consecutively 1-16, list the numbers of the people who are eliminated in each
|
|
round if every second person is eliminated and the elimination process continues
|
|
until only one person remains. Assume that the starting point is person #1.
|
|
|
|
Omitted.
|
|
|
|
b. Use ordinary mathematical induction to prove that for every integer
|
|
$n \geq 1$, given any set of $2^n$ people arranged in a circle and numbered
|
|
consecutively $1$ through $2^n$, if one starts from person #1 and goes
|
|
repeatedly around the circle successively eliminating every second person,
|
|
eventually only person #1 will remain.
|
|
|
|
Omitted.
|
|
|
|
c. Use the result of part (b) to prove that for any nonnegative integers $n$ and
|
|
$m$ with $2^n \leq 2^n + m < 2^{n + 1}$, if $r = 2^n + m$, then given any set of
|
|
$r$ people arranged in a circle and numbered consecutively $1$ through $r$, if
|
|
one starts from person #1 and goes repeatedly around the circle successively
|
|
eliminating every second person, eventually only person #$(2m + 1)$ will remain.
|
|
|
|
Omitted.
|
|
|
|
25. Find the mistake in the following "proof" that purports to show that every
|
|
nonnegative integer power of every nonzero real number is $1$.
|
|
|
|
"**Proof:**
|
|
|
|
Let $r$ be any nonzero real number and let the property $P(n)$ be the equation
|
|
$r^n = 1$.
|
|
|
|
_Show that $P(0)$ is true:_
|
|
|
|
$P(0)$ is true because $r^0 = 1$ by definition of zeroth power.
|
|
|
|
_Show that for every integer $k \geq 0$, if $P(i)$ is true for each integer $i$
|
|
from $0$ through $k$, then $P(k + 1)$ is also true:_
|
|
|
|
Let $k$ be any integer $k \geq 0$ and suppose that $r^i = 1$ for each integer
|
|
$i$ from $0$ through $k$. This is the inductive hypothesis.
|
|
|
|
We must show that $r^{k + 1} = 1$. Now
|
|
|
|
$$ r^{k + 1} = r^{k + k - (k - 1)} $$
|
|
|
|
because $k + k - (k - 1) = k + k - k + 1 = k + 1$
|
|
|
|
$$ = \frac{r^k \cdot r^k}{r^{k - 1}} $$
|
|
|
|
by the laws of exponents
|
|
|
|
$$ = \frac{1 \cdot 1}{1} $$
|
|
|
|
by inductive hypothesis
|
|
|
|
$$ = 1 $$
|
|
|
|
Thus $r^{k + 1} = 1$ _[as was to be shown]._
|
|
|
|
_[Since we have proved both the basis and the inductive step of the strong
|
|
mathematical induction, we conclude that the given statement is true.]"_
|
|
|
|
Omitted.
|
|
|
|
26. Use the well-ordering principle for the integers to prove Theorem 4.4.4:
|
|
Every integer greater than $1$ is divisible by a prime number.
|
|
|
|
Omitted.
|
|
|
|
27. Use the well-ordering principle for the integers to prove the existence part
|
|
of the unique factorization of integers theorem. In other words, prove that
|
|
every integer greater than $1$ is either prime or a product of prime
|
|
numbers.
|
|
|
|
Omitted.
|
|
|
|
28.
|
|
|
|
a. The Archimedean property for the rational numbers states that for every
|
|
rational number $r$, there is an integer $n$ such that $n > r$. Prove this
|
|
property.
|
|
|
|
Omitted.
|
|
|
|
b. Prove that given any rational number $r$, the number $-r$ is also rational.
|
|
|
|
Omitted.
|
|
|
|
c. Use the results of parts (a) and (b) to prove that given any rational number
|
|
$r$, there is an integer $m$ such that $m < r$.
|
|
|
|
Omitted.
|
|
|
|
29. Use the results of exercise 28 and the well-ordering principle for the
|
|
integers to show that given any rational number $r$, there is an integer $m$
|
|
such that $m \leq r < m + 1$.
|
|
|
|
Omitted.
|
|
|
|
30. Use the well-ordering principle to prove that given any integer $n \geq 1$,
|
|
there exists an odd integer $m$ and a nonnegative integer $k$ such that
|
|
$n = 2^k \cdot m$.
|
|
|
|
Omitted.
|
|
|
|
31. Give examples to illustrate the proof of Theorem 5.4.1.
|
|
|
|
Omitted.
|
|
|
|
32. Suppose $P(n)$ is a property such that
|
|
|
|
1. $P(0)$, $P(1)$, $P(2)$ are all true,
|
|
|
|
Omitted.
|
|
|
|
2. for each integer $k \geq 0$, if $P(k)$ is true, then $P(3k)$ is true.
|
|
Must it follow that $P(n)$ is true for every integer $n \geq 0$? If yes,
|
|
explain why; if no, give a counterexample.
|
|
|
|
Omitted.
|
|
|
|
33. Prove that if a statement can be proved by strong mathematical induction,
|
|
then it can be proved by ordinary mathematical induction. To do this, let
|
|
$P(n)$ be a property that is defined for each integer $n$, and suppose the
|
|
following two statements are true:
|
|
|
|
1. $P(a), P(a + 1), P(a + 2) \dots, P(b)$.
|
|
|
|
Omitted.
|
|
|
|
2. For any integer $k \geq b$, if $P(i)$ is true for each integer $i$ from
|
|
$a$ through $k$, then $P(k + 1)$ is true.
|
|
|
|
Omitted.
|
|
|
|
The principle of strong mathematical induction would allow us to conclude
|
|
immediately that $P(n)$ is true for every integer $n \geq a$. Can we reach the
|
|
same conclusion using the principle of ordinary mathematical induction? Yes! To
|
|
see this, let $Q(n)$ be the property
|
|
|
|
$P(j)$ is true for each integer $j$ with $a \leq j \leq n$.
|
|
|
|
Then use ordinary mathematical induction to show that $Q(n)$ is true for every
|
|
integer $n \geq b$. That is, prove:
|
|
|
|
1. $Q(b)$ is true.
|
|
|
|
Omitted.
|
|
|
|
2. For each integer $k \geq b$, if $Q(k)$ is true then $Q(k + 1)$ is true.
|
|
|
|
Omitted.
|
|
|
|
34. It is a fact that every integer $n \geq 1$ can be written in the form
|
|
|
|
$$ c_r \cdot 3^r + c_{r - 1} \cdot 3^{r - 1} + \dots + c_2 \cdot 3^2 + c_1 \cdot 3 + c_0 $$
|
|
|
|
where $c_r = 1$ or $2$ and $c_i = 0, 1,$ or $2$ for each integer
|
|
$i = 0, 1, 2, \dots, r - 1$. Sketch a proof of this fact.
|
|
|
|
Omitted.
|
|
|
|
35. Use mathematical induction to prove the existence part of the
|
|
quotient-remainder theorem. In other words, use mathematical induction to
|
|
prove that given any integer $n$ and any positive integer $d$, there exists
|
|
integers $q$ and $r$ such that $n = dq + r$ and $0 \leq r < d$.
|
|
|
|
Omitted.
|
|
|
|
36. Prove that if a statement can be proved using ordinary mathematical
|
|
induction, then it can be proved by the well-ordering principle.
|
|
|
|
Omitted.
|
|
|
|
37. Use the principle of ordinary mathematical induction to prove the
|
|
well-ordering principle for the integers.
|
|
|
|
Omitted.
|
|
|
|
---
|
|
|
|
**Exercise Set 5.5**
|
|
|
|
Page 346
|
|
|
|
Exercises 1-5 contain a while loop and a predicate. In each case show that if
|
|
the predicate is true before entry to the loop, then it is also true after exit
|
|
from the loop.
|
|
|
|
1.
|
|
|
|
loop:
|
|
|
|
$\text{\textbf{while}} (m \geq 0 \text{ and } m \leq 100)\\ \ \ \ \ m := m + 1\\ \ \ \ \ n := n - 1\\ \text{\textbf{end while}}$
|
|
|
|
predicate: $m + n = 100$
|
|
|
|
**Proof:**
|
|
|
|
Suppose the predicate $m + n = 100$ is true before entry to the loop. Then
|
|
|
|
$$ m_{\text{old}} + n_{\text{old}} = 100 $$
|
|
|
|
After execution of the loop,
|
|
|
|
$$ m_{\text{new}} = m_{\text{old}} + 1 $$
|
|
|
|
and
|
|
|
|
$$ n_{\text{new}} = n_{\text{old}} - 1 $$
|
|
|
|
so
|
|
|
|
$$ m_{\text{new}} + n_{\text{new}} = (m_{\text{old}} + 1) + (n_{\text{old}} - 1) $$
|
|
|
|
$$ = m_{\text{old}} + n_{\text{old}} = 100 $$
|
|
|
|
Q.E.D.
|
|
|
|
2.
|
|
|
|
loop:
|
|
|
|
$\text{\textbf{while}} (m \geq 0 \text{ and } m \leq 100)\\ \ \ \ \ m := m + 4\\ \ \ \ \ n := n - 2\\ \text{\textbf{end while}}$
|
|
|
|
predicate: $m + n \text{ is odd}$
|
|
|
|
**Proof:**
|
|
|
|
Suppose the predicate $m + n \text{ is odd}$ is true before entry to the loop.
|
|
Then
|
|
|
|
$$ m_{\text{old}} + n_{\text{old}} \text{ is odd} $$
|
|
|
|
After execution of the loop,
|
|
|
|
$$ m_{\text{new}} = m_{\text{old}} + 4 $$
|
|
|
|
and
|
|
|
|
$$ n_{\text{new}} = n_{\text{old}} - 2 $$
|
|
|
|
so
|
|
|
|
$$ m_{\text{new}} + n_{\text{new}} = (m_{\text{old}} + 4) + (n_{\text{old}} - 2) $$
|
|
|
|
$$ = m_{\text{old}} + n_{\text{old}} + 2 $$
|
|
|
|
Since $m_{\text{old}} + n_{\text{old}} \text{ is odd}$, then:
|
|
|
|
$$ = 2k + 1 + 2 $$
|
|
|
|
for some integer $k$
|
|
|
|
$$ = 2k + 2 + 1 $$
|
|
|
|
$$ = 2(k + 2) + 1 $$
|
|
|
|
Now, $k + 2$ is an integer by the sum of integers. Therefore
|
|
$m_{\text{new}} + n_{\text{new}}$ is odd by the definition of odd.
|
|
|
|
Q.E.D.
|
|
|
|
3.
|
|
|
|
loop:
|
|
|
|
$\text{\textbf{while}} (m \geq 0 \text{ and } m \leq 100)\\ \ \ \ \ m := 3 \cdot m\\ \ \ \ \ n := 5 \cdot n\\ \text{\textbf{end while}}$
|
|
|
|
predicate: $m^3 > n^2$
|
|
|
|
**Proof:**
|
|
|
|
Suppose the predicate $m^3 > n^2$ is true before entry to the loop. Then
|
|
|
|
$$ (m_{\text{old}})^3 > (n_{\text{old}})^2 $$
|
|
|
|
After execution of the loop,
|
|
|
|
$$ m_{\text{new}} = 3 \cdot m_{\text{old}} $$
|
|
|
|
and
|
|
|
|
$$ n_{\text{new}} = 5 \cdot n_{\text{old}} $$
|
|
|
|
so
|
|
|
|
$$ (m_{\text{new}})^3 = (3m_{\text{old}})^3 = 27(m_{\text{old}})^3 > 27(n_{\text{old}})^2 $$
|
|
|
|
Now, since $n_{\text{new}} = 5 \cdot n_{\text{old}}$, it follows that
|
|
$\dfrac{1}{5}n_{\text{new}} = n_{\text{old}}$. Hence
|
|
|
|
$$ (m_{\text{new}})^3 > 27(n_{\text{old}})^2 = 27\left(\frac{1}{5}n_{\text{new}}\right)^2 = 27 \cdot \frac{1}{25}(n_{\text{new}})^2 $$
|
|
|
|
$$ = \frac{27}{25} \cdot (n_{\text{new}})^2 > (n_{\text{new}})^2 $$
|
|
|
|
4.
|
|
|
|
loop:
|
|
|
|
$\text{\textbf{while}} (n \geq 0 \text{ and } n \leq 100)\\ \ \ \ \ n := n + 1\\ \text{\textbf{end while}}$
|
|
|
|
predicate: $2^n < (n + 2)!$
|
|
|
|
**Proof:**
|
|
|
|
Suppose the predicate $2^n < (n + 2)!$ is true before entry to the loop. Then
|
|
|
|
$$ 2^{n_{\text{old}}} < (n_{\text{old}} + 2)! $$
|
|
|
|
After execution of the loop,
|
|
|
|
$$ n_{\text{new}} = n_{\text{old}} + 1 $$
|
|
|
|
so
|
|
|
|
$$ 2^{n_{\text{new}}} = 2^{n_{\text{old}} + 1} = 2 \cdot 2^{n_{\text{old}}} < 2(n_{\text{old}} + 2)! $$
|
|
|
|
Note that $2 \leq n_{\text{old}} + 3$
|
|
|
|
since the guard condition gives $n_{\text{old}} \geq 0$, then:
|
|
|
|
$$ 2(n_{\text{old}} + 2)! \leq (n_{\text{old}} + 3)(n_{\text{old}} + 2)! = (n_{\text{old}} + 3)! = ((n_{\text{old}} + 1) + 2)! = (n_{\text{new}} + 2)! $$
|
|
|
|
Combining these gives:
|
|
|
|
$$ 2^{n_{\text{new}}} = 2 \cdot 2^{n_{\text{old}}} < (n_{\text{old}} + 3)! = (n_{\text{new}} + 2)! $$
|
|
|
|
Q.E.D.
|
|
|
|
5.
|
|
|
|
loop:
|
|
|
|
$\text{\textbf{while}} (n \geq 3 \text{ and } n \leq 100)\\ \ \ \ \ n := n + 1\\ \text{\textbf{end while}}$
|
|
|
|
predicate: $2n + 1 \leq 2^n$
|
|
|
|
**Proof:**
|
|
|
|
Suppose the predicate $2n + 1 \leq 2^n$ is true before entry to the loop. Then
|
|
|
|
$$ 2n_{\text{old}} + 1 \leq 2^{n_{\text{old}}}$$
|
|
|
|
After execution of the loop,
|
|
|
|
$$ n_{\text{new}} = n_{\text{old}} + 1 $$
|
|
|
|
so
|
|
|
|
$$ 2n_{\text{new}} + 1 = 2(n_{\text{old}} + 1) + 1 = 2n_{\text{old}} + 3 $$
|
|
|
|
and
|
|
|
|
$$ 2^{n_{\text{new}}} = 2^{n_{\text{old}} + 1} = 2 \cdot 2^{n_{\text{old}}} $$
|
|
|
|
If we take the predicate and multiply both sides by $2$, we get:
|
|
|
|
$$ 2(2n_{\text{old}} + 1) \leq 2(2^{n_{\text{old}}}) $$
|
|
|
|
$$ 4n_{\text{old}} + 2 \leq 2^{n_{\text{old}} + 1} = 2^{n_{\text{new}}} $$
|
|
|
|
Notice that the new value for the left-hand value of the inequality is:
|
|
|
|
$$ 2n_{\text{new}} + 1 = 2n_{\text{old}} + 3 $$
|
|
|
|
And that this is less than the predicate's left hand side after multiplied by
|
|
two:
|
|
|
|
$$ 2n_{\text{new}} + 1 = 2n_{\text{old}} + 3 < 4n_{\text{old}} + 2 $$
|
|
|
|
And put together this is:
|
|
|
|
$$ 2n_{\text{new}} + 1 = 2n_{\text{old}} + 3 < 4n_{\text{old}} + 2 \leq 2^{n_{\text{old}} + 1} = 2^{n_{\text{new}}} $$
|
|
|
|
Q.E.D.
|
|
|
|
Exercises 6-9 each contain a while loop annotated with a pre-and a
|
|
post-condition and also a loop invariant. In each case, use the loop invariant
|
|
theorem to prove the correctness of the loop with respect to the pre-and
|
|
post-conditions.
|
|
|
|
6. _[Pre-condition: $m$ is a nonnegative integer, $x$ is a real number, $i = 0$,
|
|
and $\text{exp} = 1$.]_
|
|
|
|
$\text{\textbf{while}} (i \neq m)\\ \ \ \ \ 1. \text{exp} := \text{exp} \cdot x\\ \ \ \ \ 2. i := i + 1\\ \text{\textbf{end while}}$
|
|
|
|
_[Post-condition: $\text{exp} = x^m$]_
|
|
|
|
loop invariant: $I(n)$ is "$\text{exp} = x^n$ and $i = n$."
|
|
|
|
**I. Basis Property:** _[$I(0)$ is true before the first iteration of the
|
|
loop.]_
|
|
|
|
$I(0)$ is "$\text{exp} = x^0$ and $i = 0$." According to the pre-condition,
|
|
before the first iteration of the loop $\text{exp} = 1$ and $i = 0$. Since
|
|
$x^0 = 1$, $I(0)$ is evidently true.
|
|
|
|
**II. Inductive Property:** _[If $G \wedge I(k)$ is true before a loop iteration
|
|
(where $k \geq 0$), then $I(k + 1)$ is true after the loop iteration.]_
|
|
|
|
Suppose $k$ is any nonnegative integer such that $G \wedge I(k)$ is true before
|
|
an iteration of the loop. Then as execution reaches the top of the loop
|
|
$i \neq m$, $\text{exp} = x^k$, and $i = k$. Since $i \neq m$, the guard is
|
|
passed and statement 1 is executed. Now before the execution of statement 1,
|
|
|
|
$$ \text{exp}_{\text{old}} = x^k $$
|
|
|
|
so execution of statement 1 has the following effect:
|
|
|
|
$$ \text{exp}_{\text{new}} = \text{exp}_{\text{old}} \cdot x = x^k \cdot x = x^{k + 1} $$
|
|
|
|
Similarly, before statement 2 is executed,
|
|
|
|
$$ i_{\text{old}} = k $$
|
|
|
|
so after execution of statement 2,
|
|
|
|
$$ i_{\text{new}} = i_{\text{old}} + 1 = k + 1 $$
|
|
|
|
Hence after the loop iteration, the two statements $\text{exp} = x^{k + 1}$ and
|
|
$i = k + 1$ are true, and so $I(k + 1)$ is true.
|
|
|
|
**III. Eventual Falsity of Guard:** _[After a finite number of iterations of the
|
|
loop, $G$ becomes false.]_
|
|
|
|
The guard $G$ is the condition $i \neq m$, and $m$ is a nonnegative integer. By
|
|
I and II, it is known that_
|
|
|
|
for every integer $n \geq 0$, if the loop is iterated $n$ times, then
|
|
$\text{exp} = x^n$ and $i = n$.
|
|
|
|
So after $m$ iterations of the loop, $i = m$. Thus $G$ becomes false after $m$
|
|
iterations of the loop.
|
|
|
|
**IV. Correctness of the Post-Condition:** _[If $N$ is the least number of
|
|
iterations after which $G$ is false and $I(N)$ is true, then the value of the
|
|
algorithm variables will be as specified in the post-condition of the loop.]_
|
|
|
|
According to the post-condition, the value of $\text{exp}$ after execution of
|
|
the loop should be $x^m$. But when $G$ is false, $i = m$. And when $I(N)$ is
|
|
true, $i = N$ and $\text{exp} = x^N$. Since _both_ conditions ($G$ is false and
|
|
$I(N)$ is true) are satisfied, $m = i = N$ and $\text{exp} = x^m$, as required.
|
|
|
|
7. _[Pre-condition: $\text{largest} = A[1]$ and $i = 1$]_
|
|
|
|
$\text{\textbf{while}} (i \neq m)\\ \ \ \ \ 1. i := i + 1\\ \ \ \ \ 2. \text{\textbf{if}} A[i] > \text{largest \textbf{then } \text{largest}} := A[i]\\ \text{\textbf{end while}}$
|
|
|
|
_[Post-condition:
|
|
$\text{largest} = \text{maximum value of } A[1], A[2], \dots, A[m]$]_
|
|
|
|
loop invariant: $I(n)$ is
|
|
"$\text{largest} = \text{maximum value of } A[1], A[2], \dots, A[n + 1]$ and
|
|
$i = n + 1$."
|
|
|
|
**I. Basis Property:** _[$I(0)$ is true before the first iteration of the
|
|
loop.]_
|
|
|
|
$I(0)$ is "$\text{largest} = A[1]$ and $i = 1$." According to the pre-condition,
|
|
this statement is true.
|
|
|
|
**II. Inductive Property:** _[If $G \wedge I(k)$ is true before a loop iteration
|
|
(where $k \geq 0$), then $I(k + 1)$ is true after the loop iteration.]_
|
|
|
|
Suppose $k$ is any nonnegative integer such that $G \wedge I(k)$ is true before
|
|
an iteration of the loop. Then as execution reaches the top of the loop,
|
|
$i \neq m$, $\text{largest} = A[k + 1]$ and $i = k + 1$. Since $i \neq m$, the
|
|
guard is passed and statement 1 is executed. Now, before execution of statement
|
|
1:
|
|
|
|
$$ i_{\text{old}} = k + 1 $$
|
|
|
|
so after statement 1 is executed:
|
|
|
|
$$ i_{\text{new}} = i_{\text{old}} + 1 = k + 2 $$
|
|
|
|
Also, before statement 2 is executed:
|
|
|
|
$$ \text{largest}_{\text{old}} = \max(A[1], \dots, A[k + 1]) $$
|
|
|
|
so after statement 2 is executed:
|
|
|
|
$$
|
|
\text{largest}_{\text{new}} =
|
|
\begin{cases}
|
|
A[k + 2] & \text{if } A[k + 2] > \text{largest}_{\text{old}} \\
|
|
\text{largest}_{\text{old}} & \text{if } A[k + 2] \leq \text{largest}_{\text{old}}
|
|
\end{cases}
|
|
$$
|
|
|
|
Thus, after the loop iteration, $I(k + 1)$ is true.
|
|
|
|
**III. Eventual Falsity of Guard:** _[After a finite number of iterations of the
|
|
loop, $G$ becomes false.]_
|
|
|
|
The guard $G$ is the condition $i \neq m$. By I and II, it is known that for
|
|
every integer $n \geq 1$, after $n$ iterations of the loop, $I(n)$ is true.
|
|
Hence after $m - 1$ iterations of the loop, $i = m$ and $G$ is false.
|
|
|
|
**IV. Correctness of the Post-Condition:** _[If $N$ is the least number of
|
|
iterations after which $G$ is false and $I(N)$ is true, then the value of the
|
|
algorithm variables will be as specified in the post-condition of the loop.]_
|
|
|
|
Suppose that $N$ is the least number of iterations after which $G$ is false and
|
|
$I(N)$ is true. Then (since $G$ is false) $i = m$ and (since $I(N)$ is true)
|
|
$i = N + 1$ and $\text{largest} = \max(A[1], \dots A[N + 1])$. Putting these
|
|
together gives $m = N + 1$, and so $\text{largest} = \max(A[1], \dots A[m])$,
|
|
which is the post-condition.
|
|
|
|
Q.E.D.
|
|
|
|
8. _[Pre-condition: $\text{sum} = A[1]$ and $i = 1$]_
|
|
|
|
$\text{\textbf{while}} (i \neq m)\\ \ \ \ \ 1. i := i + 1\\ \ \ \ \ 2. \text{sum} := \text{sum} + A[i]\\ \text{\textbf{end while}}$
|
|
|
|
_[Post condition: $\text{sum} = A[1] + A[2] + \dots + A[m]$]_
|
|
|
|
loop invariant: $I(n)$ is "$i = n + 1$ and
|
|
$\text{sum} = A[1] + A[2] + \dots + A[n + 1]$."
|
|
|
|
**I. Basis Property:** _[$I(0)$ is true before the first iteration of the
|
|
loop.]_
|
|
|
|
$I(0)$ is "$i = 1$ and $\text{sum} = A[1]$." According to the pre-condition,
|
|
this statement is true.
|
|
|
|
**II. Inductive Property:** _[If $G \wedge I(k)$ is true before a loop iteration
|
|
(where $k \geq 0$), then $I(k + 1)$ is true after the loop iteration.]_
|
|
|
|
Suppose $k$ is a nonnegative integer such that $G \wedge I(k)$ is true before
|
|
the iteration of the loop. Then as execution reaches the top of the loop,
|
|
$i \neq m$, $i = k + 1$, and $\text{sum} = A[1] + A[2] + \dots + A[k + 1]$.
|
|
Since $i \neq m$, the guard is passed and statement 1 is executed. Now before
|
|
execution of statement 1, $i_{\text{old}} = k + 1$. So after execution of
|
|
statement 1, $i_{\text{new}} = i_{\text{old}} + 1 = (k + 1) + 1 = k + 2$. Also
|
|
before statement 2 is executed
|
|
$\text{sum}_{\text{old}} = A[1] + A[2] + \dotts + A[k + 1]$. Execution of
|
|
statement 2 adds $A[k + 2]$ to this sum, and so after statement 2 is executed,
|
|
$\text{sum}_{\text{new}} = A[1] + A[2] + \dots + A[k + 1] + A[k + 2]$. Thus
|
|
after the loop iteration, $I(k + 1)$ is true.
|
|
|
|
**III. Eventual Falsity of Guard:** _[After a finite number of iterations of the
|
|
loop, $G$ becomes false.]_
|
|
|
|
The guard is the condition $i \neq m$. By I and II, it is known that for every
|
|
integer $n \geq 1$, after $n$ iterations of the loop $I(n)$ is true. Hence,
|
|
after $m - 1$ iterations of the loop, $I(m)$ is true, which implies that $i = m$
|
|
and $G$ is false.
|
|
|
|
**IV. Correctness of the Post-Condition:** _[If $N$ is the least number of
|
|
iterations after which $G$ is false and $I(N)$ is true, then the value of the
|
|
algorithm variables will be as specified in the post-condition of the loop.]_
|
|
|
|
Suppose that $N$ is the least number of iterations after which $G$ is false and
|
|
$I(N)$ is true. Then (since $G$ is false) $i = m$ and (since $I(N)$ is true)
|
|
$i = N + 1$ and $\text{sum} = A[1] + A[2] + \dots + A[N + 1]$. Putting these
|
|
together gives $m = N + 1$, and so $\text{sum} = A[1] + A[2] + \dots A[m]$,
|
|
which is the post-condition.
|
|
|
|
Q.E.D.
|
|
|
|
9. _[Pre-condition: $a = A$ and $A$ is a positive integer.]_
|
|
|
|
$\text{\textbf{while}} (a > 0)\\ \ \ \ \ a := a - 2\\ \text{\textbf{end while}}$
|
|
|
|
_[Post-condition: $a = 0$ if $A$ is even and $a = -1$ if $A$ is odd.]_
|
|
|
|
loop invariant: $I(n)$ is "Both $a$ and $A$ are even integers or both are odd
|
|
integers and, in either case, $a \geq -1$."
|
|
|
|
**I. Basis Property:** _[$I(0)$ is true before the first iteration of the
|
|
loop.]_
|
|
|
|
$I(0)$ is "$a = A$ and $A \text{ is a positive integer}$." According to the
|
|
pre-condition, this statement is true.
|
|
|
|
**II. Inductive Property:** _[If $G \wedge I(k)$ is true before a loop iteration
|
|
(where $k \geq 0$), then $I(k + 1)$ is true after the loop iteration.]_
|
|
|
|
Suppose $k$ is a nonnegative integer such that $G \wedge I(k)$ is true before
|
|
the iteration of the loop. Then as execution reaches the top of the loop,
|
|
$a_{\text{old}} > 0$, $a_{\text{old}} \text{ has the same parity as } A$, and
|
|
$a_{\text{old}} \geq -1$.
|
|
|
|
Since $a_{\text{old}} > 0$, it follows that $a_{\text{old}} \geq 1$. The guard
|
|
condition allows the loop body to execute, and statement 1 is performed. This
|
|
results in:
|
|
|
|
$$ a_{\text{new}} = a_{\text{old}} - 1 $$
|
|
|
|
Thus $I(k + 1)$ is true.
|
|
|
|
**III. Eventual Falsity of Guard:** _[After a finite number of iterations of the
|
|
loop, $G$ becomes false.]_
|
|
|
|
The guard is the condition $a > 0$. By I and II, it is known that for every
|
|
iteration of the loop $a := a - 2$. Since the initial value of $a$ is $A$, this
|
|
means that the value for $a$ follows the following sequence:
|
|
|
|
$$ A, A - 2, A - 4, \dots $$
|
|
|
|
which eventually reaches a value at $a \leq 0$.
|
|
|
|
Hence, after a finite number of iterations of the loop, $a \leq 0$ and $G$ is
|
|
false.
|
|
|
|
**IV. Correctness of the Post-Condition:** _[If $N$ is the least number of
|
|
iterations after which $G$ is false and $I(N)$ is true, then the value of the
|
|
algorithm variables will be as specified in the post-condition of the loop.]_
|
|
|
|
Suppose that $N$ is the least number of iterations after which $G$ is false and
|
|
$I(N)$ is true. Then (since $G$ is false) $a \leq 0$ and (since $I(N)$ is true)
|
|
both $a$ and $A$ have the same parity and $a \geq -1$. This means that:
|
|
|
|
$$ -1 \leq a \leq 0 $$
|
|
|
|
Therefore, if $A$ is even, then $a = 0$, and if $A$ is odd, then $a = -1$. This
|
|
fulfills the post-condition.
|
|
|
|
10. Prove correctness of the **while** loop of Algorithm 4.10.3 (in exercise 27
|
|
of Exercise Set 4.10) with respect to the following pre- and
|
|
post-conditions:
|
|
|
|
_Pre-condition:_ $A$ and $B$ are positive integers, $a = A$, and $b = B$.
|
|
|
|
_Post-condition:_ One of $a$ or $b$ is zero and the other is nonzero. Whichever
|
|
is nonzero equals $\text{gcd}(A, B)$.
|
|
|
|
Use the loop invariant
|
|
|
|
$I(n)$
|
|
|
|
"(1) $a$ and $b$ are nonnegative integers with
|
|
$\text{gcd}(a, b) = \text{gcd}(A, B)$,
|
|
|
|
(2) at most one of $a$ and $b$ equals $0$,
|
|
|
|
(3) $0 \leq a + b \leq A + B - n$."
|
|
|
|
Omitted.
|
|
|
|
11. The following **while** loop implements a way to multiply two numbers that
|
|
was developed by the ancient Egyptians.
|
|
|
|
_[Pre-condition: $A$ and $B$ are positive integers, $x = A$, $y = B$, and
|
|
$\text{product} = 0$.]_
|
|
|
|
$\text{\textbf{while}} (y \neq 0)\\ \ \ \ \ r := y \mod 2\\ \ \ \ \ \text{\textbf{if }} r = 0\\ \ \ \ \ \ \ \ \ \text{\textbf{then do }}\\ \ \ \ \ \ \ \ \ \ \ \ \ x := 2 \cdot x\\ \ \ \ \ \ \ \ \ \ \ \ \ y := y \text{ div } 2\\ \ \ \ \ \ \ \ \ \text{\textbf{end do}}\\ \ \ \ \ \text{\textbf{if }} r = 1\\ \ \ \ \ \ \ \ \ \text{\textbf{then do }}\\ \ \ \ \ \ \ \ \ \ \ \ \ \text{product} := \text{product } + x\\ \ \ \ \ \ \ \ \ \ \ \ \ y := y - 1\\ \ \ \ \ \ \ \ \ \text{\textbf{end do}}\\ \text{\textbf{end while}}$
|
|
|
|
_[Post-condition: $\text{product } = A \cdot B$]_
|
|
|
|
a. Make a trace table to show that the algorithm gives the correct answer for
|
|
multiplying $A = 13 \text{ times } B = 18$.
|
|
|
|
Omitted.
|
|
|
|
b. Prove the correctness of this loop with respect to its pre-and
|
|
post-conditions by using the loop invariant
|
|
|
|
$I(n)$: "$xy + \text{ product} = A \cdot B$"
|
|
|
|
Omitted.
|
|
|
|
12. The following sentence could be added to the loop invariant for the
|
|
Euclidean algorithm:
|
|
|
|
There exist integers $u$, $v$, $s$, and $t$ such that $a = uA + vB$ and
|
|
$b = sA + tB$.
|
|
|
|
a. Show that this sentence is a loop invariant for
|
|
|
|
$\text{\textbf{while}} (b \neq 0)\\ \ \ \ \ r := a \mod b\\ \ \ \ \ a := b\\ \ \ \ \ b := r\\ \text{\textbf{end while}}$
|
|
|
|
Omitted.
|
|
|
|
b. Show that if initially $a = A$ and $b = B$, then sentence (5.5.12) is true
|
|
before the first iteration of the loop.
|
|
|
|
Omitted.
|
|
|
|
c. Explain how the correctness proof for the Euclidean algorithm together with
|
|
the results of (a) and (b) above allow you to conclude that given any integers
|
|
$A$ and $B$ with $A > B \geq 0$, there exist integers $u$ and $v$ so that
|
|
$\text{gcd}(A, B) = uA + vB$.
|
|
|
|
Omitted.
|
|
|
|
d. By actually calculating $u$, $v$, $s$, and $t$ at each stage of execution of
|
|
the Euclidean algorithm, find integers $u$ and $v$ so that
|
|
$\text{gcd}(330, 156) = 330u + 156v$.
|
|
|
|
Omitted.
|
|
|
|
---
|
|
|
|
**Exercise Set 5.6**
|
|
|
|
Page 360
|
|
|
|
Find the first four terms of each of the recursively defined sequences in 1-8.
|
|
|
|
1. $a_k = 2a_{k - 1} + k$, for every integer $k \geq 2$ $a_1 = 1$
|
|
|
|
$$
|
|
a_1 = 1 \\
|
|
a_2 = 2a_1 + 2 = 2(1) + 2 = 2 + 2 = 4 \\
|
|
a_3 = 2a_2 + 3 = 2(4) + 3 = 8 + 3 = 11 \\
|
|
a_4 = 2a_3 + 4 = 2(11) + 4 = 22 + 4 = 26
|
|
$$
|
|
|
|
2. $b_k = b_{k - 1} + 3k$, for every integer $k \geq 2$ $b_1 = 1$
|
|
|
|
$$
|
|
b_1 = 1 \\
|
|
b_2 = b_1 + 3(2) = 1 + 6 = 7 \\
|
|
b_3 = b_2 + 3(3) = 7 + 9 = 16 \\
|
|
b_4 = b_3 + 3(4) = 16 + 12 = 28
|
|
$$
|
|
|
|
3. $c_k = k(c_{k - 1})^2$, for every integer $k \geq 1$ $c_0 = 1$
|
|
|
|
$$
|
|
c_0 = 1 \\
|
|
c_1 = 1(c_0)^2 = 1(1)^2 = 1(1) = 1 \\
|
|
c_2 = 2(c_1)^2 = 2(1)^2 = 2(1) = 2 \\
|
|
c_3 = 3(c_2)^2 = 3(2)^2 = 3(4) = 12
|
|
$$
|
|
|
|
4. $d_k = k(d_{k - 1})^2$, for every integer $k \geq 1$ $d_0 = 3$
|
|
|
|
$$
|
|
d_0 = 3 \\
|
|
d_1 = 1(d_0)^2 = 1(3)^2 = 1(9) = 9 \\
|
|
d_2 = 2(d_1)^2 = 2(9)^2 = 2(81) = 162\\
|
|
d_3 = 3(d_2)^2 = 3(162)^2 = 3(26244) = 78732
|
|
$$
|
|
|
|
5. $s_k = s_{k - 1} + 2s_{k - 2}$, for every integer $k \geq 2$, $s_0 = 1$,
|
|
$s_1 = 1$
|
|
|
|
$$
|
|
s_0 = 1 \\
|
|
s_1 = 1 \\
|
|
s_2 = s_1 + 2s_0 = 1 + 2(1) = 1 + 2 = 3 \\
|
|
s_3 = s_2 + 2(s_1) = 3 + 2(1) = 3 + 2 = 5
|
|
$$
|
|
|
|
6. $t_k = t_{k - 1} + 2t_{k - 2}$, for every integer $k \geq 2$
|
|
$t_0 = -1, t_1 = 2$
|
|
|
|
$$
|
|
t_0 = -1 \\
|
|
t_1 = 2 \\
|
|
t_2 = t_1 + 2t_0 = 2 + 2(-1) = 2 - 2 = 0 \\
|
|
t_3 = t_2 + 2t_1 = 0 + 2(2) = 0 + 4 = 4
|
|
$$
|
|
|
|
7. $u_k = ku_{k - 1} - u_{k - 2}$, for every integer $k \geq 3$
|
|
$u_1 = 1, u_2 = 1$
|
|
|
|
$$
|
|
u_1 = 1 \\
|
|
u_2 = 1 \\
|
|
u_3 = 3(u_2) - u_1 = 3(1) - 1 = 3 - 1 = 2 \\
|
|
u_4 = 4(u_3) - u_2 = 4(2) - 1 = 8 - 1 = 7
|
|
$$
|
|
|
|
8. $v_k = v_{k - 1} + v_{k - 2} + 1$, for every integer $k \geq 3$
|
|
$v_1 = 1, v_2 = 3$
|
|
|
|
$$
|
|
v_1 = 1 \\
|
|
v_2 = 3 \\
|
|
v_3 = v_2 + v_1 + 1 = 3 + 1 + 1 = 5 \\
|
|
v_4 = v_3 + v_2 + 1 = 5 + 3 + 1 = 9
|
|
$$
|
|
|
|
9. Let $a_0, a_1, a_2, \dots$ be defined by the formula $a_n = 3n + 1$, for
|
|
every integer $n \geq 0$. Show that this sequence satisfies the recurrence
|
|
relation $a_k = a_{k - 1} + 3$, for every integer $k \geq 1$.
|
|
|
|
By definition of $a_0, a_1, a_2, \dots$ for each integer $k \geq 1$,
|
|
|
|
$$ \text{(1)} \quad a_k = 3k + 1 $$
|
|
|
|
and
|
|
|
|
$$ \text{(2)} \quad a_{k - 1} = 3(k - 1) + 1 $$
|
|
|
|
Then $a_{k - 1} + 3$:
|
|
|
|
$$ a_{k - 1} + 3 = (3(k - 1) + 1) + 3 \quad \text{ by substitution of (2)} $$
|
|
|
|
$$ = 3k - 3 + 1 + 3 $$
|
|
|
|
$$ = 3k + 1 \quad \text{ by basic algebra} $$
|
|
|
|
$$ = a_k \quad \text{ by substitution of (1)} $$
|
|
|
|
10. Let $b_0, b_1, b_2, \dots$ be defined by the formula $b_n = 4^n$, for every
|
|
integer $n \geq 0$. Show that this sequence satisfies the recurrence
|
|
relation $b_k = 4b_{k - 1}$, for every integer $k \geq 1$.
|
|
|
|
By definition of $b_0, b_1, b_2, \dots$ for each integer $k \geq 1$,
|
|
|
|
$$ \text{(1)} \quad b_k = 4^k $$
|
|
|
|
and
|
|
|
|
$$ \text{(2)} \quad b_{k - 1} = 4^{k - 1} $$
|
|
|
|
Then $4b_{k - 1}$:
|
|
|
|
$$ 4b_{k - 1} = 4(4^{k - 1}) \quad \text{ by substitution of (2)} $$
|
|
|
|
$$ = 4^k \quad \text{ by the laws of exponents} $$
|
|
|
|
$$ = b_k \quad \text{ by substitution of (1)} $$
|
|
|
|
11. Let $c_0, c_1, c_2, \dots$ be defined by the formula $c_n = 2^n - 1$ for
|
|
every integer $n \geq 0$. Show that this sequence satisfies the recurrence
|
|
relation $c_k = 2c_{k - 1} + 1$ for every integer $k \geq 1$.
|
|
|
|
By the definition of $c_0, c_1, c_2, \dots$ for each integer $k \geq 1$,
|
|
|
|
$$ \text{(1)} \quad c_k = 2^k - 1 $$
|
|
|
|
and
|
|
|
|
$$ \text{(2)} \quad c_{k - 1} = 2^{k - 1} - 1 $$
|
|
|
|
Then $2c_{k - 1} + 1$:
|
|
|
|
$$ 2c_{k - 1} + 1 = 2(2^{k - 1} - 1) + 1 \quad \text{ by substitution of (2)} $$
|
|
|
|
$$ = 2^k - 2 + 1 $$
|
|
|
|
$$ = 2^k - 1 $$
|
|
|
|
$$ = c_k \quad \text{ by substitution of (1)} $$
|
|
|
|
12. Let $s_0, s_1, s_2, \dots$ be defined by the formula
|
|
$s_n = \dfrac{(-1)^n}{n!}$ for every integer $n \geq 0$. Show that this
|
|
sequence satisfies the following recurrence relation for every integer
|
|
$k \geq 1$:
|
|
|
|
$$ s_k = \frac{-s_{k - 1}}{k} $$
|
|
|
|
By the definition of $s_0, s_1, s_2, \dots$ for each integer $k \geq 1$,
|
|
|
|
$$ \text{(1)} \quad s_k = \frac{(-1)^k}{k!} $$
|
|
|
|
and
|
|
|
|
$$ \text{(2)} \quad s_{k - 1} = \frac{(-1)^{k - 1}}{(k - 1)!} $$
|
|
|
|
Then $\dfrac{-s_{k - 1}}{k}$:
|
|
|
|
$$ \frac{-s_{k - 1}}{k} = \frac{-1\left(\dfrac{(-1)^{k - 1}}{(k - 1)!}\right)}{k} \quad \text{ by substitution of (2)} $$
|
|
|
|
$$ = \frac{\dfrac{(-1)^k}{(k - 1)!}}{k} $$
|
|
|
|
$$ = \frac{(-1)^k}{k(k - 1)!} $$
|
|
|
|
$$ = \frac{(-1)^k}{k!} $$
|
|
|
|
$$ = s_k \quad \text{ by substitution of (1)} $$
|
|
|
|
13. Let $t_0, t_1, t_2, \dots$ be defined by the formula $t_n = 2 + n$ for every
|
|
integer $n \geq 0$. Show that this sequence satisfies the following
|
|
recurrence relation for every integer $k \geq 2$:
|
|
|
|
$$ t_k = 2t_{k - 1} - t_{k - 2} $$
|
|
|
|
By the definition of $t_0, t_1, t_2, \dots $ for each integer $k \geq 2$,
|
|
|
|
$$ \text{(1)} \quad t_k = 2 + k $$
|
|
|
|
and
|
|
|
|
$$ \text{(2)} \quad t_{k - 1} = 2 + (k - 1) = 1 + k $$
|
|
|
|
and
|
|
|
|
$$ \text{(3)} \quad t_{k - 2} = 2 + (k - 2) = k $$
|
|
|
|
Then $2t_{k - 1} - t_{k - 2}$:
|
|
|
|
$$ 2t_{k - 1} - t_{k - 2} = 2(1 + k) - k \quad \text{by substitution of (2) and (3)} $$
|
|
|
|
$$ = 2 + 2k - k $$
|
|
|
|
$$ = 2 + k $$
|
|
|
|
$$ = t_k \quad \text{ by substitution of (1)} $$
|
|
|
|
14. Let $d_0, d_1, d_2, \dots$ be defined by the formula $d_n = 3^n - 2^n$ for
|
|
every integer $n \geq 0$. Show that this sequence satisfies the following
|
|
recurrence relation for every integer $k \geq 2$:
|
|
|
|
$$ d_k = 5d_{k - 1} - 6d_{k - 2} $$
|
|
|
|
By the definition of $d_0, d_1, d_2, \dots$ for each integer $k \geq 2$,
|
|
|
|
$$ \text{(1)} \quad d_k = 3^k - 2^k $$
|
|
|
|
and
|
|
|
|
$$ \text{(2)} \quad d_{k - 1} = 3^{k - 1} - 2^{k - 1} $$
|
|
|
|
and
|
|
|
|
$$ \text{(3)} \quad d_{k - 2} = 3^{k - 2} - 2^{k - 2} $$
|
|
|
|
Then $5d_{k - 1} - 6d_{k - 2}$:
|
|
|
|
$$ 5d_{k - 1} - 6d_{k - 2} = 5(3^{k - 1} - 2^{k - 1}) - 6(3^{k - 2} - 2^{k - 2}) \quad \text{ by substitution of (2) and (3)} $$
|
|
|
|
$$ = 5(3^{k - 1} - 2^{k - 1}) - 6(3^{k - 2} - 2^{k - 2}) $$
|
|
|
|
$$ = 5 \cdot 3^{k - 1} - 5 \cdot 2^{k - 1} - 6 \cdot 3^{k - 2} + 6 \cdot 2^{k - 2} $$
|
|
|
|
$$ = 5 \cdot 3^{k - 1} - 5 \cdot 2^{k - 1} - (3 \cdot 2) \cdot 3^{k - 2} + (3 \cdot 2) \cdot 2^{k - 2} $$
|
|
|
|
$$ = 5 \cdot 3^{k - 1} - 5 \cdot 2^{k - 1} - 2 \cdot 3^{k - 1} + 3 \cdot 2^{k - 1} $$
|
|
|
|
$$ = 5 \cdot 3^{k - 1} - 2 \cdot 3^{k - 1} - 5 \cdot 2^{k - 1} + 3 \cdot 2^{k - 1} $$
|
|
|
|
$$ = 3 \cdot 3^{k - 1} - 2 \cdot 2^{k - 1} $$
|
|
|
|
$$ = 3^k - 2^k $$
|
|
|
|
$$ = d_k \quad \text{ by substitution of (1)} $$
|
|
|
|
15. For the sequence of Catalan numbers defined in Example 5.6.4, prove that for
|
|
each integer $n \geq 1$,
|
|
|
|
$$ C_n = \frac{1}{4n + 2}\binom{2n + 2}{n + 1} $$
|
|
|
|
_Hint:_ Mathematical induction is not needed for the proof. Start with the
|
|
right-hand side of the equation and use algebra to transform it into the
|
|
left-hand side of the equation.
|
|
|
|
Recall that:
|
|
|
|
$$ C_n = \frac{1}{n + 1}\binom{2n}{n} $$
|
|
|
|
Then $\dfrac{1}{4n + 2}\binom{2n + 2}{n + 1}$:
|
|
|
|
$$ \dfrac{1}{4n + 2}\binom{2n + 2}{n + 1} = \frac{1}{2(2n + 1)}\left(\frac{(2n + 2)!}{(n + 1)!((2n + 2) - (n + 1))!}\right) \quad \text{ by definition of binomial} $$
|
|
|
|
$$ = \frac{1}{2(2n + 1)}\left(\frac{(2n + 2)!}{(n + 1)!(n + 1)!}\right) $$
|
|
|
|
$$ = \frac{1}{2(2n + 1)}\left(\frac{(2n + 2)(2n + 1)(2n)!}{(n + 1)(n!)(n + 1)(n!)}\right) \quad \text{ by definition of factorial} $$
|
|
|
|
$$ = \frac{1}{\cancel{2(2n + 1)}}\left(\frac{\cancel{2}(n + 1)\cancel{(2n + 1)}(2n)!}{(n + 1)(n!)(n + 1)(n!)}\right) $$
|
|
|
|
$$ = \frac{1}{1}\left(\frac{(n + 1)(2n)!}{(n + 1)(n!)(n + 1)(n!)}\right) $$
|
|
|
|
$$ = \frac{\cancel{(n + 1)}(2n!)}{\cancel{(n + 1)}(n!)(n + 1)(n!)} $$
|
|
|
|
$$ = \frac{(2n)!}{(n!)(n + 1)(n!)} $$
|
|
|
|
$$ = \frac{1}{n + 1}\left(\frac{(2n)!}{(n!)(n!)}\right) $$
|
|
|
|
$$ = \frac{1}{n + 1}\left(\frac{(2n)!}{(n!)(2n - n)!}\right) $$
|
|
|
|
$$ = \frac{1}{n + 1}\binom{2n}{n} \quad \text{ by definition of binomial} $$
|
|
|
|
$$ = C_n \quad \text{ by definition of Catalan} $$
|
|
|
|
16. Use the recurrence relation and values for the Tower of Hanoi sequence
|
|
$m_1, m_2, m_3, \dots$ discussed in Example 5.6.5 to compute $m_7$ and
|
|
$m_8$.
|
|
|
|
Recall that:
|
|
|
|
$$ m_k = 2m_{k - 1} + 1 \quad \text{ recurrence relation} $$
|
|
|
|
and
|
|
|
|
$$ m_1 = 1 \quad \text{ initial conditions} $$
|
|
|
|
In Example 5.6.5, we saw that:
|
|
|
|
$$ m_2 = 2m_1 + 1 = 2 \cdot 1 + 1 = 3 $$
|
|
|
|
$$ m_3 = 2m_2 + 1 = 2 \cdot 3 + 1 = 7 $$
|
|
|
|
$$ m_4 = 2m_3 + 1 = 2 \cdot 7 + 1 = 15 $$
|
|
|
|
$$ m_5 = 2m_4 + 1 = 2 \cdot 15 + 1 = 31 $$
|
|
|
|
$$ m_6 = 2m_5 + 1 = 2 \cdot 31 + 1 = 63 $$
|
|
|
|
Therefore, continuing the computations for $m_7$ and $m_8$:
|
|
|
|
$$ m_7 = 2m_6 + 1 = 2 \cdot 63 + 1 = 127 $$
|
|
|
|
$$ m_8 = 2m_7 + 1 = 2 \cdot 127 + 1 = 255 $$
|
|
|
|
17. _Tower of Hanoi with Adjacency Requirement:_
|
|
|
|
Suppose that in addition to the requirement that they never move a larger disk
|
|
on top of a smaller one, the priests who move the disks of the Tower of Hanoi
|
|
are also allowed only to move disks one by one from one pole to an _adjacent_
|
|
pole. Assume poles $A$ and $C$ are at the two ends of the row and pole $B$ is in
|
|
the middle. Let
|
|
|
|
$$ a_n = \left[\text{the minimum number of moves needed to transfer a tower of } n \text{ disks from pole } A \text{ to pole } C \right] $$
|
|
|
|
a. Find $a_1, a_2$, and $a_3$.
|
|
|
|
$$ a_1 = 2 $$
|
|
|
|
$$ a_2 = 2 \text{(moves to move the top disk from pole A to pole C)} $$
|
|
|
|
$$ +1 \text{(move to move the bottom disk from pole A to pole B)} $$
|
|
|
|
$$ +2 \text{(moves to move top disk from pole C to pole A)} $$
|
|
|
|
$$ +1 \text{(move to move the bottom disk from pole B to pole C)} $$
|
|
|
|
$$ +2 \text{(move to move top disk from pole A to pole C)} $$
|
|
|
|
$$ = 8 $$
|
|
|
|
$$ a_3 = 8 + 1 + 8 + 1 + 8 = 26 $$
|
|
|
|
b. Find $a_4$.
|
|
|
|
$$ a_4 = 26 + 1 + 26 + 1 + 26 = 80 $$
|
|
|
|
c. Find a recurrence relation for $a_1, a_2, a_3, \dots$. Justify your answer.
|
|
|
|
For every integer $k \geq 2$,
|
|
|
|
$$ a_k = a_{k - 1} \text{(moves to move the top } k - 1 \text{ disks from pole A to pole C)} $$
|
|
|
|
$$ +1 \text{move to move the bottom disk from pole A to pole B} $$
|
|
|
|
$$ +a_{k - 1} \text{(moves to move the top disk from pole C to pole A)} $$
|
|
|
|
$$ +1 \text{(move to move the bottom disks from pole B to pole C)} $$
|
|
|
|
$$ +a_{k - 1} \text{(moves to move the top disks from pole A to pole C)} $$
|
|
|
|
$$ = 3a_{k - 1} + 2 $$
|
|
|
|
18. _Tower of Hanoi with Adjacency Requirement:_
|
|
|
|
Suppose the same situation as in exercise 17. Let
|
|
|
|
$$ b_n = \left[\text{the minimum number of moves needed to transfer a tower of } n \text{ disks from pole } A \text{ to pole } B \right] $$
|
|
|
|
a. Find $b_1, b_2$, and $b_3$.
|
|
|
|
$$ b_1 = 1 $$
|
|
|
|
$$ b_2 = 4 $$
|
|
|
|
$$ b_3 = 13 $$
|
|
|
|
b. Find $b_4$.
|
|
|
|
$$ b_4 = 40 $$
|
|
|
|
c. Show that $b_k = a_{k - 1} + 1 + b_{k - 1}$ for each integer $k \geq 2$,
|
|
where $a_1, a_2, a_3, \dots$ is the sequence defined in exercise 17.
|
|
|
|
First move the top $k - 1$ disks from $A$ to $C$, which takes a minimum of
|
|
$a_{k - 1}$ moves.
|
|
|
|
Then move the remaining $k$th disk from $A$ to $B$, which takes a minimum of $1$
|
|
move.
|
|
|
|
Then move the $k - 1$ disks from $C$ to $B$, on top of the $k$th disk, which
|
|
takes a minimum of $b_{k - 1}$ moves. (Moving from $A$ to $B$ is the same as
|
|
moving from $C$ to $B$, the same number of moves).
|
|
|
|
These moves are minimal because, due to the adjacency requirement, the top
|
|
$k - 1$ disks (have to be) moved to $C$ first.
|
|
|
|
Therefore:
|
|
|
|
$$ b_k = a_{k - 1} + 1 + b_{k - 1} $$
|
|
|
|
d. Show that $b_k \leq 3b_{k - 1} + 1$ for each integer $k \geq 2$.
|
|
|
|
We need to show $a_{k - 1} \leq 2b_{k - 1}$ by part \(c\). This is true because
|
|
we can first move $k - 1$ disks from $A$ to $B$ which takes a minimum of
|
|
$b_{k - 1}$ moves, and then move them from $B$ to $C$, which takes a minimum of
|
|
another $b_{k - 1}$ moves. Doing this results in $k - 1$ disks being moved from
|
|
$A$ to $C$, which takes a minimum of $a_{k - 1}$ moves.
|
|
|
|
Therefore:
|
|
|
|
$$ a_{k - 1} \leq 2b_{k - 1} $$
|
|
|
|
e. Show that $b_k = 3b_{k - 1} + 1$ for each integer $k \geq 2$.
|
|
|
|
**Proof (by mathematical induction):**
|
|
|
|
Let $P(k)$ by the equation $b_k = 3b_{k - 1} + 1$.
|
|
|
|
_Basis Step:_
|
|
|
|
Prove $P(2)$. That is:
|
|
|
|
$$ b_2 = 3b_1 + 1 $$
|
|
|
|
$$ 4 = 3(1) + 1 \quad \text{ by substitution of part (a)} $$
|
|
|
|
$$ 4 = 4 $$
|
|
|
|
Therefore $P(2)$ is true.
|
|
|
|
_Inductive Step:_
|
|
|
|
Suppose $P(k)$ is true where $k$ is any integer such that $k \geq 2$. That is:
|
|
|
|
$$ b_{k} = 3b_{k - 1} + 1 $$
|
|
|
|
This is the inductive hypothesis.
|
|
|
|
Prove $P(k + 1)$. That is:
|
|
|
|
$$ b_{k + 1} = 3b_{(k + 1) - 1} + 1 = 3b_k + 1 $$
|
|
|
|
We know, by part \(c\), that:
|
|
|
|
$$ b_{k + 1} = a_k + 1 + b_k $$
|
|
|
|
And we know that $a_k \leq 2b_k$ by part (d).
|
|
|
|
We need to show $a_k \geq 2b_k$.
|
|
|
|
When moving $k$ disks from $A$ to $C$, consider the largest disk. Due to the
|
|
adjacency requirement, it has to move to $B$ first. So the top $k- 1$ disks must
|
|
have moved to $C$ before that. Then for the largest disk to finally move from
|
|
$B$ to $C$, the top $k - 1$ disks must have first moved from $C$ to $A$ to get
|
|
out of the way. In the same way, the top $k - 1$ disks, on their way from $C$
|
|
back to $B$, must have been moved to $B$ (on top of the largest disk) first,
|
|
before reaching $A$ (This shows that at some point all the disks are on the
|
|
middle pole.) This takes a minimum of $b_k$ moves. Then moving all the disks
|
|
from $B$ to $C$ takes a minimum of $b_k$ moves. Therefore $a_k \geq 2b_k$.
|
|
|
|
Thus:
|
|
|
|
$$ b_{k + 1} = a_k + 1 + b_k = 2b_k + 1 + b_k = 3b_k + 1 $$
|
|
|
|
Q.E.D.
|
|
|
|
19. _Four-Pole Tower of Hanoi:_
|
|
|
|
Suppose that the Tower of Hanoi problem has four poles in a row instead of
|
|
three. Disks can be transferred one by one from one pole to any other pole, but
|
|
at no time may a larger disk be placed on top of a smaller disk. Let $s_n$ be
|
|
the minimum number of moves needed to transfer the entire tower of $n$ disks
|
|
from the left-most to the right-most pole.
|
|
|
|
a. Find $s_1, s_2$, and $s_3$.
|
|
|
|
$$ s_1 = 1, s_2 = 1 + 1 + 1 = 3, s_3 = s_1 + (1 + 1 + 1) + s_1 = 5 $$
|
|
|
|
b. Find $s_4$.
|
|
|
|
$$ s_4 = s_2 + (1 + 1 + 1) + s_2 = 9 $$
|
|
|
|
c. Show that $s_k \leq 2s_{k - 2} + 3$ for every integer $k \geq 3$.
|
|
|
|
**Proof:**
|
|
|
|
Let's label the poles A-B-C-D, from left to right.
|
|
|
|
First notice that, since there is no adjacency requirement, the number of moves
|
|
to move A to D is equal to the number of moves from any pole to any other pole.
|
|
So, moving $k$ disks from A to, say, B, still takes $s_k$ moves.
|
|
|
|
First move the top $k - 2$ disks from A to B, in $s_{k - 2}$ moves. Then move
|
|
the second largest disk from A to C. Then move the largest dis to D. Then move
|
|
the second largest disk from C to D, on top of the largest. Finally, move
|
|
$k - 2$ disks from B to D. This takes $s_{k - 2} + 1 + 1 + 1 + s_{k - 2}$ moves.
|
|
|
|
This procedure gives us the minimum number of moves, because there is no
|
|
adjacency requirement and we are taking advantage of the free space in all 4
|
|
poles. (Notice that this is faster than moving the top $k - 1$ disks somewhere
|
|
else first, then moving the largest disk to D, then moving the $k - 2$ disks.
|
|
Similarly it's faster than moving $k - 3$ disks first, then moving the bottom 3,
|
|
since there are not enough empty poles to make that efficient.)
|
|
|
|
20. _Tower of Hanoi Poles in a Circle:_
|
|
|
|
Suppose that instead of being lined up in a row, the three poles for the
|
|
original Tower of Hanoi are placed in a circle. The monks move the disks one by
|
|
one from one pole to another, but they may only move disks one over in a
|
|
clockwise direction and they may never move a larger disk on top of a smaller
|
|
one. Let $c_n$ be the minimum number of moves needed to transfer a pile of $n$
|
|
disks from one pole to the next adjacent pole in the clockwise direction.
|
|
|
|
a. Justify the inequality $c_k \leq 4c_{k - 1} + 1$ for each integer $k \geq 2$.
|
|
|
|
**Proof:**
|
|
|
|
Label the poles A, B, C, in clockwise order $A \to B \to C \to A$.
|
|
|
|
To move $k$ disks from $A$ to $B$, first move the top $k - 1$ disks from $A$ to
|
|
$B$ (which takes $c_{k - 1}$), then from $B$ to $C$ (which takes $c_{k - 1}$),
|
|
then move the largest disk from $A$ to $B$ (which takes 1 move), then move the
|
|
$k - 1$ disks from $C$ to $A$ (which takes $c_{k - 1}$), then from $A$ to $B$ on
|
|
top of the largest disk (which takes $c_{k - 1}$).
|
|
|
|
So the total moves made are $4c_{k - 1} + 1$. This shows that moving $k$ disks
|
|
from $A$ to $B$ can be accomplished in $4c_{k - 1} + 1$ moves, so
|
|
$c_k \leq 4c_{k - 1} + 1$.
|
|
|
|
b. The expression $4c_{k - 1} + 1$ is not the minimum number of moves needed to
|
|
transfer a pile of $k$ disks from one pole to another. Explain, for example, why
|
|
$c_3 \neq 4c_2 + 1$.
|
|
|
|
**Proof:**
|
|
|
|
$$
|
|
c_2 = \\
|
|
1 \text{(move to transfer the top disk from A to B)} \\
|
|
+1 \text{(move to transfer the top disk from B to C)} \\
|
|
+1 \text{(move to transfer the bottom disk from A to B)} \\
|
|
+1 \text{(move to transfer the top disk from C to A)} \\
|
|
+1 \text{(move to transfer the top disk from A to B)} \\
|
|
$$
|
|
|
|
$$
|
|
c_3 = \\
|
|
1 \text{(move to transfer the top disk from A to B)} \\
|
|
+1 \text{(move to transfer the top disk from B to C)} \\
|
|
+1 \text{(move to transfer the middle disk from A to B)} \\
|
|
+1 \text{(move to transfer the top disk from C to A)} \\
|
|
+1 \text{(move to transfer the middle disk from B to C)} \\
|
|
+1 \text{(move to transfer the top disk from A to B)} \\
|
|
+1 \text{(move to transfer the top disk from B to C)} \\
|
|
$$
|
|
|
|
After these 7 steps have been completed, the bottom disk can be transferred from
|
|
A to B. At that point the top two disks are on C, and a modified version of the
|
|
initial seven steps can be used to transfer them from C to B. Thus the total
|
|
number of steps is $7 + 1 + 7 = 15$, and $15 < 21 = 4c_2 + 1$.
|
|
|
|
21. _Double Tower of Hanoi:_
|
|
|
|
In this variation of the Tower of Hanoi there are three poles in a row and $2n$
|
|
disks, two each of $n$ different sizes, where $n$ is any positive integer.
|
|
Initially one of the poles contains all the disks placed on top of each other in
|
|
pairs of decreasing size. Disks are transferred one by one from one pole to
|
|
another, but at no time may a larger disk be placed on top of a smaller disk.
|
|
However, a disk may be placed on top of one of the same size. Let $t_n$ be the
|
|
minimum number of moves needed to transfer a tower of $2n$ disks from one pole
|
|
to another.
|
|
|
|
a. Find $t_1$ and $t_2$.
|
|
|
|
Suppose the poles are labeled A, B, and C such that $A \to B \to C$.
|
|
|
|
Let $s_1 = \text{small disk 1}$, and $s_2 = \text{small disk 2}$.
|
|
|
|
$$
|
|
t_1 = \\
|
|
1 & s_1 \to B \\
|
|
+1 & s_2 \to B \\
|
|
= 2
|
|
$$
|
|
|
|
Let $m_1 = \text{medium disk 1}$, and $m_2 = \text{medium disk 2}$.
|
|
|
|
$$
|
|
t_2 =
|
|
1 & s_1 \to B \\
|
|
+1 & s_2 \to B \\
|
|
+1 & s_1 \to C \\
|
|
+1 & s_2 \to C \\
|
|
+1 & m_1 \to B \\
|
|
+1 & m_2 \to B \\
|
|
+1 & s_1 \to B \\
|
|
+1 & s_2 \to B \\
|
|
= 8
|
|
$$
|
|
|
|
b. Find $t_3$.
|
|
|
|
Let $l_1 = \text{large disk 1}$, and $l_2 = \text{large disk 2}$.
|
|
|
|
$$
|
|
t_3 =
|
|
1 & s_1 \to B \\
|
|
+1 & s_2 \to B \\
|
|
+1 & s_2 \to C \\
|
|
+1 & s_1 \to C \\
|
|
+1 & m_1 \to B \\
|
|
+1 & m_2 \to B \\
|
|
+1 & s_1 \to B \\
|
|
+1 & s_2 \to B \\
|
|
+1 & s_2 \to A \\
|
|
+1 & s_1 \to A \\
|
|
+1 & m_2 \to C \\
|
|
+1 & m_1 \to C \\
|
|
+1 & s_1 \to B \\
|
|
+1 & s_2 \to B \\
|
|
+1 & s_2 \to C \\
|
|
+1 & s_1 \to C \\
|
|
+1 & l_1 \to B \\
|
|
+1 & l_2 \to B \\
|
|
+1 & s_1 \to B \\
|
|
+1 & s_2 \to B \\
|
|
+1 & s_2 \to A \\
|
|
+1 & s_1 \to A \\
|
|
+1 & m_1 \to B \\
|
|
+1 & m_2 \to B \\
|
|
+1 & s_1 \to B \\
|
|
+1 & s_2 \to B \\
|
|
= 26
|
|
$$
|
|
|
|
c. Find a recurrence relation for $t_1, t_2, t_3, \dots$.
|
|
|
|
$$ t_1 = 2, t_2 = 8, t_3 = 26 $$
|
|
|
|
$$ t_n = 3t_{n - 1} + 2 \quad n \geq 2 $$
|
|
|
|
22. _Fibonacci Variation:_
|
|
|
|
A single pair of rabbits (male and female) is born at the beginning of a year.
|
|
Assume the following conditions (which are somewhat more realistic than
|
|
Fibonacci's):
|
|
|
|
(1) Rabbit pairs are not fertile during their first months of life but
|
|
thereafter give birth to four new male/female pairs at the end of every month.
|
|
|
|
(2) No rabbits die.
|
|
|
|
a. Let
|
|
$r_n = \text{ the number of pairs of rabbits alive at the end of month } n$, for
|
|
each integer $n \geq 1$, and let $r_0 = 1$. Find a recurrence relation for
|
|
$r_0, r_1, r_2, \dots$. Justify your answer.
|
|
|
|
This is similar to example 5.6.6, but instead of giving birth to one new pair,
|
|
each male/female pair of rabbits gives birth to two new pairs after the first
|
|
month of life.
|
|
|
|
**Proof:**
|
|
|
|
At $r_0 = 1$, as there is only 1 pair of rabbits and they are not yet fertile.
|
|
$r_1 = 1$, as they are no yet fertile until month 2. At month 2, this pair has
|
|
four pairs, resulting in five pairs of rabbits. $r_2 = 4 + 1 = 5$.
|
|
|
|
The four new pairs can only come from the fertile pairs, which become fertile at
|
|
$n - 2$ months, where $n \in \mathbb{Z}^+ \wedge n \geq 0$. The total of
|
|
infertile pairs can be calculated simply by looking at $r_{n - 1}$. Therefore
|
|
the total number of pairs of rabbits at $n$ months can be expressed by the
|
|
recurrence relation:
|
|
|
|
$$ r_n = 4(r_{n - 2}) + r_{n - 1} $$
|
|
|
|
b. Compute $r_0, r_1, r_2, r_3, r_4, r_5$, and $r_6$.
|
|
|
|
$$
|
|
r_0 = 1 \\
|
|
r_1 = 1 \\
|
|
r_2 = 4(1) + 1 = 5 \\
|
|
r_3 = 4(1) + 5 = 9 \\
|
|
r_4 = 4(5) + 9 = 29 \\
|
|
r_5 = 4(9) + 29 = 65 \\
|
|
r_6 = 4(29) + 65 = 181 \\
|
|
$$
|
|
|
|
c. How many rabbits will there be at the end of the year?
|
|
|
|
$$
|
|
r_0 = 1 \\
|
|
r_1 = 1 \\
|
|
r_2 = 4(1) + 1 = 5 \\
|
|
r_3 = 4(1) + 5 = 9 \\
|
|
r_4 = 4(5) + 9 = 29 \\
|
|
r_5 = 4(9) + 29 = 65 \\
|
|
r_6 = 4(29) + 65 = 181 \\
|
|
r_7 = 4(65) + 181 = 441 \\
|
|
r_8 = 4(181) + 441 = 1165 \\
|
|
r_9 = 4(441) + 1165 = 2929 \\
|
|
r_{10} = 4(1165) + 2929 = 7589 \\
|
|
r_{11} = 4(2929) + 7589 = 19305 \\
|
|
r_{12} = 4(7589) + 19305 = 49661 \\
|
|
$$
|
|
|
|
23. _Fibonacci Variation:_
|
|
|
|
A single pair of rabbits (male and female) is born at the beginning of a year.
|
|
Assume the following conditions:
|
|
|
|
(1) Rabbit pairs are not fertile during their first _two_ months of life but
|
|
thereafter give birth to three new male/female pairs at the end of every month.
|
|
|
|
(2) No rabbits die.
|
|
|
|
a. Let
|
|
$s_n = \text{ the number of pairs of rabbits alive at the end of month } n$, for
|
|
each integer $n \geq 1$, and let $s_0 = 1$. Find a recurrence relation for
|
|
$s_0, s_1, s_2, \dots$. Justify your answer.
|
|
|
|
**Proof:**
|
|
|
|
We are given that in the beginning, there is only a single pair of rabbits, so
|
|
$s_0 = 1$. Since the rabbits are not fertile for the first _two_ months of life,
|
|
this means that $s_1 = 1$ and $s_2 = 1$. Afterwards which the pair of rabbits is
|
|
fertile and will give birth to three pairs of rabbits. So $s_3 = 3s_0 + s_2$.
|
|
|
|
The amount of given rabbits at $n$ months would be $3$ times the rabbits that
|
|
are fertile, which are any rabbits that exist at $n - 3$ months ($s_{n - 3}$)
|
|
plus the amount of infertile rabbits, which is just $s_{n - 1}$ rabbits. This
|
|
gives the recurrence relation:
|
|
|
|
$$ s_n = 3s_{n - 3} + s_{n - 1} $$
|
|
|
|
b. Compute $s_0, s_1, s_2, s_3, s_4$, and $s_5$.
|
|
|
|
$$
|
|
s_0 = 1 \\
|
|
s_1 = 1 \\
|
|
s_2 = 1 \\
|
|
s_3 = 3(1) + (1) = 4 \\
|
|
s_4 = 3(1) + (4) = 7 \\
|
|
s_5 = 3(1) + (7) = 10 \\
|
|
$$
|
|
|
|
c. How many rabbits will there be at the end of the year?
|
|
|
|
$$
|
|
s_0 = 1 \\
|
|
s_1 = 1 \\
|
|
s_2 = 1 \\
|
|
s_3 = 3(1) + (1) = 4 \\
|
|
s_4 = 3(1) + (4) = 7 \\
|
|
s_5 = 3(1) + (7) = 10 \\
|
|
s_6 = 3(4) + (10) = 22 \\
|
|
s_7 = 3(7) + (22) = 43 \\
|
|
s_8 = 3(10) + (43) = 73 \\
|
|
s_9 = 3(22) + (73) = 139 \\
|
|
s_{10} = 3(43) + (139) = 268 \\
|
|
s_{11} = 3(73) + (268) = 487 \\
|
|
s_{12} = 3(139) + (487) = 904 \\
|
|
$$
|
|
|
|
In 24-34, $F_0, F_1, F_2, \dots$ is the Fibonacci sequence.
|
|
|
|
24. Use the recurrence relation and values for $F_0, F_1, F_2, \dots$ given in
|
|
Example 5.6.6 to compute $F_{13}$ and $F_{14}$.
|
|
|
|
The recurrence relation and values given from Example 5.6.6 are:
|
|
|
|
$$ F_k = F_{k - 1} + F{k - 2} \quad \text{ recurrence relation} $$
|
|
|
|
$$ F_0 = 1, F_1 = 1 \quad \text{ initial conditions} $$
|
|
|
|
Luckily, 5.6.6 also gives us $F_2$ through $F_{12}$, so now to calculate
|
|
$F_{13}$ and $F_{14}$:
|
|
|
|
$$
|
|
F_2 = F_1 + F_0 = 1 + 1 = 2 \\
|
|
F_3 = F_2 + F_1 = 2 + 1 = 3 \\
|
|
F_4 = F_3 + F_2 = 3 + 2 = 5 \\
|
|
F_5 = F_4 + F_3 = 5 + 3 = 8 \\
|
|
F_6 = F_5 + F_4 = 8 + 5 = 13 \\
|
|
F_7 = F_6 + F_5 = 13 + 8 = 21 \\
|
|
F_8 = F_7 + F_6 = 21 + 13 = 34 \\
|
|
F_9 = F_8 + F_7 = 34 + 21 = 55 \\
|
|
F_{10} = F_9 + F_8 = 55 + 34 = 89 \\
|
|
F_{11} = F_{10} + F_9 = 89 + 55 = 144 \\
|
|
F_{12} = F_{11} + F_{10} = 144 + 89 = 233 \\
|
|
F_{13} = F_{12} + F_{11} = 233 + 144 = 377 \\
|
|
F_{14} = F_{13} + F_{12} = 377 + 233 = 610 \\
|
|
$$
|
|
|
|
25. The Fibonacci sequence satisfies the recurrence relation
|
|
$F_k = F_{k - 1} + F_{k - 2}$, for every integer $k \geq 2$.
|
|
|
|
a. Explain why the following is true:
|
|
|
|
$$ F_{k + 1} = F_k + F_{k - 1} \text{ for each integer } k \geq 1 $$
|
|
|
|
Each term of the Fibonacci sequence beyond the second equals the sum of the
|
|
previous two. For any integer $k \geq 1$, the two terms previous to $F_{k + 1}$
|
|
are $F_k$ and $F_{k - 1}$. Hence for every integer $k \geq 1$,
|
|
$F_{k + 1} = F_k + F_{k - 1}$.
|
|
|
|
b. Write an equation expressing $F_{k + 2}$ in terms of $F_{k + 1}$ and $F_k$.
|
|
|
|
The Fibonacci sequence satisfies the recurrence relation:
|
|
|
|
$$ F_{k + 2} = F_{k + 1} + F_k $$
|
|
|
|
for each integer $k \geq 0$.
|
|
|
|
c. Write an equation expressing $F_{k + 3}$ in terms of $F_{k + 2}$ and
|
|
$F_{k + 1}$.
|
|
|
|
The Fibonacci sequence satisfies the recurrence relation:
|
|
|
|
$$ F_{k + 3} = F_{k + 2} + F_{k + 1} $$
|
|
|
|
for each integer $k \geq -1$.
|
|
|
|
26. Prove that $F_k = 3F_{k - 3} + 2F_{k - 4}$ for every integer $k \geq 4$.
|
|
|
|
**Proof:**
|
|
|
|
Since we are trying to express this in terms of $F_k$, we must look recursively
|
|
at the definitions of it's preceding two terms until we see them as expressions
|
|
of $F_{k - 3}$ and $F_{k - 4}$ instead of $F_{k - 1}$ and $F_{k - 2}$.
|
|
|
|
$$ F_k = F_{k - 1} + F_{k - 2} $$
|
|
|
|
$$ = (F_{k - 2} + F_{k - 3}) + (F_{k - 3} + F_{k - 4}) $$
|
|
|
|
$$ = ((F_{k - 3} + F_{k - 4}) + F_{k - 3}) + (F_{k - 3} + F_{k - 4}) $$
|
|
|
|
$$ = F_{k - 3} + F_{k - 4} + F_{k - 3} + F_{k - 3} + F_{k - 4} $$
|
|
|
|
$$ = 3F_{k - 3} + 2F_{k - 4} $$
|
|
|
|
27. Prove that $F_k^2 - F_{k - 1}^2 = F_kF_{k + 1} - F_{k - 1}F_{k + 1}$, for
|
|
every integer $k \geq 1$.
|
|
|
|
The standard Fibonacci sequence from 5.6.6 is:
|
|
|
|
$$ F_k = F_{k - 1} + F_{k - 2} $$
|
|
|
|
To find the given equation to be true, we must convert the left-hand side to the
|
|
right hand-side. Meaning we must express the given Fibonacci recurrence relation
|
|
in terms of $F_{k}$, $F_{k + 1}$, and $F_{k - 1}$.
|
|
|
|
$$ F_k^2 - F_{k - 1}^2 = (F_k - F_{k - 1})(F_k + F_{k - 1}) \quad \text{ by algebra of the difference between two squares} $$
|
|
|
|
$$ = (F_k - F_{k - 1})F_{k + 1} \quad \text{ by the definition of the Fibonacci sequence} $$
|
|
|
|
$$ = F_kF_{k + 1} - F_{k - 1}F_{k + 1} \quad \text{ by distribution} $$
|
|
|
|
28. Prove that $F_{k + 1}^2 - F_k^2 - F_{k - 1}^2 = 2F_kF_{k - 1}$, for each
|
|
integer $k \geq 1$.
|
|
|
|
$$ F_{k + 1} = F_k + F_{k - 1} $$
|
|
|
|
$$ F_{k + 1}^2 = (F_k + F_{k - 1})^2 $$
|
|
|
|
$$ = (F_k + F_{k - 1})(F_k + F_{k - 1}) $$
|
|
|
|
$$ = F_k^2 + 2F_{k}F_{k - 1} + F_{k - 1}^2 $$
|
|
|
|
$$ (F_{k + 1}^2) - F_k^2 - F_{k - 1}^2 = (F_k^2 + 2F_{k}F_{k - 1} + F_{k - 1}^2) - F_k^2 - F_{k - 1}^2 $$
|
|
|
|
$$ = 2F_kF_{k - 1} $$
|
|
|
|
29. Prove that $F_{k + 1}^2 - F_k^2 = F_{k - 1}F_{k + 2}$, for every integer
|
|
$k \geq 1$.
|
|
|
|
$$ F_{k + 1} = F_k + F_{k - 1} $$
|
|
|
|
$$ F_{k + 1}^2 = (F_k + F_{k - 1})^2 $$
|
|
|
|
$$ = (F_k + F_{k - 1})(F_k + F_{k - 1}) $$
|
|
|
|
$$ = F_k^2 + 2F_kF_{k - 1} + F_{k - 1}^2 $$
|
|
|
|
$$ (F_{k + 1}^2) - F_k^2 = (F_k^2 + 2F_kF_{k - 1} + F_{k - 1}^2 ) - F_k^2 $$
|
|
|
|
$$ = 2F_kF_{k - 1} + F_{k - 1}^2 $$
|
|
|
|
$$ = F_{k - 1}(2F_k + F_{k - 1}) $$
|
|
|
|
$$ = F_{k - 1}(F_{k - 1} + F_k + F_k) $$
|
|
|
|
$$ = F_{k - 1}((F_k + F_{k - 1}) + F_k) $$
|
|
|
|
$$ = F_{k - 1}((F_{k + 1}) + F_k) $$
|
|
|
|
$$ = F_{k - 1}(F_{k + 1} + F_k) $$
|
|
|
|
$$ = F_{k - 1}(F_{k + 2}) $$
|
|
|
|
$$ = F_{k - 1}F_{k + 2} $$
|
|
|
|
30. Use mathematical induction to prove that for each integer $n \geq 0$,
|
|
$F_{n + 2}F_n - F_{n + 1}^2 = (-1)^n$.
|
|
|
|
**Proof (by mathematical induction):**
|
|
|
|
Let $P(n)$ be the equation $F_{n + 2}F_n - F_{n + 1}^2 = (-1)^n$ for each
|
|
integer $n \geq 0$.
|
|
|
|
_Basis Step:_
|
|
|
|
Prove $P(0)$. That is:
|
|
|
|
$$ F_{0 + 2}F_0 - F_{0 + 1}^2 = (-1)^0 $$
|
|
|
|
$$ F_{2}F_0 - F_{1}^2 = 1 $$
|
|
|
|
$$ (2)(1) - (1)^2 = 1 $$
|
|
|
|
$$ 2 - 1 = 1 $$
|
|
|
|
$$ 1 = 1 $$
|
|
|
|
Therefore $P(0)$ is true.
|
|
|
|
_Inductive Step:_
|
|
|
|
Suppose $P(k)$ for any integer $k \geq 0$. That is:
|
|
|
|
$$ F_{k + 2}F_k - F_{k + 1}^2 = (-1)^k $$
|
|
|
|
This is the inductive hypothesis.
|
|
|
|
Note that we might need the inductive hypothesis in this form:
|
|
|
|
$$ F_{k + 1}^2 = F_{k + 2}F_k - (-1)^k $$
|
|
|
|
Prove $P(k + 1)$, that is:
|
|
|
|
$$ F_{(k + 1) + 2}F_{k + 1} - F_{(k + 1) + 1}^2 = (-1)^{k + 1} $$
|
|
|
|
Alternatively:
|
|
|
|
$$ F_{k + 3}F_{k + 1} - F_{k + 2}^2 = (-1)^{k + 1} $$
|
|
|
|
Let's evaluate the left-hand side of this equality:
|
|
|
|
$$ F_{k + 3}F_{k + 1} - F_{k + 2}^2 $$
|
|
|
|
$$ = (F_{k + 2} + F_{k + 1})F_{k + 1} - F_{k + 2}^2 $$
|
|
|
|
$$ = F_{k + 2}F_{k + 1} + (F_{k + 1}^2) - F_{k + 2}^2 $$
|
|
|
|
By the inductive hypothesis, we can substitute thus:
|
|
|
|
$$ = F_{k + 2}F_{k + 1} + (F_{k + 2}F_k - (-1)^k) - F_{k + 2}^2 $$
|
|
|
|
$$ = F_{k + 1}(F_{k + 1} + F_k - F_{k + 2}) - (-1)^k $$
|
|
|
|
$$ = F_{k + 1}((F_{k + 1} + F_k) - F_{k + 2}) - (-1)^k $$
|
|
|
|
$$ = F_{k + 1}((F_{k + 2}) - F_{k + 2}) - (-1)^k $$
|
|
|
|
$$ = F_{k + 1}(F_{k + 2} - F_{k + 2}) - (-1)^k $$
|
|
|
|
$$ = F_{k + 1}(0) - (-1)^k $$
|
|
|
|
$$ = -(-1)^k $$
|
|
|
|
$$ = (-1) \cdot (-1)^k $$
|
|
|
|
$$ = (-1)^{k + 1} $$
|
|
|
|
Q.E.D.
|
|
|
|
31. Use strong mathematical induction to prove that $F_n < 2^n$ for every
|
|
integer $n \geq 1$.
|
|
|
|
Omitted.
|
|
|
|
32. Prove that for each integer $n \geq 0$, $\text{gcd}(F_{n + 1}, F_n) = 1$.
|
|
(The definition of $\text{gcd}$ is given in Section 4.10.)
|
|
|
|
Omitted.
|
|
|
|
33. It turns out that the Fibonacci sequence satisfies the following explicit
|
|
formula: For every integer $F_n \geq 0$,
|
|
|
|
$$ F_n = \frac{1}{\sqrt{5}}\left[\left(\frac{1 + \sqrt{5}}{2}\right)^{n + 1} - \left(\frac{1 - \sqrt{5}}{2}\right)^{n + 1}\right] $$
|
|
|
|
Verify that the sequence defined by this formula satisfies the recurrence
|
|
relation $F_k = F_{k - 1} + F_{k - 2}$ for every integer $k \geq 2$.
|
|
|
|
**Proof:**
|
|
|
|
Let $x = \left(\dfrac{1 + \sqrt{5}}{2}\right)$ and
|
|
$y = \left(\dfrac{1 - \sqrt{5}}{2}\right)$.
|
|
|
|
Note that:
|
|
|
|
$$ x^2 = \left(\frac{1 + \sqrt{5}}{2}\right)^2 $$
|
|
|
|
$$ = \frac{(1 + \sqrt{5})(1 + \sqrt{5})}{4} $$
|
|
|
|
$$ = \frac{1 + 2\sqrt{5} + 5 }{4} $$
|
|
|
|
$$ = \frac{6 + 2\sqrt{5}}{4} $$
|
|
|
|
Similarly, note that:
|
|
|
|
$$ y^2 = \left(\frac{1 - \sqrt{5}}{2}\right)^2 $$
|
|
|
|
$$ y^2 = \frac{(1 - \sqrt{5})(1 - \sqrt{5})}{4} $$
|
|
|
|
$$ y^2 = \frac{1 - 2\sqrt{5} + 5}{4} $$
|
|
|
|
$$ y^2 = \frac{6 - 2\sqrt{5}}{4} $$
|
|
|
|
Also notice that:
|
|
|
|
$$ x + 1 = \left(\frac{1 + \sqrt{5}}{2}\right) + 1 $$
|
|
|
|
$$ x + 1 = \frac{1 + \sqrt{5}}{2} + \frac{2}{2} $$
|
|
|
|
$$ x + 1 = \frac{3 + \sqrt{5}}{2} $$
|
|
|
|
$$ x + 1 = \left(\frac{3 + \sqrt{5}}{2}\right)\left(\frac{2}{2}\right) $$
|
|
|
|
$$ x + 1 = \frac{6 + 2\sqrt{5}}{4} $$
|
|
|
|
$$ x + 1 = \frac{6 + 2\sqrt{5}}{4} = x^2 $$
|
|
|
|
Similarly:
|
|
|
|
$$ y + 1 = \left(\frac{1 - \sqrt{5}}{2}\right) + 1 $$
|
|
|
|
$$ = \left(\frac{1 - \sqrt{5}}{2}\right) + \frac{2}{2} $$
|
|
|
|
$$ = \frac{3 - \sqrt{5}}{2} $$
|
|
|
|
$$ = \left(\frac{3 - \sqrt{5}}{2}\right)\left(\frac{2}{2}\right) $$
|
|
|
|
$$ = \frac{6 - 2\sqrt{5}}{4} = y^2 $$
|
|
|
|
Suppose $k \in \mathbb{Z}$ and $k \geq 2$.
|
|
|
|
We are trying to prove that:
|
|
|
|
$$ \frac{1}{\sqrt{5}}[x^k - y^k] = \frac{1}{\sqrt{5}}(x^{k - 1} - y^{k - 1}) + \frac{1}{\sqrt{5}}(x^{k - 2} - y^{k - 2}) $$
|
|
|
|
Since we know that $x^2 = x + 1$ and $y^2 = y + 1$, it follows that:
|
|
|
|
$$ x^k - y^k $$
|
|
|
|
$$ = x^2x^{k - 2} - y^2y^{k - 2} $$
|
|
|
|
$$ = (x + 1)x^{k - 2} - (y + 1)y^{k - 2} $$
|
|
|
|
$$ = ((x \cdot x^{k - 2}) + (1 \cdot x^{k - 2})) - ((y \cdot y^{k - 2}) + (1 \cdot y^{k - 2})) $$
|
|
|
|
$$ = x^{k - 1} + x^{k - 2} - y^{k - 1} - y^{k - 2} $$
|
|
|
|
$$ = x^{k - 1} - y^{k - 1} + x^{k - 2} - y^{k - 2} $$
|
|
|
|
Q.E.D.
|
|
|
|
34. (For students who have studied calculus) Find
|
|
$\lim\limits_{n \to \infty}\left(\dfrac{F_{n + 1}}{F_n}\right)$, assuming
|
|
that the limit exists.
|
|
|
|
Omitted.
|
|
|
|
35. (For students who have studied calculus) Prove that
|
|
$\lim\limits_{n \to \infty}\left(\dfrac{F_{n + 1}}{F_n}\right)$ exists.
|
|
|
|
Omitted.
|
|
|
|
36. (For students who have studied calculus) Define $x_0, x_1, x_2, \dots$ as
|
|
follows:
|
|
|
|
$$ x_k = \sqrt{2 + x_{k - 1}} \quad \text{ for each integer } k \geq 1 $$
|
|
|
|
$$ x_0 = 0 $$
|
|
|
|
Find $\lim\limits_{n \to \infty}x_n$. (Assume that the limit exists.)
|
|
|
|
Omitted.
|
|
|
|
37. _Compound Interest:_
|
|
|
|
Suppose a certain amount of money is deposited in an account paying 4% annual
|
|
interest compounded quarterly. For each positive integer $n$, let
|
|
$R_n = \text{ the amount on deposit at the end of the }$ $n$<sup>th</sup>
|
|
quarter, assuming no additional deposits or withdrawals, and let $R_0$ be the
|
|
initial amount deposited.
|
|
|
|
a. Find a recurrence relation for $R_0, R_1, R_2, \dots$. Justify your answer.
|
|
|
|
Since the account pays 4% annual interest compounded quarterly, the total
|
|
interest is $\left(\frac{0.04}{4}\right) = 0.01$ or 1%.
|
|
|
|
Let $k \in \mathbb{Z}$ such that $k \geq 0$. The recurrence relation can be
|
|
expressed as:
|
|
|
|
$$ R_k = R_{k - 1}+ 0.01(R_{k - 1}) = 1.01R_{k - 1} $$
|
|
|
|
b. If $R_0 = \$5,000$, find the am,ount of money on deposit at the end of one
|
|
year.
|
|
|
|
$$
|
|
R_0 = 5000 \\
|
|
R_1 = 1.01(5000) = 5050 \\
|
|
R_2 = 1.01(5050) = 5100.5 \\
|
|
R_3 = 1.01(5100.5) \approx 5151.51 \\
|
|
R_4 = 1.01(5151.51) \approx 5203.03 \\
|
|
$$
|
|
|
|
c. Find the APY for the account.
|
|
|
|
$$ \frac{5203.03 - 5000}{5000} = 0.040606 \text{ or } 4.0606\% $$
|
|
|
|
38. _Compound Interest:_
|
|
|
|
Suppose a certain amount of money is deposited in an account paying 3% annual
|
|
interest compounded monthly. For each positive integer $n$, let
|
|
$S_n = \text{ the amount on deposit at the end of the }$ $n$<sup>th</sup>
|
|
month, and let $S_0$ be the initial amount deposited.
|
|
|
|
a. Find a recurrence relation for $S_0, S_1, S_2, \dots$, assuming no additional
|
|
deposits or withdrawals during the year. Justify your answer.
|
|
|
|
Since the account pays 3% annual interest compounded monthly, the total interest
|
|
is $\left(\frac{0.03}{12}\right) = 0.0025$ or 0.25%.
|
|
|
|
Let $k \in \mathbb{Z}$ such that $k \geq 0$. The recurrence relation can be
|
|
expressed as:
|
|
|
|
$$ S_k = S_{k - 1}+ 0.0025(S_{k - 1}) = 1.0025S_{k - 1} $$
|
|
|
|
b. If $S_0 = \$10,000$, find the amount of money on deposit at the end of one
|
|
year.
|
|
|
|
$$
|
|
S_0 = 10000 \\
|
|
S_1 = 1.0025(10000) = 10025 \\
|
|
S_2 = 1.0025(10025) \approx 10050.06 \\
|
|
S_3 = 1.0025(10050.06) \approx 10075.19 \\
|
|
S_4 = 1.0025(10075.19) \approx 10100.38 \\
|
|
S_5 = 1.0025(10100.38) \approx 10125.63 \\
|
|
S_6 = 1.0025(10125.63) \approx 10150.94 \\
|
|
S_7 = 1.0025(10150.94) \approx 10176.32 \\
|
|
S_8 = 1.0025(10176.32) \approx 10201.76 \\
|
|
S_9 = 1.0025(10201.76) \approx 10227.26 \\
|
|
S_{10} = 1.0025(10227.26) \approx 10252.83 \\
|
|
S_{11} = 1.0025(10252.83) \approx 10278.46 \\
|
|
S_{12} = 1.0025(10278.46) \approx 10304.16 \\
|
|
$$
|
|
|
|
c. Find the APY for the account.
|
|
|
|
$$ \frac{10304.16 - 10000}{10000} = 0.030416 \text{ or } 3.0416\% $$
|
|
|
|
39. With each step you take when climbing a staircase, you can move up either
|
|
one stair or two stairs. As a result, you can climb the entire staircase
|
|
taking one stair at a time, taking two at a time, or taking a combination of
|
|
one-and two-stair increments. For each integer $n \geq 1$, if the staircase
|
|
conssits of $n$ stairs, let $c_n$ be the number of different ways to climb
|
|
the staircase. Find a recurrence relation for $c_1, c_2, c_3, \dots$.
|
|
Justify your answer.
|
|
|
|
Since $c_1 = 1$ and $c_2 = 2$, we know that if one climbs to the end of the
|
|
staircase and there is one step left, then that is $n - 1$ stairs climbed. If
|
|
there are two steps left, then that is $n - 2$ steps climbed. Therefore the
|
|
recurrence relation can be expressed as:
|
|
|
|
$$ c_n = c_{n - 1} + c_{n - 2} $$
|
|
|
|
40. A set of blocks contains blocks of heights $1$, $2$, and $4$ centimeters.
|
|
Imagine constructing towers by piling blocks of different heights directly
|
|
on top of one another. (A tower of height $6$ cm could be obtained using six
|
|
$1$-cm blocks, three $2$-cm blocks one $2$-cm block with one $4$-cm block on
|
|
top, one $4$-cm block with one $2$-cm block on top, and so forth.) Let $t_n$
|
|
be the number of ways to construct a tower of height $n$ cm using blocks
|
|
from the set. (Assume an unlimited supply of blocks of each size.) Find a
|
|
recurrence relation for $t_1, t_2, t_3, \dots$. Justify your answer.
|
|
|
|
Let's establish some initial conditions:
|
|
|
|
$$
|
|
t_1 = 1 \text{ 1 1cm block} \\
|
|
t_2 = 2 \text{ 2 1cm blocks or 1 2cm block} \\
|
|
t_3 = 3 \text{ 3 1cm blocks, 1 1cm block and 1 2cm block, or 1 2cm block and 1 1cm block} \\
|
|
$$
|
|
|
|
The recurrence relation for $n$ cm blocks then is:
|
|
|
|
$$ t_n = t_{n - 1} + t_{n - 2} + t_{n - 4} $$
|
|
|
|
41. Assume the truth of the distributive law (Appendix A, F3), and use the
|
|
recursive definition of summation, together with mathematical induction, to
|
|
prove the generalized distributive law that for every positive integer $n$,
|
|
if $a_1, a_2, \dots, a_n$ and $c$ are real numbers, then
|
|
|
|
$$ \sum_{i = 1}^{n}{ca_i} = c\left(\sum_{i = 1}^{n}{a_i}\right) $$
|
|
|
|
For reference the distributive law states:
|
|
|
|
For all real numbers $a$, $b$, and $c$A
|
|
|
|
$$ a(b + c) = ab + ac \quad \text{ and } \quad (b + c)a = ba + ca $$
|
|
|
|
**Proof (by mathematical induction):**
|
|
|
|
Let $P(n)$ be the equality:
|
|
|
|
$$ \sum_{i = 1}^{n}{ca_i} = c\left(\sum_{i = 1}^{n}{a_i}\right) $$
|
|
|
|
_Basis Step:_
|
|
|
|
Prove $P(1)$, that is:
|
|
|
|
$$ \sum_{i = 1}^{1}{ca_i} = c\left(\sum_{i = 1}^{1}{a_i}\right) $$
|
|
|
|
Evaluating the left-hand side:
|
|
|
|
$$ \sum_{i = 1}^{1}{ca_i} $$
|
|
|
|
$$ = ca_1 $$
|
|
|
|
Evaluating the right-hand side:
|
|
|
|
$$ c\left(\sum_{i = 1}^{1}{a_i}\right) $$
|
|
|
|
$$ = ca_1 $$
|
|
|
|
Therefore, since the left-hand and right-hand sides of the equality hold, $P(1)$
|
|
is true.
|
|
|
|
_Inductive Step:_
|
|
|
|
Let $k \in \mathbb{Z}$ such that $k \geq 1$.
|
|
|
|
Suppose $P(k)$, that is:
|
|
|
|
$$ \sum_{i = 1}^{k}{ca_i} = c\left(\sum_{i = 1}^{k}{a_i}\right) $$
|
|
|
|
This is the inductive hypothesis.
|
|
|
|
Prove $P(k + 1)$. That is:
|
|
|
|
$$ \sum_{i = 1}^{k + 1}{ca_i} = c\left(\sum_{i = 1}^{k + 1}{a_i}\right) $$
|
|
|
|
By the recursive definition of summation:
|
|
|
|
$$ \sum_{i = 1}^{k + 1}{ca_i} = \left(\sum_{i = 1}^{k}{ca_i}\right) + ca_{k + 1} $$
|
|
|
|
Then by the inductive hypothesis, we can substitute the first term:
|
|
|
|
$$ = c\left(\sum_{i = 1}^{k}{a_i}\right) + ca_{k + 1} $$
|
|
|
|
By the distributive law:
|
|
|
|
$$ = c\left(\sum_{i = 1}^{k}{a_i} + a_{k + 1}\right) $$
|
|
|
|
And then by the recursive definition of summation again:
|
|
|
|
$$ = c\left(\sum_{i = 1}^{k + 1}{a_i}\right) $$
|
|
|
|
Therefore $P(k + 1)$ is true.
|
|
|
|
Q.E.D.
|
|
|
|
42. Assume the truth of the commutative and associative laws (Appendix A, F1 and
|
|
F2), and use the recursive definition of product, together with mathematical
|
|
induction, to prove that for every positive integer $n$, if
|
|
$a_1, a_2, \dots, a_n$ and $b_1, b_2, \dots, b_n$ are real numbers, then
|
|
|
|
$$ \prod_{i = 1}^{n}{(a_ib_i)} = \left(\prod_{i = 1}^{n}{a_i}\right)\left(\prod_{i = 1}^{n}{b_i}\right) $$
|
|
|
|
For reference the commutative laws state:
|
|
|
|
For all real numbers $a$ and $b$,
|
|
|
|
$$ a + b = b + a \quad \text{ and } ab = ba $$
|
|
|
|
And the associative laws state:
|
|
|
|
For all real numbers $a$, $b$, and $c$,
|
|
|
|
$$ (a + b) + c = a + (b + c) \quad \text{ and } (ab)c = a(bc) $$
|
|
|
|
**Proof (by mathematical induction):**
|
|
|
|
Let $P(n)$ be the equatility:
|
|
|
|
$$ \prod_{i = 1}^{n}{(a_ib_i)} = \left(\prod_{i = 1}^{n}{a_i}\right)\left(\prod_{i = 1}^{n}{b_i}\right) $$
|
|
|
|
where $n \in \mathbb{Z}^+$.
|
|
|
|
_Basis Step:_
|
|
|
|
Prove $P(1)$. That is:
|
|
|
|
$$ \prod_{i = 1}^{1}{(a_ib_i)} = \left(\prod_{i = 1}^{1}{a_i}\right)\left(\prod_{i = 1}^{1}{b_i}\right) $$
|
|
|
|
Evaluating the left-hand side:
|
|
|
|
$$ \prod_{i = 1}^{1}{(a_ib_i)} $$
|
|
|
|
By the definition of product:
|
|
|
|
$$ = a_1 \cdot b_1 $$
|
|
|
|
Evaluating the right-hand side:
|
|
|
|
$$ \left(\prod_{i = 1}^{1}{a_i}\right)\left(\prod_{i = 1}^{1}{b_i}\right) $$
|
|
|
|
By the recusive definition of product:
|
|
|
|
$$ \prod_{i = 1}^{1}{a_i} = a_1 \quad \text{ and } \prod_{i = 1}^{1}{b_i} = b_1 $$
|
|
|
|
Therefore:
|
|
|
|
$$ = a_1 \cdot b_1 $$
|
|
|
|
Therefore, since both sides of the equality hold, $P(1)$ is true.
|
|
|
|
_Inductive Step:_
|
|
|
|
Let $k \in \mathbb{Z}^+$.
|
|
|
|
Suppose $P(k)$, that is:
|
|
|
|
$$ \prod_{i = 1}^{k}{(a_ib_i)} = \left(\prod_{i = 1}^{k}{a_i}\right)\left(\prod_{i = 1}^{k}{b_i}\right) $$
|
|
|
|
This is the inductive hypothesis.
|
|
|
|
Prove $P(k + 1)$. That is:
|
|
|
|
$$ \prod_{i = 1}^{k + 1}{(a_ib_i)} = \left(\prod_{i = 1}^{k + 1}{a_i}\right)\left(\prod_{i = 1}^{k + 1}{b_i}\right) $$
|
|
|
|
Evaluate the left-hand side:
|
|
|
|
$$ \prod_{i = 1}^{k + 1}{(a_ib_i)} $$
|
|
|
|
By the recursive definition of product:
|
|
|
|
$$ = \left(\prod_{i = 1}^{k}{(a_ib_i)}\right) \cdot a_{k + 1}b_{k + 1} $$
|
|
|
|
By the inductive hypothesis, the first term can be substituted:
|
|
|
|
$$ = \left(\prod_{i = 1}^{k}{a_i}\right)\left(\prod_{i = 1}^{k}{b_i}\right) \cdot a_{k + 1}b_{k + 1} $$
|
|
|
|
By the associative laws:
|
|
|
|
$$ = \left(\prod_{i = 1}^{k}{a_i}\right) \cdot a_{k + 1} \cdot \left(\prod_{i = 1}^{k}{b_i}\right) \cdot b_{k + 1} $$
|
|
|
|
By the recursive definition of product again:
|
|
|
|
$$ = \left(\prod_{i = 1}^{k + 1}{a_i}\right)\left(\prod_{i = 1}^{k + 1}{b_i}\right) $$
|
|
|
|
Which is the right-hand side of the equality. Therefore $P(k + 1)$ is true.
|
|
|
|
Q.E.D.
|
|
|
|
43. Assume the truth of the commutative and associative laws (Appendix A, F1 and
|
|
F2), and use the recursive definition of product, together with mathematical
|
|
induction, to prove that for each positive integer $n$, if
|
|
$a_1, a_2, \dots, a_n$ and $c$ are real numbers, then
|
|
|
|
$$ \prod_{i = 1}^{n}{(ca_i)} = c^n\left(\prod_{i = 1}^{n}{a_i}\right) $$
|
|
|
|
For reference the commutative laws state:
|
|
|
|
For all real numbers $a$ and $b$,
|
|
|
|
$$ a + b = b + a \quad \text{ and } ab = ba $$
|
|
|
|
And the associative laws state:
|
|
|
|
For all real numbers $a$, $b$, and $c$,
|
|
|
|
$$ (a + b) + c = a + (b + c) \quad \text{ and } (ab)c = a(bc) $$
|
|
|
|
**Proof (by mathematical induction):**
|
|
|
|
Let $P(n)$ be the equality:
|
|
|
|
$$ \prod_{i = 1}^{n}{(ca_i)} = c^n\left(\prod_{i = 1}^{n}{a_i}\right) $$
|
|
|
|
where $n \in \mathbb{Z}^+$.
|
|
|
|
_Basis Step:_
|
|
|
|
Prove $P(1)$. That is:
|
|
|
|
$$ \prod_{i = 1}^{1}{(ca_i)} = c^1\left(\prod_{i = 1}^{1}{a_i}\right) $$
|
|
|
|
Evaluate the left-hand side:
|
|
|
|
$$ \prod_{i = 1}^{1}{(ca_i)} $$
|
|
|
|
By the definition of product:
|
|
|
|
$$ = ca_1 $$
|
|
|
|
$$ = c^1a_1 $$
|
|
|
|
Evaluate the right-hand side:
|
|
|
|
$$ c^1\left(\prod_{i = 1}^{1}{a_i}\right) $$
|
|
|
|
By the definition of product:
|
|
|
|
$$ = c^1a_1 $$
|
|
|
|
Therefore, since the two sides of the equality hold, $P(1)$ is true.
|
|
|
|
_Inductive Step:_
|
|
|
|
Let $k \in \mathbb{Z}^+$.
|
|
|
|
Suppose $P(k)$. That is:
|
|
|
|
$$ \prod_{i = 1}^{k}{(ca_i)} = c^k\left(\prod_{i = 1}^{k}{a_i}\right) $$
|
|
|
|
This is the inductive hypothesis.
|
|
|
|
Prove $P(k + 1)$. That is:
|
|
|
|
$$ \prod_{i = 1}^{k + 1}{(ca_i)} = c^{k + 1}\left(\prod_{i = 1}^{k + 1}{a_i}\right) $$
|
|
|
|
Evaluating the left-hand side:
|
|
|
|
$$ \prod_{i = 1}^{k + 1}{(ca_i)} $$
|
|
|
|
By the definition of recursive product:
|
|
|
|
$$ = \left(\prod_{i = 1}^{k}{(ca_i)}\right) \cdot ca_{k + 1} $$
|
|
|
|
By the inductive hypothesis:
|
|
|
|
$$ = c^k\left(\prod_{i = 1}^{k}{a_i}\right) \cdot ca_{k + 1} $$
|
|
|
|
By the commutative laws:
|
|
|
|
$$ = ca_{k + 1} \cdot c^k\left(\prod_{i = 1}^{k}{a_i}\right) $$
|
|
|
|
By associative laws:
|
|
|
|
$$ = c \cdot c^k \cdot \left(\prod_{i = 1}^{k}{a_i}\right) \cdot a_{k + 1} $$
|
|
|
|
By the laws of exponents and by the recursive definition of product:
|
|
|
|
$$ = c^{k + 1}\left(\prod_{i = 1}^{k + 1}{a_i}\right) $$
|
|
|
|
This is the right-hand side of our equality. Therefore, $P(k + 1)$ is true.
|
|
|
|
Q.E.D.
|
|
|
|
44. The triangle inequality for absolute value states that for all real numbers
|
|
$a$ and $b$, $|a + b| \leq |a| + |b|$. Use the recursive definition of
|
|
summation, the triangle inequality, the definition of absolute value, and
|
|
mathematical induction to prove that for each p ositive integer $n$, if
|
|
$a_1, a_2, \dots, a_n$ are real numbers, then
|
|
|
|
$$ \left| \sum_{i = 1}^{n}{a_i} \right| \leq \sum_{i = 1}^{n}{|a_i|} $$
|
|
|
|
Omitted.
|
|
|
|
45. Prove that any sum of even integers is even.
|
|
|
|
Omitted.
|
|
|
|
46. Prove that any sum of an odd number of odd integers is odd.
|
|
|
|
Omitted.
|
|
|
|
47. Deduce from exercise 46 that for any positive integer $n$ if there is a sum
|
|
of $n$ odd integers that is even, then $n$ is even.
|
|
|
|
Omitted.
|
|
|
|
---
|
|
|
|
Page 373
|
|
|
|
**Exercise Set 5.7**
|
|
|
|
1. The formula
|
|
|
|
$$ 1 + 2 + 3 + \dots + n = \frac{n(n + 1)}{2} $$
|
|
|
|
is true for every integer $n \geq 1$. Use this fact to solve each of the
|
|
following problems:
|
|
|
|
a. If $k$ is an integer and $k \geq 2$, find a formula for the expression
|
|
$1 + 2 + 3 + \dots + (k - 1)$.
|
|
|
|
$n = k - 1$
|
|
|
|
$$ 1 + 2 + 3 + \dots + (k - 1) = \frac{(k - 1)((k - 1) + 1)}{2} $$
|
|
|
|
$$ = \frac{(k - 1)(k)}{2} $$
|
|
|
|
b. If $n$ is an integer and $n \geq 1$, find a formula for the expression
|
|
$5 + 2 + 4 + 6 + 8 + \dots + 2n$.
|
|
|
|
$$ 5 + 2 + 4 + 6 + 8 + \dots + 2n = 5 + 2\left(\frac{(n)(n + 1)}{2}\right) $$
|
|
|
|
$$ = 5 + n^2 + n $$
|
|
|
|
$$ = n^2 + n + 5 $$
|
|
|
|
c. If $n$ is an integer and $n \geq 1$, find a formula for the expression
|
|
$3 + 3 \cdot 2 + 3 \cdot 3 + \dots + 3 \cdot n + n$.
|
|
|
|
$$ 3 + 3 \cdot 2 + 3 \cdot 3 + \dots + 3 \cdot n + n = 3(1 + 2 + 3 + \dots + n) + n $$
|
|
|
|
$$ = 3\left(\frac{n(n + 1)}{2}\right) + n $$
|
|
|
|
2. The formula
|
|
|
|
$$ 1 + r + r^2 + \dots + r^n = \frac{r^{n + 1} - 1}{r - 1} $$
|
|
|
|
is true for every real number $r$ except $r = 1$ and for every integer
|
|
$n \geq 0$. Use this fact to solve each of the following problems:
|
|
|
|
a. If $i$ is an integer and $i \geq 1$, find a formula for the expression
|
|
$1 + 2 + 2^2 + \dots + 2^{i - 1}$.
|
|
|
|
$$ 1 + 2 + 2^2 + \dots + 2^{i - 1} = \frac{2^{i - 1 + 1} - 1}{2 - 1} $$
|
|
|
|
$$ = \frac{2^i - 1}{1} $$
|
|
|
|
$$ = 2^i - 1 $$
|
|
|
|
b. If $n$ is an integer and $n \geq 1$, find a formula for the expression
|
|
$3^{n - 1} + 3^{n - 2} + \dots + 3^2 + 3 + 1$.
|
|
|
|
$$ 3^{n - 1} + 3^{n - 2} + \dots + 3^2 + 3 + 1 = \frac{3^{n - 1 + 1} - 1}{3 - 1} $$
|
|
|
|
$$ = \frac{3^n - 1}{2} $$
|
|
|
|
c. If $n$ is an integer and $n \geq 2$, find a formula for the expression
|
|
$2^n + 2^{n - 2} \cdot 3 + 2^{n - 3} \cdot 3 + \dots + 2^2 \cdot 3 + 2 \cdot 3 + 3$.
|
|
|
|
$$ 3 + 3 \cdot 2 + 3 \cdot 2^2 + \dots + 3 \cdot 2^{n - 3} + 3 \cdot 2^{n - 2} + 2^n $$
|
|
|
|
$$ = 3(2^0 + 2^1 + 2^2 + \dots + 2^{n - 3} + 2^{n - 2}) + 2^n $$
|
|
|
|
$$ = 2^n + 3\left(\frac{2^{(n - 2) + 1} - 1}{2 - 1}\right) $$
|
|
|
|
$$ = 2^n + 3\left(\frac{2^{n - 1} - 1}{1}\right) $$
|
|
|
|
$$ = (2^n) + 3(2^{n - 1} - 1) $$
|
|
|
|
$$ = (2 \cdot 2^{n - 1}) + 3(2^{n - 1} - 1) $$
|
|
|
|
$$ = 2 \cdot 2^{n - 1} + 3 \cdot 2^{n - 1} - 3 $$
|
|
|
|
$$ = 5 \cdot 2^{n - 1} - 3 $$
|
|
|
|
d. If $n$ is an integer and $n \geq 1$, find a formula for the expression
|
|
|
|
Omitted.
|
|
|
|
In each of 3-15 a sequence is defined recursively. Use iteration to guess an
|
|
explicit formula for the sequence. Use formulas from Section 5.2 to simplify
|
|
your answers whenever possible.
|
|
|
|
3. $a_k = ka_{k - 1}$, for each integer $k \geq 1$ $a_0 = 1$.
|
|
|
|
$$ a_0 = 1 $$
|
|
|
|
$$ a_1 = 1 \cdot a_0 = 1 \cdot 1 = 1 $$
|
|
|
|
$$ a_2 = 2 \cdot a_1 = 2 \cdot 1 $$
|
|
|
|
$$ a_3 = 3 \cdot a_2 = 3 \cdot (2 \cdot 1) = 3 \cdot 2 \cdot 1 $$
|
|
|
|
$$ a_4 = 4 \cdot a_3 = 4 \cdot (3 \cdot 2 \cdot 1) = 4 \cdot 3 \cdot 2 \cdot 1 $$
|
|
|
|
Guess:
|
|
|
|
$$ a_n = n! $$
|
|
|
|
4. $b_k = \dfrac{b_{k - 1}}{1 + b_{k - 1}}$, for each integer $k \geq 1$
|
|
$b_0 = 1$.
|
|
|
|
$$ b_0 = 1 $$
|
|
|
|
$$ b_1 = \frac{b_0}{1 + b_0} = \frac{(1)}{1 + (1)} = \frac{1}{1 + 1} = \frac{1}{2} $$
|
|
|
|
$$ b_2 = \frac{b_1}{1 + b_1} = \frac{\dfrac{1}{2}}{1 + \left(\dfrac{1}{2}\right)} = \frac{1}{3} $$
|
|
|
|
$$ b_3 = \frac{b_2}{1 + b_2} = \frac{\dfrac{1}{3}}{1 + \left(\dfrac{1}{3}\right)} = \frac{1}{4} $$
|
|
|
|
$$ b_4 = \frac{b_3}{1 + b_3} = \frac{\dfrac{1}{4}}{1 + \left(\dfrac{1}{4}\right)} = \frac{1}{5} $$
|
|
|
|
Guess:
|
|
|
|
$$ b_n = \frac{1}{n + 1} $$
|
|
|
|
5. $c_k = 3c_{k - 1} + 1$, for each integer $k \geq 2$ $c_1 = 1$.
|
|
|
|
$$ c_1 = 1 $$
|
|
|
|
$$ c_2 = 3c_1 + 1 = 3(1) + 1 = 3 + 1 $$
|
|
|
|
$$ c_3 = 3c_2 + 1 = 3(3 + 1) + 1 = (3^2 + 3) + 1 $$
|
|
|
|
$$ c_4 = 3c_3 + 1 = 3(((3^2 + 3) + 1) + 1) + 1 = (3^3 + 3^2 + 3) + 1 $$
|
|
|
|
Guess:
|
|
|
|
$$ c_n = 3^{n - 1} + 3^{n - 2} + 3^{n - 3} + \dots + 3^3 + 3^2 + 3 + 1 $$
|
|
|
|
This is a geometric sequence (Theorem 5.2.2).
|
|
|
|
$$ = \frac{3^{(n - 1) + 1} - 1}{3 - 1} $$
|
|
|
|
$$ = \frac{3^n - 1}{2} $$
|
|
|
|
6. $d_k =2d_{k - 1} + 3$, for each integer $k \geq 2$, $d_1 = 2$.
|
|
|
|
$$ d_1 = 2 $$
|
|
|
|
$$ d_2 = 2d_1 + 3 = 2(2) + 3 = 2^2 + 3 $$
|
|
|
|
$$ d_3 = 2d_2 + 3 = 2(2^2 + 3) + 3 = 2^3 + 2 \cdot 3 + 3 $$
|
|
|
|
$$ d_4 = 2d_3 + 3 = 2(2^3 + 2 \cdot 3 + 3) + 3 = 2^4 + 2^2 \cdot 3 + 2 \cdot 3 + 3 $$
|
|
|
|
$$ d_5 = 2d_4 + 3 = 2(2^4 + 2^2 \cdot 3 + 2 \cdot 3 + 3) + 3 = 2^5 + 2^3 \cdot 3 + 2^2 \cdot 3 + 2 \cdot 3 + 3 $$
|
|
|
|
$$ d_5 = 2^5 + 3(2^3 + 2^2 + 2^1 + 2^0) $$
|
|
|
|
$$ d_5 = 2^5 + 3\sum_{i = 0}^{3}{2^i} $$
|
|
|
|
This is a geometric sequence (Theorem 5.2.2).
|
|
|
|
Guess:
|
|
|
|
$$ d_n = 2^n + 3\sum_{i = 0}^{n - 2}{2^i} $$
|
|
|
|
$$ d_n = 2^n + 3\frac{2^{(n - 2) + 1} - 1}{2 - 1} $$
|
|
|
|
$$ = 2^n + 3\frac{2^{n - 1} - 1}{1} $$
|
|
|
|
$$ = 2^n + 3(2^{n - 1} - 1) $$
|
|
|
|
$$ = 2^n + 3(2^{n - 1} - 1) $$
|
|
|
|
$$ = 2^n + 3 \cdot 2^{n - 1} - 3 $$
|
|
|
|
$$ = 2 \cdot 2^{n - 1} + 3 \cdot 2^{n - 1} - 3 $$
|
|
|
|
$$ = 5 \cdot 2^{n - 1} - 3 $$
|
|
|
|
7. $e_k = 4e_{k - 1} + 5$, for each integer $k \geq 1$ $e_0 = 2$.
|
|
|
|
$$ e_0 = 2 $$
|
|
|
|
$$ e_1 = 4e_0 + 5 = 4 \cdot 2 + 5 $$
|
|
|
|
$$ e_2 = 4e_1 + 5 = 4(4 \cdot 2 + 5) + 5 = 4^2 \cdot 2 + 4 \cdot 5 + 5 $$
|
|
|
|
$$ e_3 = 4e_2 + 5 = 4(4^2 \cdot 2 + 4 \cdot 5 + 5) + 5 = 4^3 \cdot 2 + 4^2 \cdot 5 + 4 \cdot 5 + 5 $$
|
|
|
|
$$ e_4 = 4e_3 + 5 = 4(4^3 \cdot 2 + 4^2 \cdot 5 + 4 \cdot 5 + 5) + 5 = 4^4 \cdot 2 + 4^3 \cdot 5 + 4^2 \cdot 5 + 4 \cdot 5 + 5 $$
|
|
|
|
Guess:
|
|
|
|
$$ e_n = 4^n \cdot 2 + 4^{n - 1} \cdot 5 + 4^{n - 2} \cdot 5 + \dots + 4 \cdot 5 + 5 $$
|
|
|
|
$$ = 4^n \cdot 2 + 5(4^{n - 1} + 4^{n - 2} + \dots + 4 + 1) $$
|
|
|
|
$$ = 4^n \cdot 2 + 5\sum_{i = 0}^{n - 1}{4^i} $$
|
|
|
|
$$ = 4^n \cdot 2 + 5\left(\frac{4^{(n - 1) + 1} - 1}{4 - 1}\right) $$
|
|
|
|
$$ = 4^n \cdot 2 + 5\left(\frac{4^n - 1}{3}\right) $$
|
|
|
|
$$ = \frac{3(4^n \cdot 2)}{3} + \left(\frac{5(4^n - 1)}{3}\right) $$
|
|
|
|
$$ = \frac{3(4^n \cdot 2) + 5(4^n - 1)}{3} $$
|
|
|
|
$$ = \frac{(6 \cdot 4^n) + (5 \cdot 4^n - 5)}{3} $$
|
|
|
|
$$ = \frac{6 \cdot 4^n + 5 \cdot 4^n - 5}{3} $$
|
|
|
|
$$ = \frac{11 \cdot 4^n - 5}{3} $$
|
|
|
|
8. $f_k = f_{k - 1} + 2^k$, for each integer $k \geq 2$ $f_1 = 1$.
|
|
|
|
$$ f_1 = 1 $$
|
|
|
|
$$ f_2 = f_1 + 2^2 = (1) + 2^2 = 1 + 2^2 $$
|
|
|
|
$$ f_3 = f_2 + 2^3 = (1 + 2^2) + 2^3 = 1 + 2^2 + 2^3 $$
|
|
|
|
$$ f_4 = f_3 + 2^4 = (1 + 2^2 + 2^3) + 2^4 = 1 + 2^2 + 2^3 + 2^4 $$
|
|
|
|
Guess:
|
|
|
|
$$ f_n = 1 + \sum_{i = 2}^{n}{2^i} $$
|
|
|
|
$$ = 1 + \left(\sum_{i = 0}^{n}{2^i} - \sum_{i = 0}^{1}{2^i}\right) $$
|
|
|
|
$$ = 1 + \frac{2^{n + 1} - 1}{2 - 1} - (2^0 + 2^1) $$
|
|
|
|
$$ = 1 + 2^{n + 1} - 1 - (1 + 2) $$
|
|
|
|
$$ = 2^{n + 1} - 3 $$
|
|
|
|
9. $g_k = \dfrac{g_{k - 1}}{g_{k - 1} + 2}$, for each integer $k \geq 2$
|
|
$g_1 = 1$.
|
|
|
|
$$ g_1 = 1 $$
|
|
|
|
$$ g_2 = \frac{g_1}{g_1 + 2} = \frac{1}{1 + 2} = \frac{1}{3} = \frac{1}{2^2 - 1} $$
|
|
|
|
$$ g_3 = \frac{g_2}{g_2 + 2} = \frac{\dfrac{1}{3}}{\dfrac{1}{3} + 2} = \frac{1}{7} = \frac{1}{2^3 - 1} $$
|
|
|
|
$$ g_4 = \frac{g_3}{g_3 + 2} = \frac{\dfrac{1}{7}}{\dfrac{1}{7} + 2} = \frac{1}{15} = \frac{1}{2^4 - 1} $$
|
|
|
|
Guess:
|
|
|
|
$$ g_n = \frac{1}{2^n - 1} $$
|
|
|
|
10. $h_k = 2^k - h_{k - 1}$, for each integer $k \geq 1$ $h_0 = 1$.
|
|
|
|
$$ h_0 = 1 $$
|
|
|
|
$$ h_1 = 2^1 - h_0 = 2 - 1 = 2^1 - 2^0 $$
|
|
|
|
$$ h_2 = 2^2 - h_1 = 2^2 - (2^1 - 1) = 2^2 - 2^1 + 2^0 $$
|
|
|
|
$$ h_3 = 2^3 - h_2 = 2^3 - (2^2 - 2^2 + 2^0) = 2^3 - 2^2 + 2^1 - 2^0 $$
|
|
|
|
$$ h_4 = 2^4 - h_3 = 2^4 - (2^3 - 2^2 + 2^1 - 2^0) = 2^4 - 2^3 + 2^2 - 2^1 + 2^0 $$
|
|
|
|
Guess:
|
|
|
|
$$ h_n = 2^n - 2^{n - 1} + \dots + (-1)^{n - 2} \cdot 2^2 + (-1)^{n - 1} \cdot 2^1 + (-1)^n \cdot 2^0 $$
|
|
|
|
$$ = (-1)^n[(-1)^n \cdot 2^n + \dots + (-1)^2 \cdot 2^2 + (-1)^1 \cdot 2^1 + (-1)^n \cdot 2^0] $$
|
|
|
|
$$ = (-1)^n[(-2)^n + (-2)^{n - 1} + \dots + (-2)^2 + (-2)^1 + (-2)^0] $$
|
|
|
|
By the definition of a geometric sequence:
|
|
|
|
$$ = (-1)^n\left(\frac{(-2)^{n + 1} - 1}{(-2) - 1}\right) $$
|
|
|
|
$$ = (-1)^n\left(\frac{(-2)^{n + 1} - 1}{-3}\right) $$
|
|
|
|
$$ = \frac{(-1)^{n + 1}((-2)^{n + 1} - 1)}{(-1)(-3)} $$
|
|
|
|
$$ = \frac{2^{n + 1} - (-1)^{n + 1}}{3} $$
|
|
|
|
11. $p_k = p_{k - 1} + 2 \cdot 3^k$, for each integer $k \geq 2$ $p_1 = 2$.
|
|
|
|
$$ p_1 = 2 $$
|
|
|
|
$$ p_2 = p_1 + 2 \cdot 3^2 = 2 + 2 \cdot 3^2 $$
|
|
|
|
$$ p_3 = p_2 + 2 \cdot 3^3 = (2 + 2 \cdot 3^2) + 2 \cdot 3^3 = 2 + 2 \cdot 3^2 + 2 \cdot 3^3 $$
|
|
|
|
Guess:
|
|
|
|
$$ p_n = 2 + 2(3^2 + 3^3 + \dots + 3^n) $$
|
|
|
|
$$ = 2 + 2(3^0 + 3^1 + 3^2 + 3^3 + \dots + 3^n - 1 - 3^1) $$
|
|
|
|
$$ = 2 + 2\left(\sum_{i = 0}^{n}{3^i} - 1 - 3\right) $$
|
|
|
|
$$ = 2 + 2\left(\frac{3^{n + 1} - 1}{3 - 1} - 1 - 3\right) $$
|
|
|
|
$$ = 2 + 2\left(\frac{3^{n + 1} - 1}{2} - 4\right) $$
|
|
|
|
$$ = 2 + 3^{n + 1} - 1 - 8 $$
|
|
|
|
$$ = 2 + 3^{n + 1} - 9 $$
|
|
|
|
$$ = 3^{n + 1} - 7 $$
|
|
|
|
12. $s_k = s_{k - 1} + 2k$, for each integer $k \geq 1$ $s_0 = 3$.
|
|
|
|
$$ s_0 = 3 $$
|
|
|
|
$$ s_1 = s_0 + 2(1) = 3 + 2 = 5 $$
|
|
|
|
$$ s_2 = s_1 + 2(2) = (3 + 2) + 2(2) = 3 + 2 + 4 = 9 $$
|
|
|
|
$$ s_3 = s_2 + 2(3) = (3 + 2 + 4) + 2(3) = 3 + 2 + 4 + 6 = 15 $$
|
|
|
|
$$ s_4 = s_3 + 2(4) = (3 + 2 + 4 + 6) + 2(4) = 3 + 2 + 4 + 6 + 8 = 23 $$
|
|
|
|
Guess:
|
|
|
|
$$ s_n = 3 + 2(1 + 2 + 3 + 4 + \dots + n) $$
|
|
|
|
By Theorem 5.2.1:
|
|
|
|
$$ = 3 + 2\left(\frac{n(n + 1)}{2}\right) $$
|
|
|
|
$$ = 3 + n(n + 1) $$
|
|
|
|
$$ = 3 + n^2 + n $$
|
|
|
|
$$ = n^2 + n + 3 $$
|
|
|
|
13. $t_k = t_{k - 1} + 3k + 1$, for each integer $k \geq 1$ $t_0 = 0$.
|
|
|
|
$$ t_0 = 0 $$
|
|
|
|
$$ t_1 = t_0 + 3(1) + 1 = 0 + 3 + 1 = 3 + 1 = 3 \cdot 1 + 1 $$
|
|
|
|
$$ t_2 = t_1 + 3(2) + 1 = (3 \cdot 1 + 1) + 3 \cdot 2 + 1 = 3 \cdot 1 + 1 + 3 \cdot 2 + 1 $$
|
|
|
|
$$ t_3 = t_2 + 3(3) + 1 = (3 \cdot 1 + 1 + 3 \cdot 2 + 1) + 3 \cdot 3 + 1 = 3 \cdot 1 + 1 + 3 \cdot 2 + 1 + 3 \cdot 3 + 1 $$
|
|
|
|
$$ t_4 = t_3 + 3(4) + 1 = (3 \cdot 1 + 1 + 3 \cdot 2 + 1 + 3 \cdot 3 + 1) + 3 \cdot 4 + 1 $$
|
|
|
|
Guess:
|
|
|
|
$$ t_n = 3(1 + 2 + 3 + \dots + n) + n $$
|
|
|
|
$$ = 3\left(\frac{n(n + 1)}{2}\right) + n $$
|
|
|
|
$$ = \frac{3(n^2 + n)}{2} + \frac{2n}{2} $$
|
|
|
|
$$ = \frac{3n^2 + 3n + 2n}{2} $$
|
|
|
|
$$ = \frac{3n^2 + 5n}{2} $$
|
|
|
|
14. $x_k = 3x_{k - 1} + k$, for each integer $k \geq 2$ $x_1 = 1$.
|
|
|
|
Omitted.
|
|
|
|
15. $y_k = y_{k - 1} + k^2$, for each integer $k \geq 2$ $y_1 = 1$.
|
|
|
|
$$ y_1 = 1 $$
|
|
|
|
$$ y_2 = y_1 + (2)^2 = 1 + 2^2 $$
|
|
|
|
$$ y_3 = y_2 + (3)^2 = (1 + 2^2) + 3^2 = 1 + 2^2 + 3^2 $$
|
|
|
|
$$ y_4 = y_3 + (4)^2 = (1 + 2^2 + 3^2) + 4^2 = 1 + 2^2 + 3^2 + 4^2 $$
|
|
|
|
Guess:
|
|
|
|
$$ y_n = 1^2 + 2^2 + 3^2 + \dots + n^2 $$
|
|
|
|
By Exercise 5.2.10:
|
|
|
|
$$ = \frac{n(n + 1)(2n + 1)}{6} $$
|
|
|
|
16. Solve the recurrence relation obtained as the answer to exercise 17\(c\) of
|
|
Section 5.6.
|
|
|
|
The recurrence relation in question is:
|
|
|
|
$$ 3a_{k - 1} + 2 $$
|
|
|
|
For reference:
|
|
|
|
$$ a_1 = 2 $$
|
|
|
|
Solving:
|
|
|
|
$$ a_1 = 2 $$
|
|
|
|
$$ a_2 = 3a_1 + 2 = 3 \cdot 2 + 2 $$
|
|
|
|
$$ a_3 = 3a_2 + 2 = 3 \cdot (3 \cdot 2 + 2) + 2 = 3^2 \cdot 2 + 3 \cdot 2 + 2 $$
|
|
|
|
$$ a_4 = 3a_3 + 2 = 3 \cdot (3^2 \cdot 2 + 3 \cdot 2 + 2) + 2 = 3^3 \cdot 2 + 3^2 \cdot 2 + 3 \cdot 2 + 2 $$
|
|
|
|
Guess:
|
|
|
|
$$ a_n = 2(3^n + 3^{n - 1} + 3^{n - 2} + \dots + 3^1 + 3^0) $$
|
|
|
|
By the definition of a geometric sequence:
|
|
|
|
$$ = 2\left(\frac{3^n - 1}{3 - 1}\right) $$
|
|
|
|
$$ = 2\left(\frac{3^n - 1}{2}\right) $$
|
|
|
|
$$ = 3^n - 1 $$
|
|
|
|
17. Solve the recurrence relation obtained as the answer to exercise 21\(c\) of
|
|
Section 5.6.
|
|
|
|
The recurrence relation in question is:
|
|
|
|
$$ t_n = 3t_{n - 1} + 2 \quad n \geq 2 $$
|
|
|
|
For reference:
|
|
|
|
$$ t_1 = 2 $$
|
|
|
|
$$ t_2 = 3t_1 + 2 = 3 \cdot 2 + 2 $$
|
|
|
|
$$ t_3 = 3t_2 + 2 = 3 \cdot (3 \cdot 2 + 2) + 2 = 3^2 \cdot 2 + 3 \cdot 2 + 2 $$
|
|
|
|
$$ t_4 = 3t_3 + 2 = 3 \cdot (3^2 \cdot 2 + 3 \cdot 2 + 2) + 2 = 3^3 \cdot 2 + 3^2 \cdot 2 + 3 \cdot 2 + 2 $$
|
|
|
|
Guess:
|
|
|
|
$$ t_n = 2(3^{n - 1} + 3^{n - 2} + \dots + 3^1 + 3^0) $$
|
|
|
|
By the definition of a geometric sequence (Theorem 5.2.2):
|
|
|
|
$$ = 2\left(\frac{3^{(n - 1) + 1} - 1}{3 - 1}\right) $$
|
|
|
|
$$ = 2\left(\frac{3^n - 1}{2}\right) $$
|
|
|
|
$$ = 3^n - 1 $$
|
|
|
|
18. Suppose $d$ is a fixed constant and $a_0, a_1, a_2, \dots$ is a sequence
|
|
that satisfies the recurrence relation $a_k = a_{k - 1} + d$, for each
|
|
integer $k \geq 1$. Use mathematical induction to prove that
|
|
$a_n = a_0 + nd$, for every integer $n \geq 0$.
|
|
|
|
**Proof (by mathematical induction):**
|
|
|
|
Let $d$ be any fixed constant, and let $a_0, a_1, a_2, \dots$ be the sequence
|
|
defined recursively by $a_k = a_{k - 1} + d$ for each integer $k \geq 1$.
|
|
|
|
Let $P(n)$ be the equation:
|
|
|
|
$$ a_n = a_0 + nd $$
|
|
|
|
We must show by mathematical induction that $P(n)$ is true for every integer
|
|
$n \geq 0$.
|
|
|
|
_Basis Step:_
|
|
|
|
Prove that $P(0)$ is true. That is:
|
|
|
|
$$ a_0 = a_0 + (0)d $$
|
|
|
|
$$ a_0 = a_0 $$
|
|
|
|
This equality holds, and therefore $P(0)$ is true.
|
|
|
|
_Inductive Step:_
|
|
|
|
Let $k$ be any integer such that $k \geq 0$.
|
|
|
|
Suppose $P(k)$. That is:
|
|
|
|
$$ a_k = a_0 + kd $$
|
|
|
|
This is the inductive hypothesis.
|
|
|
|
Prove $P(k + 1)$. That is:
|
|
|
|
$$ a_{k + 1} = a_0 + (k + 1)d $$
|
|
|
|
By the definition of the given sequence:
|
|
|
|
$$ a_{k + 1} = a_k + d $$
|
|
|
|
By substitution of the inductive hypothesis:
|
|
|
|
$$ a_{k + 1} = (a_0 + kd) + d $$
|
|
|
|
By algebra:
|
|
|
|
$$ a_{k + 1} = a_0 + (k + 1)d $$
|
|
|
|
Which is the equality that was to be shown. Therefore $P(k + 1)$ is true.
|
|
|
|
Q.E.D.
|
|
|
|
19. A worker is promised a bonus if he can increase his productivity by 2 units
|
|
a day for a period of 30 days. If on day 0 he produces 170 units, how many
|
|
units must he produce on day 30 to qualify for the bonus?
|
|
|
|
Let $U_n$ be the number of units produced on day $n$. Then:
|
|
|
|
$$ U_k = U_{k - 1} + 2 $$
|
|
|
|
for every integer $k \geq 1$, and:
|
|
|
|
$$ U_0 = 170 $$
|
|
|
|
Hence $U_0, U_1, U_2, \dots$ is an arithmetic sequence with a fixed constant
|
|
$2$. It then follows that when $n = 30$:
|
|
|
|
$$ U_n = U_0 + n \cdot 2 $$
|
|
|
|
$$ U_{30} = 170 + (30) \cdot 2 = 170 + 60 = 230 \text{ units} $$
|
|
|
|
Thus, in order to qualify for the bonus, the worker must produce 230 units on
|
|
day 30.
|
|
|
|
20. A runner targets herself to improve her time on a certain course by 3
|
|
seconds a day. If on day 0 she runs the course in 3 minutes, how fast must
|
|
she run it on day 14 to stay on target?
|
|
|
|
First, let's convert 3 minutes to seconds for ease of evaluation:
|
|
|
|
$$ 3 \text{ minutes } \cdot 60 \frac{\text{seconds}}{\text{minute}} = 180 \text{ seconds} $$
|
|
|
|
Let $R_n$ be the number of seconds the runner ran on day $n$. Then:
|
|
|
|
$$ R_k = R_{k - 1} - 3 $$
|
|
|
|
for every integer $k \geq 1$, and:
|
|
|
|
$$ R_0 = 180 $$
|
|
|
|
Hence $R_0, R_1, R_2, \dots$ is an arithmetic sequence with a fixed constant
|
|
$3$. It follows then that when $n = 14$:
|
|
|
|
$$ U_n = U_0 - n \cdot 3 $$
|
|
|
|
$$ U_{14} = (180) - 14 \cdot 3 = 180 - 42 = 138 \text{ seconds} $$
|
|
|
|
Therefore, the runner must run the certain course in 138 seconds (approximately
|
|
2.3 minutes) on day 14 in order to stay on target.
|
|
|
|
21. Suppose $r$ is a fixed constant and $a_0, a_1, a_2, \dots$ is a sequence
|
|
that satisfies the recurrence relation $a_k = ra_{k - 1}$, for each integer
|
|
$k \geq 1$ and $a_0 = a$. Use mathematical induction to prove that
|
|
$a_n = ar^n$, for every integer $n \geq 0$.
|
|
|
|
**Proof (by mathematical induction):**
|
|
|
|
Let $P(n)$ be the equation:
|
|
|
|
$$ a_n = ar^n $$
|
|
|
|
_Basis Step:_
|
|
|
|
Prove $P(0)$, that is:
|
|
|
|
$$ a_0 = ar^0 $$
|
|
|
|
$$ a_0 = a(1) $$
|
|
|
|
$$ a_0 = a $$
|
|
|
|
The given problem statement tells us that $a_0 = a$. Since this matches the
|
|
equality found for $P(0)$, we can conclude therefore that $P(0)$ is true.
|
|
|
|
_Inductive Step:_
|
|
|
|
Let $k$ be any integer such that $k \geq 1$.
|
|
|
|
Suppose $P(k)$, that is:
|
|
|
|
$$ a_k = ar^k $$
|
|
|
|
This is the inductive hypothesis.
|
|
|
|
Prove $P(k + 1)$, that is:
|
|
|
|
$$ a_{k + 1} = ar^{k + 1} $$
|
|
|
|
By the given recurrence relation:
|
|
|
|
$$ a_{k + 1} = r \cdot a_k $$
|
|
|
|
By substitution with the inductive hypothesis:
|
|
|
|
$$ a_{k + 1} = r \cdot (a \cdot r^k) $$
|
|
|
|
By algebra:
|
|
|
|
$$ a_{k + 1} = ar^{k + 1} $$
|
|
|
|
This equality is what was to be shown, therefore $P(k + 1)$ is true.
|
|
|
|
Q.E.D.
|
|
|
|
22. As shown in Example 5.6.8, if a bank pays interest at a rate of $i$
|
|
compounded $m$ times a year, then the amount of money $P_k$ at the end of
|
|
$k$ time periods (where one time period = $\dfrac{1}{m}$<sup>th</sup> of a
|
|
year) satisfies the recurrence relation
|
|
$P_k = \left[1 + \left(\dfrac{1}{m}\right)\right]P_{k - 1}$ with initial
|
|
condition $P_0 = \text{ the initial amount deposited}$. Find an explicit
|
|
formula for $P_n$.
|
|
|
|
$$ P_0 = P_0 $$
|
|
|
|
$$ P_1 = \left[1 + \frac{i}{m}\right] \cdot P_0 = \left[1 + \frac{i}{m}\right]^1 \cdot P_0 $$
|
|
|
|
$$ P_2 = \left[1 + \frac{i}{m}\right] \cdot P_1 = \left[1 + \frac{i}{m}\right] \cdot \left[1 + \frac{i}{m}\right] \cdot P_0 = \left[1 + \frac{i}{m}\right]^2 + P_0 $$
|
|
|
|
Guess:
|
|
|
|
$$ P_n = \left[1 + \frac{i}{m}\right]^n \cdot P_0 $$
|
|
|
|
23. Suppose the population of a country increases at a steady rate of 3% per
|
|
year. If the population is 50 million at a certain time, what will it be 25
|
|
years later?
|
|
|
|
Let $P_n$ be the population of the country at year $n$. Then:
|
|
|
|
$$ P_{k + 1} = 1.03 \cdot P_k $$
|
|
|
|
for every integer $k \geq 1$, and:
|
|
|
|
$$ P_0 = 50000000 $$
|
|
|
|
The explicit formula then is:
|
|
|
|
$$ P_n = (1.03)^n \cdot P_0 $$
|
|
|
|
Then:
|
|
|
|
$$ P_{25} = (1.03)^{25} \cdot 50000000 $$
|
|
|
|
$$ \approx 104688896 $$
|
|
|
|
Therefore, the population of the country 25 years later will be approximately
|
|
104,688896.
|
|
|
|
24. A chain letter works as follows: One person sends a copy of the letter to
|
|
five friends, each of whom sends a copy to five friends, each of whom sends
|
|
a copy to five friends, each of whom sends a copy to five friends, and so
|
|
forth. How many people will have received copies of the letter after the
|
|
twentieth reception of this process, assuming no person receives more than
|
|
one copy?
|
|
|
|
$$ \sum_{k = 0}^{20}{5^k} = \frac{5^{21} - 1}{4} $$
|
|
|
|
$$ \approx 1.192092896 \cdot 10^{14} \text{ people} $$
|
|
|
|
25. A certain computer algorithm executes twice as many operations when it is
|
|
run with an input size $k$ as when it is run with an input size $k - 1$
|
|
(where $k$ is an integer that is greater than $1$). When the algorithm is
|
|
run with an input size $1$, it executes seven operations. How many
|
|
operations does it execute when it is run with an input size of $25$?
|
|
|
|
Let $P_k$ be the number of operations the algorithm when the input size is $k$,
|
|
and $P_0 = 7$. The recurrence relation is:
|
|
|
|
$$ P_k = 2P_{k - 1} $$
|
|
|
|
So:
|
|
|
|
$$ P_n = 2^n \cdot P_0 $$
|
|
|
|
$$ P_{25} = 2^{25} \cdot 7 $$
|
|
|
|
$$ = 234881024 $$
|
|
|
|
So the algorithm executes 234881024 operations when it is run with an input size
|
|
of 25.
|
|
|
|
26. A person saving for retirement makes an initial deposit of $1,000 to a bank
|
|
account earning interest at a rate of 3% per year compounded monthly, and
|
|
each month she adds an addition $200 to the account.
|
|
|
|
a. For each nonnegative integer $n$, let $A_n$ be the amount in the account at
|
|
the end of $n$ months. Find the recurrence relation relating $A_k$ to
|
|
$A_{k - 1}$.
|
|
|
|
$$ A_k = \left[1 + \left(\frac{1}{12}\right)\left(\frac{3}{100}\right)\right] \cdot A_{k - 1} + 200 $$
|
|
|
|
$$ = \left[1 + \left(\frac{3}{1200}\right)\right] \cdot A_{k - 1} + 200 $$
|
|
|
|
$$ = \left[1 + \left(\frac{1}{400}\right)\right] \cdot A_{k - 1} + 200 $$
|
|
|
|
$$ = \frac{401}{400} \cdot A_{k - 1} + 200 $$
|
|
|
|
$$ = 1.0025 \cdot A_{k - 1} + 200 $$
|
|
|
|
b. Use iteration to find an explicit formula for $A_n$.
|
|
|
|
$$ A_0 = 1000 $$
|
|
|
|
$$ A_1 = 1.0025 \cdot A_0 + 200 = 1.0025 \cdot (1000) + 200 $$
|
|
|
|
$$ A_2 = 1.0025 \cdot A_1 + 200 = 1.0025 \cdot (1.0025 \cdot (1000) + 200) + 200 $$
|
|
|
|
$$ A_3 = 1.0025 \cdot A_3 + 200 = 1.0025 \cdot (1.0025 \cdot (1.0025 \cdot (1000) + 200) + 200) + 200 $$
|
|
|
|
Guess:
|
|
|
|
$$ A_n = 200 + 200(1.0025) + 200(1.0025)^2 + \cdots + 200(1.0025)^{n - 1} + 1000(1.0025)^n $$
|
|
|
|
Explicit formula:
|
|
|
|
$$ A_n = 200 \cdot \left(\frac{1.0025^n - 1}{1.0025 - 1}\right) + 1000(1.0025)^n $$
|
|
|
|
c. Use mathematical induction to prove the correctness of the formula you
|
|
obtained in part (b).
|
|
|
|
**Proof (by mathematical induction):**
|
|
|
|
Let $P(n)$ be the equation:
|
|
|
|
$$ A_n = 200 \cdot \left(\frac{1.0025^n - 1}{1.0025 - 1}\right) + 1000(1.0025)^n $$
|
|
|
|
_Basis Step:_
|
|
|
|
Prove $P(0)$, that is:
|
|
|
|
$$ A_0 = 200 \cdot \left(\frac{1.0025^0 - 1}{1.0025 - 1}\right) + 1000(1.0025)^0 $$
|
|
|
|
$$ A_0 = 200 \cdot \left(\frac{1 - 1}{0.0025}\right) + 1000(1) $$
|
|
|
|
$$ A_0 = 200 \cdot 0 + 1000 $$
|
|
|
|
$$ A_0 = 0 + 1000 $$
|
|
|
|
$$ A_0 = 1000 $$
|
|
|
|
This equality holds as $A_0$ was established as being equal to $1000$ in the
|
|
given problem statement. Therefore $P(0)$ is true.
|
|
|
|
_Inductive Step:_
|
|
|
|
Let $k$ be any integer such that $k \geq 0$.
|
|
|
|
Suppose $P(k)$, that is:
|
|
|
|
$$ A_k = 200 \cdot \left(\frac{1.0025^k - 1}{1.0025 - 1}\right) + 1000(1.0025)^k $$
|
|
|
|
This is the inductive hypothesis.
|
|
|
|
Prove $P(k + 1)$, that is:
|
|
|
|
$$ A_{k + 1} = 200 \cdot \left(\frac{1.0025^{k + 1} - 1}{1.0025 - 1}\right) + 1000(1.0025)^{k + 1} $$
|
|
|
|
By the recurrence relation in part (a), we have:
|
|
|
|
$$ A_{k + 1} = 1.0025 \cdot A_k + 200 $$
|
|
|
|
By substitution with the inductive hypothesis:
|
|
|
|
$$ = 1.0025 \cdot (200 \cdot \left(\frac{1.0025^k - 1}{1.0025 - 1}\right) + 1000(1.0025)^k) + 200 $$
|
|
|
|
$$ = 200 \cdot \left(\frac{1.0025^{k + 1} - 1.0025}{1.0025 - 1}\right) + 1000(1.0025)^{k + 1} + 200 $$
|
|
|
|
$$ = 200 \cdot \left(\frac{1.0025^{k + 1} - 1.0025}{0.0025}\right) + 1000(1.0025)^{k + 1} + \frac{0.5}{0.0025} $$
|
|
|
|
$$ = 200 \cdot \left(\frac{1.0025^{k + 1} - 1.0025 + 0.0025}{0.0025}\right) + 1000(1.0025)^{k + 1} $$
|
|
|
|
$$ = 200 \cdot \left(\frac{1.0025^{k + 1} - 1}{0.0025}\right) + 1000(1.0025)^{k + 1} $$
|
|
|
|
Q.E.D.
|
|
|
|
d. How much will the account be worth at the end of 20 years? At the end of 40
|
|
years?
|
|
|
|
We can just use the explicit formula:
|
|
|
|
$$ A_n = 200 \cdot \left(\frac{1.0025^n - 1}{1.0025 - 1}\right) + 1000(1.0025)^n $$
|
|
|
|
$$ A_{20} = 200 \cdot \left(\frac{1.0025^{20} - 1}{1.0025 - 1}\right) + 1000(1.0025)^{20} $$
|
|
|
|
$$ \approx \$5147.65 $$
|
|
|
|
$$ A_{40} = 200 \cdot \left(\frac{1.0025^{40} - 1}{1.0025 - 1}\right) + 1000(1.0025)^{40} $$
|
|
|
|
$$ \approx \$9507.67 $$
|
|
|
|
e. In how many years will the account be worth $10,000?
|
|
|
|
$$ A_n = 200 \cdot \left(\frac{1.0025^n - 1}{1.0025 - 1}\right) + 1000(1.0025)^n $$
|
|
|
|
$$ 10000 = 200 \cdot \left(\frac{1.0025^n - 1}{0.0025}\right) + 1000(1.0025)^n $$
|
|
|
|
$$ 10000 = \left(\frac{200 \cdot (1.0025^n - 1)}{0.0025}\right) + 1000(1.0025)^n $$
|
|
|
|
$$ 10000 = 80000(1.0025^n - 1) + 1000(1.0025)^n $$
|
|
|
|
$$ 10000 = 80000(1.0025^n) - 80000 + 1000(1.0025)^n $$
|
|
|
|
$$ 10000 = 81000(1.0025^n) - 80000 $$
|
|
|
|
$$ 90000 = 81000(1.0025^n) $$
|
|
|
|
$$ \frac{10}{9} = 1.0025^n $$
|
|
|
|
$$ \ln\left(\frac{10}{9}\right) = \ln(1.0025^n) $$
|
|
|
|
$$ \ln\left(\frac{10}{9}\right) = n\ln(1.0025) $$
|
|
|
|
$$ \frac{\ln\left(\frac{10}{9}\right)}{\ln(1.0025)} = n $$
|
|
|
|
$$ n \approx 42 \text{ months} $$
|
|
|
|
$$ \frac{42}{12} = 3.5 \text{ years} $$
|
|
|
|
27. A person borrows $3,000 on a bank credit card at a nominal rate of 18% per
|
|
year, which is actually charged at a rate of 1.5% per month.
|
|
|
|
a. What is the annual percentage yield (APY) for the card? (See Example 5.6.8
|
|
for a definition of APY.)
|
|
|
|
$$ \text{APY} = \left(1 + \frac{r}{n}\right)^n - 1 $$
|
|
|
|
$$ = \left(1 + \frac{0.18}{12}\right)^{12} - 1 $$
|
|
|
|
$$ \approx 0.1956181715 $$
|
|
|
|
$$ \approx 19.6\% $$
|
|
|
|
b. Assume that the person does not place any additional charges on the card and
|
|
pays the bank $150 each month to pay off the loan. Let $B_n$ be the balance owed
|
|
on the card after $n$ months. Find an explicit formula for $B_n$.
|
|
|
|
$$ B_0 = 3000 $$
|
|
|
|
$$ B_n = 1.015 \cdot B_{n - 1} - 150 \quad n \geq 1 $$
|
|
|
|
$$ B_1 = 1.015 \cdot B_0 - 150 = 1.015 \cdot 3000 - 150 $$
|
|
|
|
$$ B_2 = 1.015 \cdot B_1 - 150 = 1.015 \cdot (1.015 \cdot 3000 - 150) - 150 $$
|
|
|
|
$$ B_3 = 1.015 \cdot B_2 - 150 = 1.015 \cdot (1.015 \cdot (1.015 \cdot 3000 - 150) - 150) - 150 $$
|
|
|
|
Guess:
|
|
|
|
$$ B_n = 1.015^n \cdot B_0 - 150 \cdot \left(\frac{1.015^n - 1}{1.015 - 1}\right) $$
|
|
|
|
$$ = 1.015^n \cdot 3000 - 150 \cdot \left(\frac{1.015^n - 1}{0.015}\right) $$
|
|
|
|
$$ = 1.015^n \cdot 3000 - 10000(1.015^n - 1) $$
|
|
|
|
$$ = 1.015^n \cdot 3000 - 10000(1.015^n) + 10000 $$
|
|
|
|
$$ = 10000 - 7000(1.015^n) $$
|
|
|
|
c. How long will be required to pay off the debt?
|
|
|
|
$$ 0 = 10000 - 7000(1.015^n) $$
|
|
|
|
$$ 7000(1.015^n) = 10000 $$
|
|
|
|
$$ 1.015^n = \frac{10000}{7000} $$
|
|
|
|
$$ 1.015^n = \frac{10}{7} $$
|
|
|
|
$$ \ln(1.015^n) = \ln\left(\frac{10}{7}\right) $$
|
|
|
|
$$ n\ln(1.015) = \ln\left(\frac{10}{7}\right) $$
|
|
|
|
$$ n = \frac{\ln\left(\frac{10}{7}\right)}{\ln(1.015)} $$
|
|
|
|
$$ n \approx 24 \text{ months } = 2 \text{ years} $$
|
|
|
|
d. What is the total amount of money the person will have paid for the loan?
|
|
|
|
$$ 24 \cdot 150 = \$3600 $$
|
|
|
|
In 28-42 use mathematical induction to verify the correctness of the formula you
|
|
obtained in the referenced exercise.
|
|
|
|
28. Exercise 3
|
|
|
|
Let $a_0, a_1, a_2, \dots$ be the sequence defined recursively by $a_0 = 1$ and
|
|
$a_k = ka_{k - 1}$ for each integer $k \geq 1$.
|
|
|
|
Let the property $P(n)$ be the equation $a_n = n!$.
|
|
|
|
**Proof by mathematical induction:**
|
|
|
|
Prove $P(n)$ for every integer $n \geq 0$.
|
|
|
|
_Basis Step:_
|
|
|
|
Prove $P(0)$, that is:
|
|
|
|
$$ a_0 = 0! $$
|
|
|
|
$$ a_0 = 1 $$
|
|
|
|
Since $0! = 1$, and since by definition of the given sequence, $a_0 = 1$, the
|
|
equality holds and therefore $P(0)$ is true.
|
|
|
|
_Inductive Step:_
|
|
|
|
Let $k$ be any integer such that $k \geq 1$.
|
|
|
|
Suppose $P(k)$, that is:
|
|
|
|
$$ a_k = k! $$
|
|
|
|
This is the inductive hypothesis.
|
|
|
|
Prove $P(k + 1)$. That is:
|
|
|
|
$$ a_{k + 1} = (k + 1)! $$
|
|
|
|
By the definition of the given sequence:
|
|
|
|
$$ a_k = ka_{k - 1} $$
|
|
|
|
Then:
|
|
|
|
$$ a_{k + 1} = (k + 1) \cdot a_k $$
|
|
|
|
By substitution of the inductive hypothesis:
|
|
|
|
$$ a_{k + 1} = (k + 1) \cdot k! $$
|
|
|
|
By definition of factorial:
|
|
|
|
$$ a_{k + 1} = (k + 1)! $$
|
|
|
|
This is what was to be shown, therefore $P(k + 1)$ is true.
|
|
|
|
Q.E.D.
|
|
|
|
29. Exercise 4
|
|
|
|
Let $b_0, b_1, b_2, \dots$ be the sequence defined recursively by $b_0 = 1$ and
|
|
$b_k = \dfrac{b_{k - 1}}{1 + b_{k - 1}}$ for each integer $k \geq 1$.
|
|
|
|
Let the property $P(n)$ be the equation $b_n = \dfrac{1}{n + 1}$.
|
|
|
|
**Proof by mathematical induction:**
|
|
|
|
_Basis Step:_
|
|
|
|
Prove $P(0)$, that is:
|
|
|
|
$$ b_0 = \frac{1}{0 + 1} $$
|
|
|
|
$$ = \frac{1}{1} $$
|
|
|
|
$$ = 1 $$
|
|
|
|
This equality matches the given value of $b_0 = 1$, therefore $P(0)$ is true.
|
|
|
|
_Inductive Step:_
|
|
|
|
Let $k$ be any integer such that $k \geq 1$.
|
|
|
|
Suppose $P(k)$, that is:
|
|
|
|
$$ b_k = \dfrac{1}{k + 1} $$
|
|
|
|
This is the inductive hypothesis.
|
|
|
|
Prove $P(k + 1)$, that is:
|
|
|
|
$$ b_{k + 1} = \dfrac{1}{(k + 1) + 1} $$
|
|
|
|
Alternatively:
|
|
|
|
$$ b_{k + 1} = \dfrac{1}{k + 2} $$
|
|
|
|
By the definition of the given sequence:
|
|
|
|
$$ b_k = \dfrac{b_{k - 1}}{1 + b_{k - 1}} $$
|
|
|
|
Then:
|
|
|
|
$$ b_{k + 1} = \dfrac{b_k}{1 + b_k} $$
|
|
|
|
By substitution of the inductive hypothesis:
|
|
|
|
$$ b_{k + 1} = \frac{\dfrac{1}{k + 1}}{1 + \dfrac{1}{k + 1}} $$
|
|
|
|
$$ b_{k + 1} = \frac{1}{(k + 1)\left(1 + \dfrac{1}{k + 1}\right)} $$
|
|
|
|
$$ b_{k + 1} = \frac{1}{k + 1 + 1} $$
|
|
|
|
$$ b_{k + 1} = \frac{1}{k + 2} $$
|
|
|
|
This is what was to be shown. Therefore $P(k + 1)$ is true.
|
|
|
|
Q.E.D.
|
|
|
|
30. Exercise 5
|
|
|
|
Let $c_1, c_2, c_3, \dots$ be the sequence defined recursively by $c_1 = 1$ and
|
|
$c_k = 3c_{k - 1} + 1$ for each integer $k \geq 2$.
|
|
|
|
Let the property $P(n)$ be the equation $c_n = \frac{3^n - 1}{2}$.
|
|
|
|
**Proof by mathematical induction:**
|
|
|
|
_Basis Step:_
|
|
|
|
Prove $P(1)$, that is:
|
|
|
|
$$ c_1 = \frac{3^1 - 1}{2} $$
|
|
|
|
$$ = \frac{3 - 1}{2} $$
|
|
|
|
$$ = \frac{2}{2} $$
|
|
|
|
$$ = 1 $$
|
|
|
|
This matches the definition of the given sequence with $c_1 = 1$. Therefore
|
|
$P(1)$ is true.
|
|
|
|
_Inductive Step:_
|
|
|
|
Let $k$ be any integer such that $k \geq 2$.
|
|
|
|
Suppose $P(k)$, that is:
|
|
|
|
$$ c_k = \frac{3^k - 1}{2} $$
|
|
|
|
This is the inductive hypothesis.
|
|
|
|
Prove $P(k + 1)$, that is:
|
|
|
|
$$ c_{k + 1} = \frac{3^{k + 1} - 1}{2} $$
|
|
|
|
By the given sequence:
|
|
|
|
$$ c_k = 3c_{k - 1} + 1 $$
|
|
|
|
Then:
|
|
|
|
$$ c_{k + 1} = 3c_k + 1 $$
|
|
|
|
By substitution of the inductive hypothesis:
|
|
|
|
$$ = 3\left(\frac{3^k - 1}{2}\right) + 1 $$
|
|
|
|
$$ = \frac{3(3^k - 1)}{2} + 1 $$
|
|
|
|
$$ = \frac{3^{k + 1} - 3)}{2} + 1 $$
|
|
|
|
$$ = \frac{3^{k + 1} - 3}{2} + \frac{2}{2} $$
|
|
|
|
$$ = \frac{3^{k + 1} - 3 + 2}{2} $$
|
|
|
|
$$ = \frac{3^{k + 1} - 1}{2} $$
|
|
|
|
This is what was to be shown. Therefore $P(k + 1)$ is true.
|
|
|
|
Q.E.D.
|
|
|
|
31. Exercise 6
|
|
|
|
Let $d_1, d_2, d_3, \dots$ be the sequence defined recursively by $d_1 = 2$ and
|
|
$d_k = 2d_{k - 1} + 3$ for each integer $k \geq 2$.
|
|
|
|
Let the property $P(n)$ be the equation $d_n = 5 \cdot 2^{n - 1} - 3$.
|
|
|
|
**Proof by mathematical induction:**
|
|
|
|
_Basis Step:_
|
|
|
|
Prove $P(1)$, that is:
|
|
|
|
$$ d_1 = 5 \cdot 2^{1 - 1} - 3 $$
|
|
|
|
$$ = 5 \cdot 2^0 - 3 $$
|
|
|
|
$$ = 5 \cdot 1 - 3 $$
|
|
|
|
$$ = 5 - 3 $$
|
|
|
|
$$ = 2 $$
|
|
|
|
This equality matches the given value of $d_1 = 2$, therefore $P(1)$ is true.
|
|
|
|
_Inductive Step:_
|
|
|
|
Let $k$ be any integer such that $k \geq 2$.
|
|
|
|
Suppose $P(k)$, that is:
|
|
|
|
$$ d_k = 5 \cdot 2^{k - 1} - 3 $$
|
|
|
|
This is the inductive hypothesis.
|
|
|
|
Prove $P(k + 1)$, that is:
|
|
|
|
$$ d_{k + 1} = 5 \cdot 2^k - 3 $$
|
|
|
|
By the definition of the given sequence:
|
|
|
|
$$ d_k = 2d_{k - 1} + 3 $$
|
|
|
|
Then:
|
|
|
|
$$ d_{k + 1} = 2d_k + 3 $$
|
|
|
|
By substitution of the inductive hypothesis:
|
|
|
|
$$ = 2(5 \cdot 2^{k - 1} - 3) + 3 $$
|
|
|
|
$$ = 10 \cdot 2^{k - 1} - 6 + 3 $$
|
|
|
|
$$ = 5 \cdot 2 \cdot 2^{k - 1} - 3 $$
|
|
|
|
$$ = 5 \cdot 2^k - 3 $$
|
|
|
|
This is what was to be shown. Therefore $P(k + 1)$ is true.
|
|
|
|
Q.E.D.
|
|
|
|
32. Exercise 7
|
|
|
|
Let $e_0, e_1, e_2, \dots$ be the sequence defined recursively by $e_0 = 2$ and
|
|
$e_k = 4e_{k - 1} + 5$ for each integer $k \geq 1$.
|
|
|
|
Let the property $P(n)$ be the equation $e_n = \dfrac{11 \cdot 4^n - 5}{3}$.
|
|
|
|
**Proof by mathematical induction:**
|
|
|
|
_Basis Step:_
|
|
|
|
Prove $P(0)$, that is:
|
|
|
|
$$ e_0 = \dfrac{11 \cdot 4^0 - 5}{3} $$
|
|
|
|
$$ = \dfrac{11 \cdot 1 - 5}{3} $$
|
|
|
|
$$ = \dfrac{11 - 5}{3} $$
|
|
|
|
$$ = \dfrac{6}{3} $$
|
|
|
|
$$ = 2 $$
|
|
|
|
This equality matches the given value of $e_0 = 2$, therefore $P(0)$ is true.
|
|
|
|
_Inductive Step:_
|
|
|
|
Let $k$ be any integer such that $k \geq 1$.
|
|
|
|
Suppose $P(k)$, that is:
|
|
|
|
$$ e_k = \frac{11 \cdot 4^k - 5}{3} $$
|
|
|
|
This is the inductive hypothesis.
|
|
|
|
Prove $P(k + 1)$, that is:
|
|
|
|
$$ e_{k + 1} = \frac{11 \cdot 4^{k + 1} - 5}{3} $$
|
|
|
|
By the given sequence:
|
|
|
|
$$ e_k = 4e_{k - 1} + 5 $$
|
|
|
|
It follows that:
|
|
|
|
$$ e_{k + 1} = 4e_k + 5 $$
|
|
|
|
By substitution of the inductive hypothesis:
|
|
|
|
$$ = 4\left(\frac{11 \cdot 4^k - 5}{3}\right) + 5 $$
|
|
|
|
$$ = \frac{4(11 \cdot 4^k - 5)}{3} + 5 $$
|
|
|
|
$$ = \frac{44 \cdot 4^k - 20}{3} + 5 $$
|
|
|
|
$$ = \frac{11 \cdot 4 \cdot 4^k - 20}{3} + 5 $$
|
|
|
|
$$ = \frac{11 \cdot 4^{k + 1} - 20}{3} + 5 $$
|
|
|
|
$$ = \frac{11 \cdot 4^{k + 1} - 20}{3} + \frac{15}{3} $$
|
|
|
|
$$ = \frac{11 \cdot 4^{k + 1} - 20 + 15}{3} $$
|
|
|
|
$$ = \frac{11 \cdot 4^{k + 1} - 5}{3} $$
|
|
|
|
This is what was to be shown. Therefore $P(k + 1)$ is true.
|
|
|
|
Q.E.D.
|
|
|
|
33. Exercise 8
|
|
|
|
Let $f_1, f_2, f_3, \dots$ be the sequence defined recursively by $f_1 = 1$ and
|
|
$f_k = f_{k - 1} + 2^k$ for each integer $k \geq 2$.
|
|
|
|
Let the property $P(n)$ be the equation $f_n = 2^{n + 1} - 3$.
|
|
|
|
**Proof by mathematical induction:**
|
|
|
|
_Basis Step:_
|
|
|
|
Prove $P(1)$, that is:
|
|
|
|
$$ f_1 = 2^{1 + 1} - 3 $$
|
|
|
|
$$ = 2^2 - 3 $$
|
|
|
|
$$ = 4 - 3 $$
|
|
|
|
$$ = 1 $$
|
|
|
|
This equality matches the give value of $f_1 = 1$, therefore $P(1)$ is true.
|
|
|
|
_Inductive Step:_
|
|
|
|
Let $k$ be any integer such that $k \geq 2$.
|
|
|
|
Suppose $P(k)$, that is:
|
|
|
|
$$ f_k = 2^{k + 1} - 3 $$
|
|
|
|
This is the inductive hypothesis.
|
|
|
|
Prove $P(k + 1)$, that is:
|
|
|
|
$$ f_{k + 1} = 2^{k + 2} - 3 $$
|
|
|
|
By the given sequence:
|
|
|
|
$$ f_k = f_{k - 1} + 2^k $$
|
|
|
|
It follows that:
|
|
|
|
$$ f_{k + 1} = f_k + 2^{k + 1} $$
|
|
|
|
By substitution of the inductive hypothesis:
|
|
|
|
$$ = (2^{k + 1} - 3) + 2^{k + 1} $$
|
|
|
|
$$ = 2^{k + 2} - 3 $$
|
|
|
|
This is what was to be shown. Therefore $P(k + 1)$ is true.
|
|
|
|
Q.E.D.
|
|
|
|
34. Exercise 9
|
|
|
|
Let $g_1, g_2, g_3, \dots$ be the sequence defined recursively by $g_1 = 1$ and
|
|
$g_k = \dfrac{g_{k - 1}}{g_{k - 1} + 2}$ for each integer $k \geq 2$.
|
|
|
|
Let the property $P(n)$ be the equation $g_n = \frac{1}{2^n - 1}$.
|
|
|
|
**Proof by mathematical induction:**
|
|
|
|
_Basis Step:_
|
|
|
|
Prove $P(1)$, that is:
|
|
|
|
$$ g_1 = \frac{1}{2^1 - 1} $$
|
|
|
|
$$ = \frac{1}{2 - 1} $$
|
|
|
|
$$ = \frac{1}{1} $$
|
|
|
|
$$ = 1 $$
|
|
|
|
This equality matches the given value of $g_1 = 1$. Therefore $P(1)$ is true.
|
|
|
|
_Inductive Step:_
|
|
|
|
Let $k$ be any integer such that $k \geq 2$.
|
|
|
|
Suppose $P(k)$, that is:
|
|
|
|
$$ g_k = \frac{1}{2^k - 1} $$
|
|
|
|
This is the inductive hypothesis.
|
|
|
|
Prove $P(k + 1)$, that is:
|
|
|
|
$$ g_{k + 1} = \frac{1}{2^{k + 1} - 1} $$
|
|
|
|
By the given sequence:
|
|
|
|
$$ g_k = \dfrac{g_{k - 1}}{g_{k - 1} + 2} $$
|
|
|
|
It follows that:
|
|
|
|
$$ g_{k + 1} = \dfrac{g_k}{g_k + 2} $$
|
|
|
|
By substitution of the inductive hypothesis:
|
|
|
|
$$ = \dfrac{\dfrac{1}{2^k - 1}}{\left(\dfrac{1}{2^k - 1}\right) + 2} $$
|
|
|
|
$$ = \dfrac{1}{(2^k - 1)\left(\dfrac{1}{2^k - 1} + 2\right)} $$
|
|
|
|
$$ = \dfrac{1}{1 + 2(2^k - 1)} $$
|
|
|
|
$$ = \dfrac{1}{1 + 2^{k + 1} - 2} $$
|
|
|
|
$$ = \dfrac{1}{2^{k + 1} - 1} $$
|
|
|
|
This is what was to be shown. Therefore $P(k + 1)$ is true.
|
|
|
|
Q.E.D.
|
|
|
|
35. Exercise 10
|
|
|
|
Let $h_0, h_1, h_2, \dots$ be the sequence defined recursively by $h_0 = 1$ and
|
|
$h_k = 2^k - h_{k - 1}$ for each integer $k \geq 1$.
|
|
|
|
Let the property $P(n)$ be the equation
|
|
$h_n = \dfrac{2^{n + 1} - (-1)^{n + 1}}{3}$.
|
|
|
|
**Proof by mathematical induction:**
|
|
|
|
_Basis Step:_
|
|
|
|
Prove $P(0)$, that is:
|
|
|
|
$$ h_0 = \frac{2^{0 + 1} - (-1)^{0 + 1}}{3} $$
|
|
|
|
$$ = \frac{2^1 - (-1)^1}{3} $$
|
|
|
|
$$ = \frac{2 - (-1)}{3} $$
|
|
|
|
$$ = \frac{2 + 1}{3} $$
|
|
|
|
$$ = \frac{3}{3} $$
|
|
|
|
$$ = 1 $$
|
|
|
|
This equality matches the given value of $h_0 = 1$. Therefore $P(0)$ is true.
|
|
|
|
_Inductive Step:_
|
|
|
|
Let $k$ be any integer such that $k \geq 1$.
|
|
|
|
Suppose $P(k)$, that is:
|
|
|
|
$$ h_k = \frac{2^{k + 1} - (-1)^{k + 1}}{3} $$
|
|
|
|
This is the inductive hypothesis.
|
|
|
|
Prove $P(k + 1)$, that is:
|
|
|
|
$$ h_{k + 1} = \frac{2^{k + 2} - (-1)^{k + 2}}{3} $$
|
|
|
|
By the given sequence:
|
|
|
|
$$ h_k = 2^k - h_{k - 1} $$
|
|
|
|
It follows that:
|
|
|
|
$$ h_{k + 1} = 2^{k + 1} - h_k $$
|
|
|
|
By substitution of the inductive hypothesis:
|
|
|
|
$$ = 2^{k + 1} - \left(\frac{2^{k + 1} - (-1)^{k + 1}}{3}\right) $$
|
|
|
|
$$ = 2^{k + 1} - \frac{2^{k + 1} - (-1)^{k + 1}}{3} $$
|
|
|
|
$$ = \frac{3 \cdot 2^{k + 1}}{3} - \frac{2^{k + 1} - (-1)^{k + 1}}{3} $$
|
|
|
|
$$ = \frac{3 \cdot 2^{k + 1} - (2^{k + 1} - (-1)^{k + 1})}{3} $$
|
|
|
|
$$ = \frac{3 \cdot 2^{k + 1} - 2^{k + 1} + (-1)^{k + 1}}{3} $$
|
|
|
|
$$ = \frac{2 \cdot 2^{k + 1} + (-1)^{k + 1}}{3} $$
|
|
|
|
$$ = \frac{2^{k + 2} + (-1)^{k + 1}}{3} $$
|
|
|
|
$$ = \frac{2^{k + 2} + (-1)(-1)^{k + 2}}{3} $$
|
|
|
|
$$ = \frac{2^{k + 2} - (-1)^{k + 2}}{3} $$
|
|
|
|
This is what was to be shown. Therefore $P(k + 1)$ is true.
|
|
|
|
Q.E.D.
|
|
|
|
36. Exercise 11
|
|
|
|
Let $p_1, p_2, p_3, \dots$ be the sequence defined recursively by $p_1 = 2$ and
|
|
$p_k = p_{k - 1} + 2 \cdot 3^k$ for each integer $k \geq 2$.
|
|
|
|
Let the property $P(n)$ be the equation $p_n = 3^{n + 1} - 7$.
|
|
|
|
**Proof by mathematical induction:**
|
|
|
|
_Basis Step:_
|
|
|
|
Prove $P(1)$, that is:
|
|
|
|
$$ p_1 = 3^{1 + 1} - 7 $$
|
|
|
|
$$ = 3^2 - 7 $$
|
|
|
|
$$ = 9 - 7 $$
|
|
|
|
$$ = 2 $$
|
|
|
|
This equality matches the given value of $p_1 = 2$. Therefore $P(1)$ is true.
|
|
|
|
_Inductive Step:_
|
|
|
|
Let $k$ be any integer such that $k \geq 2$.
|
|
|
|
Suppose $P(k)$, that is:
|
|
|
|
$$ p_k = 3^{k + 1} - 7 $$
|
|
|
|
This is the inductive hypothesis.
|
|
|
|
Prove $P(k + 1)$, that is:
|
|
|
|
$$ p_{k + 1} = 3^{k + 2} - 7 $$
|
|
|
|
By the given sequence:
|
|
|
|
$$ p_k = p_{k - 1} + 2 \cdot 3^k $$
|
|
|
|
It follows that:
|
|
|
|
$$ p_{k + 1} = p_k + 2 \cdot 3^{k + 1} $$
|
|
|
|
By substitution of the inductive hypothesis:
|
|
|
|
$$ = (3^{k + 1} - 7) + 2 \cdot 3^{k + 1} $$
|
|
|
|
$$ = -7 + 3 \cdot 3^{k + 1} $$
|
|
|
|
$$ = -7 + 3^{k + 2} $$
|
|
|
|
$$ = 3^{k + 2} - 7 $$
|
|
|
|
This is what was to be shown. Therefore $P(k + 1)$ is true.
|
|
|
|
Q.E.D.
|
|
|
|
37. Exercise 12
|
|
|
|
Let $s_0, s_1, s_2, \dots$ be the sequence defined recursively by $s_0 = 3$ and
|
|
$s_k = s_{k - 1} + 2k$ for each integer $k \geq 1$.
|
|
|
|
Let the property $P(n)$ be the equation $s_n = n^2 + n + 3$.
|
|
|
|
**Proof by mathematical induction:**
|
|
|
|
_Basis Step:_
|
|
|
|
Prove $P(0)$, that is:
|
|
|
|
$$ s_0 = 0^2 + 0 + 3 $$
|
|
|
|
$$ = 0 + 0 + 3 $$
|
|
|
|
$$ = 3 $$
|
|
|
|
This equality matches the given value of $s_0 = 3$. Therefore $P(0)$ is true.
|
|
|
|
_Inductive Step:_
|
|
|
|
Let $k$ be any integer such that $k \geq 1$.
|
|
|
|
Suppose $P(k)$, that is:
|
|
|
|
$$ s_k = k^2 + k + 3 $$
|
|
|
|
This is the inductive hypothesis.
|
|
|
|
Prove $P(k + 1)$, that is:
|
|
|
|
$$ s_{k + 1} = (k + 1)^2 + (k + 1) + 3 $$
|
|
|
|
Alternatively:
|
|
|
|
$$ s_{k + 1} = (k + 1)(k + 1) + k + 4 $$
|
|
|
|
$$ s_{k + 1} = k^2 + 2k + 1 + k + 4 $$
|
|
|
|
$$ s_{k + 1} = k^2 + 3k + 5 $$
|
|
|
|
By the given sequence:
|
|
|
|
$$ s_k = s_{k - 1} + 2k $$
|
|
|
|
It follows that:
|
|
|
|
$$ s_{k + 1} = s_k + 2(k + 1) $$
|
|
|
|
$$ s_{k + 1} = s_k + 2k + 2 $$
|
|
|
|
By substitution of the inductive hypothesis:
|
|
|
|
$$ = (k^2 + k + 3) + 2k + 2 $$
|
|
|
|
$$ = k^2 + 3k + 5 $$
|
|
|
|
This is what was to be shown. Therefore $P(k + 1)$ is true.
|
|
|
|
Q.E.D.
|
|
|
|
38. Exercise 13
|
|
|
|
Let $t_0, t_1, t_2, \dots$ be the sequence defined recursively by $t_0 = 0$ and
|
|
$t_k = t_{k - 1} + 3k + 1$ for each integer $k \geq 1$.
|
|
|
|
Let the property $P(n)$ be the equation $t_n = \frac{3n^2 + 5n}{2}$.
|
|
|
|
**Proof by mathematical induction:**
|
|
|
|
_Basis Step:_
|
|
|
|
Prove $P(0)$, that is:
|
|
|
|
$$ t_0 = \frac{3(0)^2 + 5(0)}{2} $$
|
|
|
|
$$ = \frac{3(0) + 0}{2} $$
|
|
|
|
$$ = \frac{0 + 0}{2} $$
|
|
|
|
$$ = \frac{0}{2} $$
|
|
|
|
$$ = 0 $$
|
|
|
|
This equality matches the given value of $t_0 = 0$. Therefore $P(0)$ is true.
|
|
|
|
_Inductive Step:_
|
|
|
|
Let $k$ be any integer such that $k \geq 1$.
|
|
|
|
Suppose $P(k)$, that is:
|
|
|
|
$$ t_k = \frac{3k^2 + 5k}{2} $$
|
|
|
|
This is the inductive hypothesis.
|
|
|
|
Prove $P(k + 1)$, that is:
|
|
|
|
$$ t_{k + 1} = \frac{3(k + 1)^2 + 5(k + 1)}{2} $$
|
|
|
|
Alternatively:
|
|
|
|
$$ t_{k + 1} = \frac{3(k + 1)(k + 1) + 5k + 5}{2} $$
|
|
|
|
$$ t_{k + 1} = \frac{3(k^2 + 2k + 1) + 5k + 5}{2} $$
|
|
|
|
$$ t_{k + 1} = \frac{3k^2 + 6k + 3 + 5k + 5}{2} $$
|
|
|
|
$$ t_{k + 1} = \frac{3k^2 + 11k + 8}{2} $$
|
|
|
|
By the given sequence:
|
|
|
|
$$ t_k = t_{k - 1} + 3k + 1 $$
|
|
|
|
It follows that:
|
|
|
|
$$ t_{k + 1} = t_k + 3(k + 1) + 1 $$
|
|
|
|
$$ t_{k + 1} = t_k + 3k + 3 + 1 $$
|
|
|
|
$$ t_{k + 1} = t_k + 3k + 4 $$
|
|
|
|
By substitution of the inductive hypothesis:
|
|
|
|
$$ = \left(\frac{3k^2 + 5k}{2}\right) + 3k + 4 $$
|
|
|
|
$$ = \frac{3k^2 + 5k}{2} + \frac{6k}{2} + \frac{8}{2} $$
|
|
|
|
$$ = \frac{3k^2 + 5k + 6k + 8}{2} $$
|
|
|
|
$$ = \frac{3k^2 + 11k + 8}{2} $$
|
|
|
|
This is what was to be shown. Therefore $P(k + 1)$ is true.
|
|
|
|
Q.E.D.
|
|
|
|
39. Exercise 14
|
|
|
|
Let $x_1, x_2, x_3, \dots$ be the sequence defined recursively by $x_1 = 1$ and
|
|
$x_k = 3x_{k - 1} + k$ for each integer $k \geq 2$.
|
|
|
|
Let the property $P(n)$ be the equation
|
|
$x_n = \frac{1}{4}\left[3^{n + 1} - 3 - 2n\right]$.
|
|
|
|
**Proof by mathematical induction:**
|
|
|
|
_Basis Step:_
|
|
|
|
Prove $P(1)$, that is:
|
|
|
|
$$ x_1 = \frac{1}{4}\left[3^{1 + 1} - 3 - 2(1)\right] $$
|
|
|
|
$$ = \frac{1}{4}\left[3^2 - 3 - 2\right] $$
|
|
|
|
$$ = \frac{1}{4}\left[9 - 5\right] $$
|
|
|
|
$$ = \frac{1}{4}(4) $$
|
|
|
|
$$ = 1 $$
|
|
|
|
This equality matches the given value of $x_1 = 1$. Therefore $P(1)$ is true.
|
|
|
|
_Inductive Step:_
|
|
|
|
Let $k$ be any integer such that $k \geq 2$.
|
|
|
|
Suppose $P(k)$, that is:
|
|
|
|
$$ x_k = \frac{1}{4}\left[3^{k + 1} - 3 - 2k\right] $$
|
|
|
|
This is the inductive hypothesis.
|
|
|
|
Prove $P(k + 1)$, that is:
|
|
|
|
$$ x_{k + 1} = \frac{1}{4}\left[3^{k + 2} - 3 - 2(k + 1)\right] $$
|
|
|
|
Alternatively:
|
|
|
|
$$ x_{k + 1} = \frac{1}{4}\left[3^{k + 2} - 3 - 2k - 2\right] $$
|
|
|
|
$$ x_{k + 1} = \frac{1}{4}\left[3^{k + 2} - 2k - 5\right] $$
|
|
|
|
By the given sequence:
|
|
|
|
$$ x_k = 3x_{k - 1} + k $$
|
|
|
|
It follows that:
|
|
|
|
$$ x_{k + 1} = 3x_k + k + 1 $$
|
|
|
|
By substitution of the inductive hypothesis:
|
|
|
|
$$ = 3\left[\frac{1}{4}\left(3^{k + 1} - 3 - 2k\right)\right] + k + 1 $$
|
|
|
|
$$ = \frac{1}{4}\left(3(3^{k + 1} - 3 - 2k)\right) + k + 1 $$
|
|
|
|
$$ = \frac{1}{4}(3^{k + 2} - 9 - 6k) + k + 1 $$
|
|
|
|
$$ = \frac{1}{4}(3^{k + 2} - 9 - 6k) + \frac{4k}{4} + \frac{4}{4} $$
|
|
|
|
$$ = \frac{1}{4}(3^{k + 2} - 9 - 6k + 4k + 4) $$
|
|
|
|
$$ = \frac{1}{4}(3^{k + 2} - 5 - 2k) $$
|
|
|
|
$$ = \frac{1}{4}(3^{k + 2} - 2k - 5) $$
|
|
|
|
This is what was to be shown. Therefore $P(k + 1)$ is true.
|
|
|
|
Q.E.D.
|
|
|
|
40. Exercise 15
|
|
|
|
Let $y_1, y_2, y_3, \dots$ be the sequence defined recursively by $y_1 = 1$ and
|
|
$y_k = y_{k - 1} + k^2$ for each integer $k \geq 2$.
|
|
|
|
Let the property $P(n)$ be the equation $y_n = \frac{n(n + 1)(2n + 1)}{6}$.
|
|
|
|
**Proof by mathematical induction:**
|
|
|
|
_Basis Step:_
|
|
|
|
Prove $P(1)$, that is:
|
|
|
|
$$ y_1 = \frac{1(1 + 1)(2(1) + 1)}{6} $$
|
|
|
|
$$ = \frac{1(2)(2 + 1)}{6} $$
|
|
|
|
$$ = \frac{(2)(3)}{6} $$
|
|
|
|
$$ = \frac{6}{6} $$
|
|
|
|
$$ = 1 $$
|
|
|
|
This equality matches the given value of $y_1 = 1$. Therefore $P(1)$ is true.
|
|
|
|
_Inductive Step:_
|
|
|
|
Let $k$ be any integer such that $k \geq 2$.
|
|
|
|
Suppose $P(k)$, that is:
|
|
|
|
$$ y_k = \frac{k(k + 1)(2k + 1)}{6} $$
|
|
|
|
This is the inductive hypothesis.
|
|
|
|
Prove $P(k + 1)$, that is:
|
|
|
|
$$ y_{k + 1} = \frac{(k + 1)(k + 2)(2(k + 1) + 1)}{6} $$
|
|
|
|
Alternatively:
|
|
|
|
$$ y_{k + 1} = \frac{(k^2 + 3k + 2)(2k + 2 + 1)}{6} $$
|
|
|
|
$$ y_{k + 1} = \frac{(k^2 + 3k + 2)(2k + 3)}{6} $$
|
|
|
|
$$ y_{k + 1} = \frac{k^2(2k + 3) + 3k(2k + 3) + 2(2k + 3)}{6} $$
|
|
|
|
$$ y_{k + 1} = \frac{2k^3 + 3k^2 + 6k^2 + 9k + 4k + 6}{6} $$
|
|
|
|
$$ y_{k + 1} = \frac{2k^3 + 9k^2 + 13k + 6}{6} $$
|
|
|
|
By the given sequence:
|
|
|
|
$$ y_k = y_{k - 1} + k^2 $$
|
|
|
|
It follows that:
|
|
|
|
$$ y_{k + 1} = y_k + (k + 1)^2 $$
|
|
|
|
By substitution of the inductive hypothesis:
|
|
|
|
$$ = \left(\frac{k(k + 1)(2k + 1)}{6}\right) + (k + 1)^2 $$
|
|
|
|
$$ = \frac{k(k + 1)(2k + 1)}{6} + \frac{6(k + 1)^2}{6} $$
|
|
|
|
$$ = \frac{k(k + 1)(2k + 1) + 6(k + 1)^2}{6} $$
|
|
|
|
$$ = \frac{k(2k^2 + 3k + 1) + 6(k^2 + 2k + 1)}{6} $$
|
|
|
|
$$ = \frac{2k^3 + 3k^2 + k + 6k^2 + 12k + 6}{6} $$
|
|
|
|
$$ = \frac{2k^3 + 9k^2 + 13k + 6}{6} $$
|
|
|
|
This is what was to be shown. Therefore $P(k + 1)$ is true.
|
|
|
|
41. Exercise 16
|
|
|
|
Let $a_1, a_2, a_3, \dots$ be the sequence defined recursively by $a_1 = 2$ and
|
|
$a_k = 3a_{k - 1} + 2$ for each integer $k \geq 2$.
|
|
|
|
Let the property $P(n)$ be the equation $a_n = 3^n - 1$.
|
|
|
|
**Proof by mathematical induction:**
|
|
|
|
_Basis Step:_
|
|
|
|
Prove $P(1)$, that is:
|
|
|
|
$$ a_1 = 3^1 - 1 $$
|
|
|
|
$$ = 3 - 1 $$
|
|
|
|
$$ = 2 $$
|
|
|
|
This equality matches the given value of $a_1 = 2$. Therefore $P(1)$ is true.
|
|
|
|
_Inductive Step:_
|
|
|
|
Let $k$ be any integer such that $k \geq 2$.
|
|
|
|
Suppose $P(k)$, that is:
|
|
|
|
$$ a_k = 3^k - 1 $$
|
|
|
|
This is the inductive hypothesis.
|
|
|
|
Prove $P(k + 1)$, that is:
|
|
|
|
$$ a_{k + 1} = 3^{k + 1} - 1 $$
|
|
|
|
By the given sequence:
|
|
|
|
$$ a_k = 3a_{k - 1} + 2 $$
|
|
|
|
It follows that:
|
|
|
|
$$ a_{k + 1} = 3a_k + 2 $$
|
|
|
|
By substitution of the inductive hypothesis:
|
|
|
|
$$ = 3(3^k - 1) + 2 $$
|
|
|
|
$$ = 3^{k + 1} - 3 + 2 $$
|
|
|
|
$$ = 3^{k + 1} - 1 $$
|
|
|
|
This is what was to be shown. Therefore $P(k + 1)$ is true.
|
|
|
|
Q.E.D.
|
|
|
|
42. Exercise 17
|
|
|
|
Let $t_1, t_2, t_3, \dots$ be the sequence defined recursively by $t_1 = 2$ and
|
|
$t_k = 3t_{k - 1} + 2$ for each integer $k \geq 2$.
|
|
|
|
Let the property $P(n)$ be the equation $t_n = 3^n - 1$.
|
|
|
|
**Proof by mathematical induction:**
|
|
|
|
_Basis Step:_
|
|
|
|
Prove $P(1)$, that is:
|
|
|
|
$$ t_1 = 3^1 - 1 $$
|
|
|
|
$$ = 3 - 1 $$
|
|
|
|
$$ = 2 $$
|
|
|
|
This equality matches the given value of $t_1 = 2$. Therefore $P(1)$ is true.
|
|
|
|
_Inductive Step:_
|
|
|
|
Let $k$ be any integer such that $k \geq 2$.
|
|
|
|
Suppose $P(k)$, that is:
|
|
|
|
$$ t_k = 3^k - 1 $$
|
|
|
|
This is the inductive hypothesis.
|
|
|
|
Prove $P(k + 1)$, that is:
|
|
|
|
$$ t_{k + 1} = 3^{k + 1} - 1 $$
|
|
|
|
By the given sequence:
|
|
|
|
$$ t_k = 3t_{k - 1} + 2 $$
|
|
|
|
It follows that:
|
|
|
|
$$ t_{k + 1} = 3t_k + 2 $$
|
|
|
|
By substitution of the inductive hypothesis:
|
|
|
|
$$ = 3(3^k - 1) + 2 $$
|
|
|
|
$$ = 3^{k + 1} - 3 + 2 $$
|
|
|
|
$$ = 3^{k + 1} - 1 $$
|
|
|
|
This is what was to be shown. Therefore $P(k + 1)$ is true.
|
|
|
|
Q.E.D.
|
|
|
|
In each of 43-49 a sequence is defined recursively. (a) Use iteration to guess
|
|
an explicit formula for the sequence. (b) Use strong mathematical induction to
|
|
verify that the formula of part (a) is correct.
|
|
|
|
43. $a_k = \dfrac{a_{k - 1}}{2a_{k - 1} - 1}$, for each integer $k \geq 1$
|
|
$a_0 = 2$.
|
|
|
|
a.
|
|
|
|
$$ a_0 = 2 $$
|
|
|
|
$$ a_1 = \frac{a_0}{2a_0 - 1} = \frac{2}{2(2) - 1} = \frac{2}{3} $$
|
|
|
|
$$ a_2 = \frac{a_1}{2a_1 - 1} = \frac{\dfrac{2}{3}}{2\left(\dfrac{2}{3}\right) - 1} = \frac{2}{4 - 3} = 2 $$
|
|
|
|
$$ a_3 = \frac{a_2}{2a_2 - 1} = \frac{2}{2(2) - 1} = \frac{2}{3} $$
|
|
|
|
Guess:
|
|
|
|
$$
|
|
a_n =
|
|
\begin{cases}
|
|
2 & \text{if } n \text{ is even} \\
|
|
\dfrac{2}{3} & n \text{ is odd}
|
|
\end{cases}
|
|
$$
|
|
|
|
b.
|
|
|
|
Let $a_0, a_1, a_2, \dots$ be the sequence defined recursively by $a_0 = 2$ and
|
|
$a_k = \dfrac{a_{k - 1}}{2a_{k - 1} - 1}$ for each integer $k \geq 1$.
|
|
|
|
Let the property $P(n)$ be the equation:
|
|
|
|
$$
|
|
a_n =
|
|
\begin{cases}
|
|
2 & \text{if } n \text{ is even} \\
|
|
\dfrac{2}{3} & \text{if } n \text{ is odd}
|
|
\end{cases}
|
|
$$
|
|
|
|
**Proof by strong mathematical induction:**
|
|
|
|
_Basis Step:_
|
|
|
|
Prove $P(0)$ and $P(1)$, that is:
|
|
|
|
$$
|
|
a_0 =
|
|
\begin{cases}
|
|
2 & \text{if } 0 \text{ is even} \\
|
|
\dfrac{2}{3} & \text{if } 0 \text{ is odd}
|
|
\end{cases}
|
|
$$
|
|
|
|
and:
|
|
|
|
$$
|
|
a_1 =
|
|
\begin{cases}
|
|
2 & \text{if } 1 \text{ is even} \\
|
|
\dfrac{2}{3} & \text{if } 1 \text{ is odd}
|
|
\end{cases}
|
|
$$
|
|
|
|
$P(0)$ is true since the piecewise function tells us that $a_0 = 2$ since $0$ is
|
|
even and this matches the given value of $a_0 = 2$.
|
|
|
|
$P(1)$ is true given the evaluation of $a_1$ in part (a).
|
|
|
|
Therefore both $P(0)$ and $P(1)$ are true.
|
|
|
|
_Inductive Step:_
|
|
|
|
Let $k$ be any integer such that $k \geq 0$. Let $i$ be some integer such that
|
|
$0 \leq i \leq k$.
|
|
|
|
Suppose $P(i)$, that is:
|
|
|
|
$$
|
|
a_i =
|
|
\begin{cases}
|
|
2 & \text{if } i \text{ is even} \\
|
|
\dfrac{2}{3} & \text{if } i \text{ is odd}
|
|
\end{cases}
|
|
$$
|
|
|
|
This is the inductive hypothesis.
|
|
|
|
Prove $P(k + 1)$, that is:
|
|
|
|
$$
|
|
a_{k + 1} =
|
|
\begin{cases}
|
|
2 & \text{if } k + 1 \text{ is even} \\
|
|
\dfrac{2}{3} & \text{if } k + 1 \text{ is odd}
|
|
\end{cases}
|
|
$$
|
|
|
|
By the definition of the sequence:
|
|
|
|
$$ a_k = \dfrac{a_{k - 1}}{2a_{k - 1} - 1} $$
|
|
|
|
It follows that:
|
|
|
|
$$ a_{k + 1} = \dfrac{a_k}{2a_k - 1} $$
|
|
|
|
By the inductive hypothesis:
|
|
|
|
$$
|
|
a_{k + 1} =
|
|
\begin{cases}
|
|
\dfrac{2}{2(2) - 1} & \text{if } k \text{ is even} \\
|
|
\dfrac{\dfrac{2}{3}}{2\left(\dfrac{2}{3}\right) - 1} & \text{if } k \text{ is odd}
|
|
\end{cases}
|
|
$$
|
|
|
|
$$
|
|
= \\
|
|
\begin{cases}
|
|
\dfrac{2}{3} & \text{if } k \text{ is even} \\
|
|
2 & \text{if } k \text{ is odd}
|
|
\end{cases}
|
|
$$
|
|
|
|
It follows that:
|
|
|
|
$$
|
|
= \\
|
|
\begin{cases}
|
|
\dfrac{2}{3} & \text{if } k + 1 \text{ is odd} \\
|
|
2 & \text{if } k + 1 \text{ is even}
|
|
\end{cases}
|
|
$$
|
|
|
|
This is what was to be shown. Therefore $P(k + 1)$ is true.
|
|
|
|
Q.E.D.
|
|
|
|
44. $b_k = \dfrac{2}{b_{k - 1}}$, for each integer $k \geq 2$ $b_1 = 1$.
|
|
|
|
a.
|
|
|
|
$$ b_1 = 1 $$
|
|
|
|
$$ b_2 = \frac{2}{b_1} = \frac{2}{1} = 2 $$
|
|
|
|
$$ b_3 = \frac{2}{b_2} = \frac{2}{2} = 1 $$
|
|
|
|
Guess:
|
|
|
|
$$
|
|
b_n =
|
|
\begin{cases}
|
|
1 & \text{if } n \text{ is odd} \\
|
|
2 & \text{if } n \text{ is even}
|
|
\end{cases}
|
|
$$
|
|
|
|
b.
|
|
|
|
Let $b_1, b_2, b_3, \dots$ be the sequence defined recursively by $b_1 = 1$ and
|
|
$b_k = \dfrac{2}{b_{k - 1}}$ for each integer $k \geq 2$.
|
|
|
|
**Proof by strong mathematical induction:**
|
|
|
|
Let the property $P(n)$ be the equation:
|
|
|
|
$$
|
|
b_n =
|
|
\begin{cases}
|
|
1 & \text{if } n \text{ is odd} \\
|
|
2 & \text{if } n \text{ is even}
|
|
\end{cases}
|
|
$$
|
|
|
|
_Basis Step:_
|
|
|
|
Prove $P(1)$ and $P(2)$.
|
|
|
|
Both $P(1)$ and $P(2)$ are proven true in part (a).
|
|
|
|
_Inductive Step:_
|
|
|
|
Let $k$ be any integer such that $k \geq 2$, and let $i$ be some integer such
|
|
that $1 \leq i \leq k$.
|
|
|
|
Suppose $P(i)$, that is:
|
|
|
|
$$
|
|
b_i =
|
|
\begin{cases}
|
|
1 & \text{if } i \text{ is odd} \\
|
|
2 & \text{if } i \text{ is even}
|
|
\end{cases}
|
|
$$
|
|
|
|
This is the inductive hypothesis.
|
|
|
|
Prove $P(k + 1)$, that is:
|
|
|
|
$$
|
|
b_{k + 1} =
|
|
\begin{cases}
|
|
1 & \text{if } k + 1 \text{ is odd} \\
|
|
2 & \text{if } k + 1 \text{ is even}
|
|
\end{cases}
|
|
$$
|
|
|
|
By the definition of the given sequence:
|
|
|
|
$$ b_k = \dfrac{2}{b_{k - 1}} $$
|
|
|
|
It follows that:
|
|
|
|
$$ b_{k + 1} = \dfrac{2}{b_k} $$
|
|
|
|
By the inductive hypothesis:
|
|
|
|
$$
|
|
b_{k + 1} =
|
|
\begin{cases}
|
|
\dfrac{2}{1} & \text{if } k \text{ is odd} \\
|
|
\dfrac{2}{2} & \text{if } k \text{ is even}
|
|
\end{cases}
|
|
$$
|
|
|
|
$$
|
|
= \\
|
|
\begin{cases}
|
|
2 & \text{if } k \text{ is odd} \\
|
|
1 & \text{if } k \text{ is even}
|
|
\end{cases}
|
|
$$
|
|
|
|
$$
|
|
= \\
|
|
\begin{cases}
|
|
2 & \text{if } k + 1 \text{ is even} \\
|
|
1 & \text{if } k + 1 \text{ is odd}
|
|
\end{cases}
|
|
$$
|
|
|
|
This is what was to be shown. Therefore $P(k + 1)$ is true.
|
|
|
|
Q.E.D.
|
|
|
|
45. $v_k = v_{\lfloor \dfrac{k}{2} \rfloor} + v_{\lfloor \dfrac{(k + 1)}{2}\rfloor} + 2$,
|
|
for each integer $k \geq 2$ $v_1 = 1$.
|
|
|
|
a.
|
|
|
|
$$ v_1 = 1 $$
|
|
|
|
$$ v_2 = v_{\lfloor \dfrac{2}{2} \rfloor} + v_{\lfloor \dfrac{(2 + 1)}{2} \rfloor} + 2 $$
|
|
|
|
$$ = v_1 + v_1 + 2 = 1 + 1 + 2 $$
|
|
|
|
$$ v_3 = v_{\lfloor \dfrac{3}{2} \rfloor} + v_{\lfloor \dfrac{(3 + 1)}{2} \rfloor} + 2 $$
|
|
|
|
$$ = v_1 + v_2 + 2 = 1 + (1 + 1 + 2) + 2 $$
|
|
|
|
$$ = 3 + 2 \cdot 2 $$
|
|
|
|
$$ v_4 = v_{\lfloor \dfrac{4}{2} \rfloor} + v_{\lfloor \dfrac{(4 + 1)}{2} \rfloor} + 2 $$
|
|
|
|
$$ = v_2 + v_2 + 2 = (1 + 1 + 2) + (1 + 1 + 2) + 2 $$
|
|
|
|
$$ = 4 + 3 \cdot 2 $$
|
|
|
|
$$ v_5 = v_{\lfloor \dfrac{5}{2} \rfloor} + v_{\lfloor \dfrac{(5 + 1)}{2} \rfloor} + 2 $$
|
|
|
|
$$ = v_2 + v_3 + 2 = (1 + 1 + 2) + (3 + 2 \cdot 2) + 2 $$
|
|
|
|
$$ = 5 + 4 \cdot 2 $$
|
|
|
|
Guess:
|
|
|
|
$$ v_n = n + 2(n - 1) = n + 2n - 2 $$
|
|
|
|
$$ = 3n - 2 $$
|
|
|
|
b.
|
|
|
|
Let $v_1, v_2, v_3, \dots$ be the sequence defined recursively by $v_1 = 1$ and
|
|
$v_k = v_{\lfloor \frac{k}{2} \rfloor} + v_{\lfloor \frac{(k + 1)}{2} \rfloor} + 2$
|
|
for each integer $k \geq 2$.
|
|
|
|
**Proof by strong mathematical induction:**
|
|
|
|
Let the property $P(n)$ be the equation $v_n = 3n - 2$.
|
|
|
|
_Basis Step:_
|
|
|
|
Prove $P(1)$, that is:
|
|
|
|
$$ v_1 = 3(1) - 2 $$
|
|
|
|
$$ = 3 - 2 $$
|
|
|
|
$$ = 1 $$
|
|
|
|
This equality matches the given value of $v_1 = 1$. Therefore $P(1)$ is true.
|
|
|
|
_Inductive Step:_
|
|
|
|
Let $k$ be any integer such that $k \geq 2$, and let $i$ be some integer such
|
|
that $1 \leq i \leq k$.
|
|
|
|
Suppose $P(i)$, that is:
|
|
|
|
$$ v_i = 3i - 2 $$
|
|
|
|
This is the inductive hypothesis.
|
|
|
|
Prove $P(k + 1)$, that is:
|
|
|
|
$$ v_{k + 1} = 3(k + 1) - 2 $$
|
|
|
|
Alternatively:
|
|
|
|
$$ v_{k + 1} = 3k + 1 $$
|
|
|
|
By the definition of the given sequence:
|
|
|
|
$$ v_k = v_{\lfloor \frac{k}{2} \rfloor} + v_{\lfloor \frac{(k + 1)}{2} \rfloor} + 2 $$
|
|
|
|
It follows that:
|
|
|
|
$$ v_{k + 1} = v_{\lfloor \frac{(k + 1)}{2} \rfloor} + v_{\lfloor \frac{(k + 2)}{2} \rfloor} + 2 $$
|
|
|
|
By the inductive hypothesis:
|
|
|
|
$$ = \left(3\lfloor \frac{k + 1}{2} \rfloor - 2\right) + \left(3\lfloor \frac{k + 2}{2}\rfloor - 2\right) + 2 $$
|
|
|
|
$$ = 3\left(\lfloor \frac{k + 1}{2} \rfloor + \lfloor \frac{k + 2}{2} \rfloor \right) - 2 $$
|
|
|
|
$$
|
|
= \\
|
|
\begin{cases}
|
|
3\left(\frac{k}{2} + \frac{k + 2}{2}\right) - 2 & \text{if } k \text{ is even} \\
|
|
3\left(\frac{k + 1}{2} + \frac{k + 1}{2}\right) - 2 & \text{if } k \text{ is odd}
|
|
\end{cases}
|
|
$$
|
|
|
|
$$ = 3\left(\frac{2k + 2}{2}\right) - 2 $$
|
|
|
|
$$ = 3(k + 1) - 2 $$
|
|
|
|
$$ = 3k + 3 - 2 $$
|
|
|
|
$$ = 3k + 1 $$
|
|
|
|
This is what was to be shown. Therefore $P(k + 1)$ is true.
|
|
|
|
Q.E.D.
|
|
|
|
46. $s_k = 2s_{k - 2}$, for each integer $k \geq 2$ $s_0 = 1$, $s_1 = 2$.
|
|
|
|
Omitted.
|
|
|
|
47. $t_k = k - t_{k - 1}$, for each integer $k \geq 1$ $t_0 = 0$.
|
|
|
|
Omitted.
|
|
|
|
48. $w_k = w_{k - 2} + k$, for each integer $k \geq 3$ $w_1 = 1$, $w_2 = 2$.
|
|
|
|
Omitted.
|
|
|
|
49. $u_k = u_{k - 2} \cdot u_{k - 1}$, for each integer $k \geq 2$
|
|
$u_0 = u_1 = 2$
|
|
|
|
Omitted.
|
|
|
|
In 50 and 51 determine whether the given recursively defined sequence satisfies
|
|
the explicit formula $a_n = (n - 1)^2$, for every integer $n \geq 1$.
|
|
|
|
50. $a_k = 2a_{k - 1} + k - 1$, for each integer $k \geq 2$ $a_1 = 0$.
|
|
|
|
Omitted.
|
|
|
|
51. $a_k = 4a_{k - 1} - k + 3$, for each integer $k \geq 2$ $a_1 = 0$.
|
|
|
|
Omitted.
|
|
|
|
52. A single line divides a plane into two regions. Two lines (by crossing) can
|
|
divide a plane into four regions; three lines can divide it into seven
|
|
regions (see the figure). Let $P_n$ be the maximum number of regions into
|
|
which $n$ lines divide a plane, where $n$ is a positive integer.
|
|
|
|
[See Page 375 for image]
|
|
|
|
a. Derive a recurrence relation for $P_k$ in terms of $P_{k - 1}$, for each
|
|
integer $k \geq 2$.
|
|
|
|
Omitted.
|
|
|
|
b. Use iteration to guess an explicit formula for $P_n$.
|
|
|
|
Omitted.
|
|
|
|
53. Compute $\left[\begin{array}{cc} 1 & 1 \\ 1 & 0 \\ \end{array}\right]^n$ for
|
|
small values of $n$ (up to about 5 or 6). Conjecture explicit formulas for
|
|
the entries in this matrix, and prove your conjecture using mathematical
|
|
induction.
|
|
|
|
Omitted.
|
|
|
|
54. In economics the behavior of an economy from one period to another is often
|
|
modeled by recurrence relations. Let $Y_k$ be the income in period $k$ and
|
|
$C_k$ be the consumption in period $k$. In one economic model, income in any
|
|
period is assumed to be the sum of consumption in that period plus
|
|
investment and government expenditures (which are assumed to be constant
|
|
from period to period), and consumption in each period is assumed to be a
|
|
linear function of the income of the preceding period. That is,
|
|
|
|
$$ Y_k = C_k + E $$
|
|
|
|
where $E$ is the sum of investment plus government expenditures.
|
|
|
|
$$ C_k = c + mY_{k - 1} $$
|
|
|
|
where $c$ and $m$ are constants.
|
|
|
|
Substituting the second equation into the first gives
|
|
$Y_k = E + c + mY_{k - 1}$.
|
|
|
|
a. Use iteration on the above recurrence relation to obtain
|
|
|
|
Omitted.
|
|
|
|
$$ Y_n = (E + c)\left(\frac{m^n - 1}{m - 1}\right) + m^nY_0 $$
|
|
|
|
for every integer $n \geq 1$.
|
|
|
|
b. (For students who have studied calculus) Show that if $0 < m < 1$, then
|
|
$\lim\limits_{m \to \infty}Y_n = \dfrac{E + c}{1 - m}$.
|
|
|
|
Omitted.
|
|
|
|
---
|
|
|
|
Page 385
|
|
|
|
**Exercise Set 5.8**
|
|
|
|
1. Which of the following are second-order linear homogeneous recurrence
|
|
relations with constant coefficients?
|
|
|
|
a. $a_k = 2a_{k - 1} - 5a_{k - 2}$
|
|
|
|
Yes, $A = 2$ and $B = -5$
|
|
|
|
b. $b_k = kb_{k - 1} + b_{k - 2}$
|
|
|
|
No, $A = k$, which is not a constant coefficient.
|
|
|
|
c. $c_k = 3c_{k - 1} \cdot c_{k - 2}^2$
|
|
|
|
No, this does not take the form of a linear homogeneous recurrence relation with
|
|
constant coefficients. The first term $3c_{k - 1}$ is fine, but then there is
|
|
multiplication instead of addition/subtraction, and $c_{k - 2}^2$ violates the
|
|
homogeneous rule (each term must have the same total degree).
|
|
|
|
d. $d_k = 3d_{k - 1} + d_{k - 2}$
|
|
|
|
Yes, $A = 3$, $B = 1$.
|
|
|
|
e. $r_k = r_{k - 1} - r_{k - 2} - 2$
|
|
|
|
No, this violates the "second-order" definition, as $r_k$ must only contain the
|
|
two previous terms $r_{k - 1}$ and $r_{k - 2}$, the $-2$ at the end of the
|
|
equation violates this definition.
|
|
|
|
f. $s_k = 10s_{k - 2}$
|
|
|
|
Yes.
|
|
|
|
2. Which of the following are second-order linear homogeneous recurrence
|
|
relations with constant coefficients?
|
|
|
|
a. $a_k = (k - 1)a_{k - 1} + 2ka_{k - 2}$
|
|
|
|
No.
|
|
|
|
b. $b_k = -b_{k - 1} + 7b_{k - 2}$
|
|
|
|
Yes.
|
|
|
|
c. $c_k = 3c_{k - 1} + 1$
|
|
|
|
No.
|
|
|
|
d. $d_k = 3d_{k - 1}^2 + d_{k - 2}$
|
|
|
|
No.
|
|
|
|
e. $r_k = r_{k - 1} + 6r_{k - 3}$
|
|
|
|
Yes.
|
|
|
|
f. $s_k = s_{k - 1} + 10s_{k - 2}$
|
|
|
|
Yes.
|
|
|
|
3. Let $a_0, a_1, a_2, \dots$ be the sequence defined by the explicit formula
|
|
|
|
$$ a_n = C \cdot 2^n + D \quad \text{ for every integer } n \geq 0 $$
|
|
|
|
where $C$ and $D$ are real numbers.
|
|
|
|
a. Find $C$ and $D$ so that $a_0 = 1$ and $a_1 = 3$. What is $a_2$ in this case?
|
|
|
|
$$ a_0 = 1 = C \cdot 2^0 + D = C(1) + D = C + D $$
|
|
|
|
$$ a_1 = 3 = C \cdot 2^1 + D = 2C + D $$
|
|
|
|
So we have:
|
|
|
|
$$ C + D = 1 \quad \text{ and } 2C + D = 3 $$
|
|
|
|
Let's solve:
|
|
|
|
$$ C = 1 - D $$
|
|
|
|
$$ 2(1 - D) + D = 3 $$
|
|
|
|
$$ 2 - 2D + D = 3 $$
|
|
|
|
$$ 2 - D = 3 $$
|
|
|
|
$$ 2 = 3 + D $$
|
|
|
|
$$ -1 = D $$
|
|
|
|
$$ C = 1 - (-1) $$
|
|
|
|
$$ C = 2 $$
|
|
|
|
So:
|
|
|
|
$$ a_n = C \cdot 2^n + D $$
|
|
|
|
$$ a_n = 2 \cdot 2^n + (-1) $$
|
|
|
|
$$ a_n = 2^{n + 1} - 1 $$
|
|
|
|
And:
|
|
|
|
$$ a_2 = 2^{2 + 1} - 1 $$
|
|
|
|
$$ a_2 = 2^{3} - 1 $$
|
|
|
|
$$ a_2 = 8 - 1 $$
|
|
|
|
$$ a_2 = 7 $$
|
|
|
|
b. Find $C$ and $D$ so that $a_0 = 0$ and $a_1 = 2$. What is $a_2$ in this case?
|
|
|
|
$$ a_0 = 0 = C \cdot 2^0 + D = C + D $$
|
|
|
|
$$ a_1 = 2 = C \cdot 2^1 + D = 2C + D $$
|
|
|
|
So we have:
|
|
|
|
$$ C + D = 0 \quad \text{ and } \quad 2C + D = 2 $$
|
|
|
|
Let's solve:
|
|
|
|
$$ C = -D $$
|
|
|
|
$$ 2(-D) + D = 2 $$
|
|
|
|
$$ -2D + D = 2 $$
|
|
|
|
$$ -D = 2 $$
|
|
|
|
$$ D = -2 $$
|
|
|
|
$$ C = -(-2) $$
|
|
|
|
$$ C = 2 $$
|
|
|
|
Then:
|
|
|
|
$$ a_n = 2 \cdot 2^n + (-2) $$
|
|
|
|
$$ a_n = 2^{n + 1} - 2 $$
|
|
|
|
Then:
|
|
|
|
$$ a_2 = 2^{2 + 1} - 2 $$
|
|
|
|
$$ a_2 = 2^3 - 2 $$
|
|
|
|
$$ a_2 = 8 - 2 $$
|
|
|
|
$$ a_2 = 6 $$
|
|
|
|
4. Let $b_0, b_1, b_2, \dots$ be the sequence defined by the explicit formula
|
|
|
|
$$ b_n = C \cdot 3^n + D(-2)^n \quad \text{ for each integer } n \geq 0 $$
|
|
|
|
where $C$ and $D$ are real numbers.
|
|
|
|
a. Find $C$ and $D$ so that $b_0 = 0$ and $b_1 = 5$. What is $b_2$ in this case?
|
|
|
|
$$ b_0 = 0 = C \cdot 3^0 + D(-2)^0 = C + D $$
|
|
|
|
$$ b_1 = 5 = C \cdot 3^1 + D(-2)^1 = 3C - 2D $$
|
|
|
|
So we have:
|
|
|
|
$$ C + D = 0 \quad \text{ and } \quad 3C - 2D = 5 $$
|
|
|
|
Let's solve:
|
|
|
|
$$ C = -D $$
|
|
|
|
$$ 3(-D) - 2D = 5 $$
|
|
|
|
$$ -3D - 2D = 5 $$
|
|
|
|
$$ -5D = 5 $$
|
|
|
|
$$ D = -1 $$
|
|
|
|
$$ C = -(-1) $$
|
|
|
|
$$ C = 1 $$
|
|
|
|
So we have:
|
|
|
|
$$ b_n = 1 \cdot 3^n + (-1)(-2)^n $$
|
|
|
|
$$ b_n = 3^n - (-2)^n $$
|
|
|
|
Then:
|
|
|
|
$$ b_2 = 3^2 - (-2)^2 $$
|
|
|
|
$$ b_2 = 9 - (4) $$
|
|
|
|
$$ b_2 = 5 $$
|
|
|
|
b. Find $C$ and $D$ so that $b_0 = 3$ and $b_1 = 4$. What is $b_2$ in this case?
|
|
|
|
$$ b_0 = 3 = C \cdot 3^0 + D(-2)^0 = C + D $$
|
|
|
|
$$ b_1 = 4 = C \cdot 3^1 + D(-2)^1 = 3C - 2D $$
|
|
|
|
So we have:
|
|
|
|
$$ C + D = 3 \quad \text{ and } \quad 3C - 2D = 4 $$
|
|
|
|
Let's solve:
|
|
|
|
$$ C = 3 - D $$
|
|
|
|
$$ 3(3 - D) - 2D = 4 $$
|
|
|
|
$$ 9 - 3D - 2D = 4 $$
|
|
|
|
$$ 9 - 5D = 4 $$
|
|
|
|
$$ 9 = 4 + 5D $$
|
|
|
|
$$ 5 = 5D $$
|
|
|
|
$$ 1 = D $$
|
|
|
|
$$ C = 3 - 1 $$
|
|
|
|
$$ C = 2 $$
|
|
|
|
Then we have:
|
|
|
|
$$ b_n = 2 \cdot 3^n + (1)(-2)^n $$
|
|
|
|
$$ b_n = 2 \cdot 3^n + (-2)^n $$
|
|
|
|
Then we have:
|
|
|
|
$$ b_2 = 2 \cdot 3^2 + (-2)^2 $$
|
|
|
|
$$ b_2 = 2 \cdot 9 + (4) $$
|
|
|
|
$$ b_2 = 18 + 4 $$
|
|
|
|
$$ b_2 = 22 $$
|
|
|
|
5. Let $a_0, a_1, a_2, \dots$ be the sequence defined by the explicit formula
|
|
|
|
$$ a_n = C \cdot 2^n + D \quad \text{ for each integer } n \geq 0 $$
|
|
|
|
where $C$ and $D$ are real numbers. Show that for any choice of $C$ and $D$,
|
|
|
|
$$ a_k = 3a_{k - 1} - 2a_{k - 2} \quad \text{ for every integer } k \geq 2 $$
|
|
|
|
**Proof:**
|
|
|
|
Let $a_0, a_1, a_2, \dots$ be the sequence defined by the explicit formula
|
|
$a_n = C \cdot 2^n + D$ for each integer $n \geq 0$, where $C$ and $D$ are real
|
|
numbers.
|
|
|
|
Let $k$ be any integer such that $k \geq 2$. It follows that:
|
|
|
|
$$ a_k = C \cdot 2^k + D $$
|
|
|
|
$$ a_{k - 1} = C \cdot 2^{k - 1} + D $$
|
|
|
|
$$ a_{k - 2} = C \cdot 2^{k - 2} + D $$
|
|
|
|
We must show that for any choice of $C$ and $D$, that:
|
|
|
|
$$ a_k = 3a_{k - 1} - 2a_{k - 2} $$
|
|
|
|
By substitution:
|
|
|
|
$$ = 3(C \cdot 2^{k - 1} + D) - 2(C \cdot 2^{k - 2} + D) $$
|
|
|
|
Then, by algebra:
|
|
|
|
$$ = 3C \cdot 2^{k - 1} + 3D - 2C \cdot 2^{k - 2} - 2D $$
|
|
|
|
$$ = 3C \cdot 2^{k - 1} - 2C \cdot 2^{k - 2} + D $$
|
|
|
|
$$ = 3C \cdot 2^{k - 1} - C \cdot 2 \cdot 2^{k - 2} + D $$
|
|
|
|
$$ = 3C \cdot 2^{k - 1} - C \cdot 2^{k - 1} + D $$
|
|
|
|
$$ = 2C \cdot 2^{k - 1} + D $$
|
|
|
|
$$ = C \cdot 2 \cdot 2^{k - 1} + D $$
|
|
|
|
$$ = C \cdot 2^k + D $$
|
|
|
|
By the definition of the given equation:
|
|
|
|
$$ = a_k $$
|
|
|
|
This is what was to be shown.
|
|
|
|
Q.E.D.
|
|
|
|
6. Let $b_0, b_1, b_2, \dots$ be the sequence defined by the explicit formula
|
|
|
|
$$ b_n = C \cdot 3^n + D(-2)^n \quad \text{ for every integer } n \geq 0 $$
|
|
|
|
where $C$ and $D$ are real numbers. Show that for any choice of $C$ and $D$,
|
|
|
|
$$ b_k = b_{k - 1} + 6b_{k - 2} \quad \text{ for each integer } k \geq 2 $$
|
|
|
|
**Proof:**
|
|
|
|
Let $b_0, b_1, b_2, \dots$ be the sequence defined by the explicit formula
|
|
$b_n = C \cdot 3^n + D(-2)^n$ for every integer $n \geq 0$, where $C$ and $D$
|
|
are real numbers.
|
|
|
|
Let $k$ be any integer where $k \geq 2$. It then follows that:
|
|
|
|
$$ b_k = C \cdot 3^k + D(-2)^k $$
|
|
|
|
$$ b_{k - 1} = C \cdot 3^{k - 1} + D(-2)^{k - 1} $$
|
|
|
|
$$ b_{k - 2} = C \cdot 3^{k - 2} + D(-2)^{k - 2} $$
|
|
|
|
We must show that for any choice $C$ and $D$:
|
|
|
|
$$ b_k = b_{k - 1} + 6b_{k - 2} $$
|
|
|
|
By substitution:
|
|
|
|
$$ b_{k - 1} + 6b_{k - 2} = (C \cdot 3^{k - 1} + D(-2)^{k - 1}) + 6(C \cdot 3^{k - 2} + D(-2)^{k - 2}) $$
|
|
|
|
By algebra:
|
|
|
|
$$ = C \cdot 3^{k - 1} + D(-2)^{k - 1} + 6C \cdot 3^{k - 2} + 6D(-2)^{k - 2} $$
|
|
|
|
$$ = C \cdot 3^{k - 1} + D(-2)^{k - 1} + 2C \cdot 3 \cdot 3^{k - 2} + (-3)D \cdot (-2) \cdot (-2)^{k - 2} $$
|
|
|
|
$$ = C \cdot 3^{k - 1} + D(-2)^{k - 1} + 2C \cdot 3^{k - 1} + (-3)D \cdot (-2)^{k - 1} $$
|
|
|
|
$$ = 3C \cdot 3^{k - 1} + (-2)D(-2)^{k - 1} $$
|
|
|
|
$$ = C \cdot 3 \cdot 3^{k - 1} + D \cdot (-2) \cdot (-2)^{k - 1} $$
|
|
|
|
$$ = C \cdot 3^k + D(-2)^k $$
|
|
|
|
$$ = b_k $$
|
|
|
|
This is what was to be shown.
|
|
|
|
Q.E.D.
|
|
|
|
7. Solve the system of equations in Example 5.8.4 to obtain
|
|
|
|
$$ C = \frac{1 + \sqrt{5}}{2\sqrt{5}} \quad \text{ and } \quad D = \frac{-(1 - \sqrt{5})}{2\sqrt{5}} $$
|
|
|
|
**Proof:**
|
|
|
|
The initial conditions are:
|
|
|
|
$$ F_0 = F_1 = 1 $$
|
|
|
|
5.8.4 has established that the Fibonacci relation is a second-order linear
|
|
homogeneous recurrence relation with constant coefficients. It also established
|
|
that the explicit formula for the Fibonacci sequence is:
|
|
|
|
$$ F_n = C\left(\frac{1 + \sqrt{5}}{2}\right)^n + D\left(\frac{1 - \sqrt{5}}{2}\right)^n $$
|
|
|
|
It follows then that:
|
|
|
|
$$ F_0 = 1 = C\left(\frac{1 + \sqrt{5}}{2}\right)^0 + D\left(\frac{1 - \sqrt{5}}{2}\right)^0 $$
|
|
|
|
$$ F_0 = 1 = C + D $$
|
|
|
|
$$ F_1 = 1 = C\left(\frac{1 + \sqrt{5}}{2}\right)^1 + D\left(\frac{1 - \sqrt{5}}{2}\right)^1 $$
|
|
|
|
$$ F_1 = 1 = C\left(\frac{1 + \sqrt{5}}{2}\right) + D\left(\frac{1 - \sqrt{5}}{2}\right) $$
|
|
|
|
So we have:
|
|
|
|
$$ C + D = 1 $$
|
|
|
|
$$ C\left(\frac{1 + \sqrt{5}}{2}\right) + D\left(\frac{1 - \sqrt{5}}{2}\right) = 1 $$
|
|
|
|
Evaluating for both $C$ and $D$:
|
|
|
|
$$ C = 1 - D $$
|
|
|
|
$$ (1 - D)\left(\frac{1 + \sqrt{5}}{2}\right) + D\left(\frac{1 - \sqrt{5}}{2}\right) = 1 $$
|
|
|
|
$$ (1 - D)\left(\frac{1 + \sqrt{5}}{2}\right) = 1 - D\left(\frac{1 - \sqrt{5}}{2}\right) $$
|
|
|
|
$$ \frac{(1 - D)(1 + \sqrt{5})}{2} = \frac{2}{2} - \left(\frac{D(1 - \sqrt{5})}{2}\right) $$
|
|
|
|
$$ \frac{(1 - D)(1 + \sqrt{5})}{2} = \left(\frac{2 - D(1 - \sqrt{5})}{2}\right) $$
|
|
|
|
$$ (1 - D)(1 + \sqrt{5}) = 2 - D(1 - \sqrt{5}) $$
|
|
|
|
$$ (1 \cdot 1) + (-D)(1) + (1)(\sqrt{5}) + (-D)(\sqrt{5}) = 2 - D(1 - \sqrt{5}) $$
|
|
|
|
$$ 1 - D + \sqrt{5} - D\sqrt{5} = 2 - D(1 - \sqrt{5}) $$
|
|
|
|
$$ (1 + \sqrt{5}) - D(1 + \sqrt{5}) = 2 - D(1 - \sqrt{5}) $$
|
|
|
|
$$ -D(1 + \sqrt{5}) + D(1 - \sqrt{5}) = 2 - (1 + \sqrt{5}) $$
|
|
|
|
$$ D(-1(1 + \sqrt{5}) + (1 - \sqrt{5})) = 2 - 1 - \sqrt{5} $$
|
|
|
|
$$ D((1 - \sqrt{5}) - (1 + \sqrt{5})) = 1 - \sqrt{5} $$
|
|
|
|
$$ D(1 - \sqrt{5} - 1 - \sqrt{5}) = 1 - \sqrt{5} $$
|
|
|
|
$$ D(-\sqrt{5} - \sqrt{5}) = 1 - \sqrt{5} $$
|
|
|
|
$$ D(-2\sqrt{5}) = 1 - \sqrt{5} $$
|
|
|
|
$$ D = \frac{1 - \sqrt{5}}{-2\sqrt{5}} $$
|
|
|
|
$$ D = \frac{\sqrt{5} - 1}{2\sqrt{5}} $$
|
|
|
|
$$ D = \frac{5 - \sqrt{5}}{2 \cdot 5} $$
|
|
|
|
$$ D = \frac{5 - \sqrt{5}}{10} $$
|
|
|
|
Then:
|
|
|
|
$$ C = 1 - \left(\frac{5 - \sqrt{5}}{10}\right) $$
|
|
|
|
$$ C = \frac{10}{10} - \left(\frac{5 - \sqrt{5}}{10}\right) $$
|
|
|
|
$$ C = \frac{10 - (5 - \sqrt{5})}{10} $$
|
|
|
|
$$ C = \frac{10 - 5 + \sqrt{5}}{10} $$
|
|
|
|
$$ C = \frac{5 + \sqrt{5}}{10} $$
|
|
|
|
The closed form solution then is:
|
|
|
|
$$ F_n = \left(\frac{5 + \sqrt{5}}{10}\right)\left(\frac{1 + \sqrt{5}}{2}\right)^n + \left(\frac{5 - \sqrt{5}}{10}\right)\left(\frac{1 - \sqrt{5}}{2}\right)^n $$
|
|
|
|
In each of 8-10: (a) suppose a sequence of the form
|
|
$1, t, t^2, t^3, \dots, t^n, \dots$ where $t \neq 0$, satisfies the given
|
|
recurrence relation (but not necessarily the initial conditions), and find all
|
|
possible values of $t$: (b) suppose a sequence satisfies the given initial
|
|
conditions as well as the recurrence relation, and find an explicit formula for
|
|
the sequence.
|
|
|
|
8. $a_k = 2a_{k - 1} + 3a_{k - 2}$, for every integer $k \geq 2$
|
|
$a_0 = 1, a_1 = 2$
|
|
|
|
a.
|
|
|
|
Since the given second-order homogeneous recurrence relation with constant
|
|
coefficient, $a_k$, is satisfied by the sequence
|
|
$1, t, t^2, t^3, \dots t^n, \dots$. $t^k$ can be expressed as:
|
|
|
|
$$ t^k = 2t^{k - 1} + 3t^{k - 2} $$
|
|
|
|
It follows that:
|
|
|
|
$$ t^2 = 2t^{2 - 1} + 3t^{2 - 2} $$
|
|
|
|
$$ t^2 = 2t^1 + 3t^0 $$
|
|
|
|
$$ t^2 = 2t + 3 $$
|
|
|
|
By algebra then:
|
|
|
|
$$ t^2 - 2t - 3 = 0 $$
|
|
|
|
$$ (t - 3)(t + 1) = 0 $$
|
|
|
|
Therefore the possible values of $t$ are:
|
|
|
|
$$ t = -1 \quad \text{ and } \quad t = 3 $$
|
|
|
|
b.
|
|
|
|
It follows from (a) and the distinct roots theorem that the explicit formula for
|
|
the given sequence follows the form of:
|
|
|
|
$$ a_n = Cr^n + Ds^n $$
|
|
|
|
where $r$ and $s$ are the roots found in part (a):
|
|
|
|
$$ a_n = C \cdot 3^n + D \cdot (-1)^n $$
|
|
|
|
for every integer $n \geq 0$.
|
|
|
|
By the Distinct roots theorem we can find $C$ and $D$ by looking at the values
|
|
of $a_0$ and $a_1$ and evaluating for them:
|
|
|
|
$$ a_0 = 1 = C \cdot 3^0 + D \cdot (-1)^0 = C + D $$
|
|
|
|
$$ a_1 = 2 = C \cdot 3^1 + D \cdot (-1)^1 = 3C - D $$
|
|
|
|
So we have:
|
|
|
|
$$ C + D = 1 \quad \text{ and } \quad 3C - D = 2 $$
|
|
|
|
Evaluating:
|
|
|
|
$$ C = 1 - D $$
|
|
|
|
$$ 3(1 - D) - D = 2 $$
|
|
|
|
$$ 3 - 3D - D = 2 $$
|
|
|
|
$$ 3 - 4D = 2 $$
|
|
|
|
$$ 3 = 2 + 4D $$
|
|
|
|
$$ 1 = 4D $$
|
|
|
|
$$ \frac{1}{4} = D $$
|
|
|
|
$$ C = 1 - \frac{1}{4} $$
|
|
|
|
$$ C = \frac{4}{4} - \frac{1}{4} $$
|
|
|
|
$$ C = \frac{3}{4} $$
|
|
|
|
So, our explicit formula for the given sequence is:
|
|
|
|
$$ a_n = \frac{3}{4} \cdot 3^n + \frac{1}{4} \cdot (-1)^n $$
|
|
|
|
9. $b_k = 7b_{k - 1} - 10b_{k - 2}$, for every integer $k \geq 2$
|
|
$b_0 = 2, b_1, = 2$
|
|
|
|
a.
|
|
|
|
Since the given second-order homogeneous recurrence relation with constant
|
|
coefficient, $b_k$, is satisfied by the sequence
|
|
$1, t, t^2, t^3, \dots t^n, \dots$. $t^k$ can be expressed as:
|
|
|
|
$$ t^k = 7t^{k - 1} - 10t^{k - 2} $$
|
|
|
|
It follows that:
|
|
|
|
$$ t^2 = 7t^{2 - 1} - 10t^{2 - 2} $$
|
|
|
|
$$ t^2 = 7t^1 - 10t^0 $$
|
|
|
|
$$ t^2 = 7t - 10 $$
|
|
|
|
By algebra:
|
|
|
|
$$ t^2 - 7t + 10 = 0 $$
|
|
|
|
$$ (t - 5)(t - 2) = 0 $$
|
|
|
|
The possible values of $t$ then are:
|
|
|
|
$$ t = 2 \quad \text{ and } \quad t = 5 $$
|
|
|
|
b.
|
|
|
|
It follows from (a) and the distinct roots theorem that the explicit formula for
|
|
the given sequence follows the form of:
|
|
|
|
$$ b_n = Cr^n + Ds^n $$
|
|
|
|
where $r$ and $s$ are the roots found in part (a):
|
|
|
|
$$ b_n = C \cdot 2^n + D \cdot 5^n $$
|
|
|
|
for every integer $n \geq 0$.
|
|
|
|
By the Distinct roots theorem we can find $C$ and $D$ by looking at the values
|
|
of $b_0$ and $b_1$ and evaluating for them:
|
|
|
|
$$ b_0 = 2 = C \cdot 2^0 + D \cdot 5^0 = C + D $$
|
|
|
|
$$ b_1 = 2 = C \cdot 2^1 + D \cdot 5^1 = 2C + 5D $$
|
|
|
|
So we have:
|
|
|
|
$$ C + D = 2 \quad \text{ and } \quad 2C + 5D = 2 $$
|
|
|
|
Evaluating:
|
|
|
|
$$ C = 2 - D $$
|
|
|
|
$$ 2(2 - D) + 5D = 2 $$
|
|
|
|
$$ 4 - 2D + 5D = 2 $$
|
|
|
|
$$ 4 + 3D = 2 $$
|
|
|
|
$$ 3D = 2 - 4 $$
|
|
|
|
$$ 3D = -2 $$
|
|
|
|
$$ D = -\frac{2}{3} $$
|
|
|
|
$$ C = 2 - \left(-\frac{2}{3}\right) $$
|
|
|
|
$$ C = \frac{6}{3} + \frac{2}{3} $$
|
|
|
|
$$ C = \frac{8}{3} $$
|
|
|
|
So the explicit formula for the given sequence is:
|
|
|
|
$$ b_n = \frac{8}{3} \cdot 2^n + \left(-\frac{2}{3}\right) \cdot 5^n $$
|
|
|
|
10. $c_k = c_{k - 1} + 6c_{k - 2}$, for every integer $k \geq 2$
|
|
$c_0 = 0, c_1 = 3$
|
|
|
|
a.
|
|
|
|
Since the given second-order homogeneous recurrence relation with constant
|
|
coefficient, $c_k$, is satisfied by the sequence
|
|
$1, t, t^2, t^3, \dots t^n, \dots$. $t^k$ can be expressed as:
|
|
|
|
$$ t^k = t^{k - 1} + 6t^{k - 2} $$
|
|
|
|
It follows that:
|
|
|
|
$$ t^2 = t^{2 - 1} + 6t^{2 - 2} $$
|
|
|
|
$$ t^2 = t^1 + 6t^0 $$
|
|
|
|
$$ t^2 = t + 6 $$
|
|
|
|
By algebra:
|
|
|
|
$$ t^2 - t - 6 = 0 $$
|
|
|
|
$$ (t - 3)(t + 2) = 0 $$
|
|
|
|
So the possible values of $t$ are:
|
|
|
|
$$ t = -2 \quad \text{ and } \quad t = 3 $$
|
|
|
|
b.
|
|
|
|
It follows from (a) and the distinct roots theorem that the explicit formula for
|
|
the given sequence follows the form of:
|
|
|
|
$$ c_n = Cr^n + Ds^n $$
|
|
|
|
where $r$ and $s$ are the roots found in part (a):
|
|
|
|
$$ c_n = C \cdot 3^n + D \cdot (-2)^n $$
|
|
|
|
for every integer $n \geq 0$.
|
|
|
|
By the Distinct roots theorem we can find $C$ and $D$ by looking at the values
|
|
of $c_0$ and $c_1$ and evaluating for them:
|
|
|
|
$$ c_0 = 0 = C \cdot 3^0 + D \cdot (-2)^0 = C + D $$
|
|
|
|
$$ c_1 = 3 = C \cdot 3^1 + D \cdot (-2)^1 = 3C - 2D $$
|
|
|
|
So we have:
|
|
|
|
$$ C + D = 0 \quad \text{ and } \quad 3C - 2D = 3 $$
|
|
|
|
Evaluating:
|
|
|
|
$$ C = -D $$
|
|
|
|
$$ 3(-D) - 2D = 3 $$
|
|
|
|
$$ -3D - 2D = 3 $$
|
|
|
|
$$ -5D = 3 $$
|
|
|
|
$$ D = -\frac{3}{5} $$
|
|
|
|
$$ C = -\left(-\frac{3}{5}\right) $$
|
|
|
|
$$ C = \frac{3}{5} $$
|
|
|
|
So our explicit formula for the given sequence is:
|
|
|
|
$$ c_n = \frac{3}{5} \cdot 3^n + \left(-\frac{3}{5}\right) \cdot (-2)^n $$
|
|
|
|
In each of 11-16 suppose a sequence satisfies the given recurrence relation and
|
|
initial conditions. Find an explicit formula for the sequence.
|
|
|
|
11. $d_k = 4d_{k - 2}$ , for each integer $k \geq 2$ $d_0 = 1, d_1 = -1$
|
|
|
|
A lot of this is similar to 8-10. Let's streamline it:
|
|
|
|
$$ t^k = 4t^{k - 2} $$
|
|
|
|
$$ t^2 = 4t^{2 - 2} $$
|
|
|
|
$$ t^2 = 4t^0 $$
|
|
|
|
$$ t^2 = 4 $$
|
|
|
|
$$ t = \pm 2 $$
|
|
|
|
$$ d_n = C \cdot r^n + D \cdot s^n $$
|
|
|
|
$$ d_n = C \cdot 2^n + D \cdot (-2)^n $$
|
|
|
|
$$ d_0 = 1 = C \cdot 2^0 + D \cdot (-2)^0 = C + D $$
|
|
|
|
$$ d_1 = -1 = C \cdot 2^1 + D \cdot (-2)^1 = 2C - 2D $$
|
|
|
|
$$ C + D = 1 \quad \text{ and } \quad 2C - 2D = -1 $$
|
|
|
|
$$ C = 1 - D $$
|
|
|
|
$$ 2(1 - D) - 2D = -1 $$
|
|
|
|
$$ 2 - 2D - 2D = -1 $$
|
|
|
|
$$ 2 - 4D = -1 $$
|
|
|
|
$$ 2 = -1 + 4D $$
|
|
|
|
$$ 3 = 4D $$
|
|
|
|
$$ \frac{3}{4} = D $$
|
|
|
|
$$ C = 1 - \frac{3}{4} $$
|
|
|
|
$$ C = \frac{4}{4} - \frac{3}{4} $$
|
|
|
|
$$ C = \frac{1}{4} $$
|
|
|
|
$$ \boxed{d_n = \frac{1}{4} \cdot 2^n + \frac{3}{4} \cdot (-2)^n} $$
|
|
|
|
12. $e_k = 9e_{k - 1}$, for each integer $k \geq 2$ $e_0 = 0, e_1 = 2$
|
|
|
|
$$ t^k = 9t^{k - 1} $$
|
|
|
|
$$ t^2 = 9t^{2 - 1} $$
|
|
|
|
$$ t^2 = 9t^1 $$
|
|
|
|
$$ t^2 = 9t $$
|
|
|
|
$$ t^2 - 9t = 0 $$
|
|
|
|
$$ t(t - 9) = 0 $$
|
|
|
|
This is a single root: $t = 9$. And so takes the form:
|
|
|
|
$$ e_n = Cr^n + Dnr^n $$
|
|
|
|
$$ e_n = C \cdot 9^n + Dn \cdot 9^n $$
|
|
|
|
$$ e_0 = 0 = C \cdot 9^0 + D(0) \cdot 9^0 = C $$
|
|
|
|
$$ e_1 = 2 = C \cdot 9^1 + D(1) \cdot 9^1 = 9C + 9D $$
|
|
|
|
$$ C = 0 \quad \text{ and } 9C + 9D = 2 $$
|
|
|
|
$$ 9(0) + 9D = 2 $$
|
|
|
|
$$ 0 + 9D = 2 $$
|
|
|
|
$$ 9D = 2 $$
|
|
|
|
$$ D = \frac{2}{9} $$
|
|
|
|
$$ e_n = 0 \cdot 9^n + \left(\frac{2}{9}\right)n \cdot 9^n $$
|
|
|
|
$$ \boxed{e_n = \frac{2}{9}n \cdot 9^n} $$
|
|
|
|
13. $r_k = 2r^{k - 1} - r^{k - 2}$, for each integer $k \geq 2$
|
|
$r_0 = 1, r_1 = 4$
|
|
|
|
$$ t^k = 2t^{k - 1} - t^{k - 2} $$
|
|
|
|
$$ t^2 = 2t^{2 - 1} - t^{2 - 2} $$
|
|
|
|
$$ t^2 = 2t^1 - t^0 $$
|
|
|
|
$$ t^2 = 2t - 1 $$
|
|
|
|
$$ t^2 - 2t + 1 = 0 $$
|
|
|
|
$$ (t - 1)(t - 1) = 0 $$
|
|
|
|
$$ t = 1 $$
|
|
|
|
$$ r_n = Cs^n + Dns^n $$
|
|
|
|
$$ r_n = C \cdot 1^n + D(n) \cdot 1^n $$
|
|
|
|
$$ r_0 = 1 = C \cdot 1^0 + D(0) \cdot 1^0 = C $$
|
|
|
|
$$ r_1 = 4 = C \cdot 1^1 + D(1) \cdot 1^1 = C + D $$
|
|
|
|
$$ C = 1 \quad \text{ and } \quad C + D = 4 $$
|
|
|
|
$$ 1 + D = 4 $$
|
|
|
|
$$ D = 3 $$
|
|
|
|
$$ r_n = 1 \cdot 1^n + 3(n) \cdot 1^n $$
|
|
|
|
$$ r_n = 1^n + 3n \cdot 1^n $$
|
|
|
|
$$ \boxed{r_n = 1 + 3n} $$
|
|
|
|
14. $s_k = -4s_{k - 1} - 4s_{k - 2}$, for every integer $k \geq 2$
|
|
$s_0 = 0, s_1 = -1$
|
|
|
|
$$ t^k = -4t^{k - 1} - 4t^{k - 2} $$
|
|
|
|
$$ t^2 = -4t^{2 - 1} - 4t^{2 - 2} $$
|
|
|
|
$$ t^2 = -4t^1 - 4t^0 $$
|
|
|
|
$$ t^2 = -4t - 4 $$
|
|
|
|
$$ t^2 + 4t + 4 = 0 $$
|
|
|
|
$$ (t + 2)(t + 2) = 0 $$
|
|
|
|
$$ t = -2 $$
|
|
|
|
$$ s_n = Cr^n + Dnr^n $$
|
|
|
|
$$ s_n = C \cdot (-2)^n + Dn \cdot (-2)^n $$
|
|
|
|
$$ s_0 = 0 = C \cdot (-2)^0 + D(0) \cdot (-2)^0 = C $$
|
|
|
|
$$ s_1 = -1 = C \cdot (-2)^1 + D(1) \cdot (-2)^1 = -2C - 2D $$
|
|
|
|
$$ C = 0 \quad \text{ and } \quad -2C - 2D = -1 $$
|
|
|
|
$$ -2(0) - 2D = -1 $$
|
|
|
|
$$ -2D = -1 $$
|
|
|
|
$$ D = \frac{1}{2} $$
|
|
|
|
$$ s_n = 0 \cdot (-2)^n + \frac{1}{2}n \cdot (-2)^n $$
|
|
|
|
$$ \boxed{s_n = \frac{1}{2}n \cdot (-2)^n} $$
|
|
|
|
15. $t_k = 6t_{k - 1} - 9t_{k - 2}$, for each integer $k \geq 2$
|
|
$t_0 = 1, t_1= 3$
|
|
|
|
$$ x^k = 6x^{k - 1} - 9x^{k - 2} $$
|
|
|
|
$$ x^2 = 6x^{2 - 1} - 9x^{2 - 2} $$
|
|
|
|
$$ x^2 = 6x^1 - 9x^0 $$
|
|
|
|
$$ x^2 = 6x - 9 $$
|
|
|
|
$$ x^2 - 6x + 9 = 0 $$
|
|
|
|
$$ (x - 3)(x - 3) = 0 $$
|
|
|
|
$$ x = 3 $$
|
|
|
|
$$ t_n = Cr^n + Dnr^n $$
|
|
|
|
$$ t_n = C \cdot 3^n + Dn \cdot 3^n $$
|
|
|
|
$$ t_0 = 1 = C \cdot 3^0 + D(0) \cdot 3^0 = C $$
|
|
|
|
$$ t_1 = 3 = C \cdot 3^1 + D(1) \cdot 3^1 = 3C + 3D $$
|
|
|
|
$$ C = 1 \quad \text{ and } \quad 3C + 3D = 3 $$
|
|
|
|
$$ 3(1) + 3D = 3 $$
|
|
|
|
$$ 3 + 3D = 3 $$
|
|
|
|
$$ 3D = 0 $$
|
|
|
|
$$ D = 0 $$
|
|
|
|
$$ t_n = 1 \cdot 3^n + (0)n \cdot 3^n $$
|
|
|
|
$$ \boxed{t_n = 3^n} $$
|
|
|
|
16. $s_k = 2s_{k - 1} + 2s_{k - 2}$, for every integer $k \geq 2$
|
|
$s_0 = 1, s_1 = 3$
|
|
|
|
$$ t^k = 2t^{k - 1} + 2t^{k - 2} $$
|
|
|
|
$$ t^2 = 2t^{2 - 1} + 2t^{2 - 2} $$
|
|
|
|
$$ t^2 = 2t^1 + 2t^0 $$
|
|
|
|
$$ t^2 = 2t + 2 $$
|
|
|
|
$$ t^2 - 2t - 2 = 0 $$
|
|
|
|
$$ t = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(1)(-2)}}{2(1)} $$
|
|
|
|
$$ t = \frac{2 \pm \sqrt{12}}{2} $$
|
|
|
|
$$ t = \frac{2 \pm 2\sqrt{3}}{2} $$
|
|
|
|
$$ t = 1 \pm \sqrt{3} $$
|
|
|
|
$$ s_n = Cx^n + Dy^n $$
|
|
|
|
$$ s_n = C \cdot \left(1 + \sqrt{3}\right)^n + D \cdot \left(1 - \sqrt{3}\right)^n $$
|
|
|
|
$$ s_0 = 1 = C \cdot \left(1 + \sqrt{3}\right)^0 + D \cdot \left(1 - \sqrt{3}\right)^0 = C + D $$
|
|
|
|
$$ s_1 = 3 = C \cdot \left(1 + \sqrt{3}\right)^1 + D \cdot \left(1 - \sqrt{3}\right)^1 = C \cdot \left(1 + \sqrt{3}\right) + D \cdot \left(1 - \sqrt{3}\right) $$
|
|
|
|
$$ C + D = 1 \quad \text{ and } \quad C \cdot \left(1 + \sqrt{3}\right) + D \cdot \left(1 - \sqrt{3}\right) = 3 $$
|
|
|
|
$$ C = 1 - D $$
|
|
|
|
$$ (1 - D) \cdot \left(1 + \sqrt{3}\right) + D \cdot \left(1 - \sqrt{3}\right) = 3 $$
|
|
|
|
$$ (1 - D) \cdot \left(1 + \sqrt{3}\right) = 3 - \left[D \cdot \left(1 - \sqrt{3}\right)\right] $$
|
|
|
|
$$ (1 - D) \cdot \left(1 + \sqrt{3}\right) = 3 - \left[D \cdot \left(1 - \sqrt{3}\right)\right] $$
|
|
|
|
$$ (1)(1) + (-D)(1) + (1)(\sqrt{3}) + (-D)(\sqrt{3}) = 3 - \left[D \cdot \left(1 - \sqrt{3}\right)\right] $$
|
|
|
|
$$ 1 + (-D)(1) + \sqrt{3} + (-D)(\sqrt{3}) = 3 - \left[D \cdot \left(1 - \sqrt{3}\right)\right] $$
|
|
|
|
$$ (1 + \sqrt{3}) + (-D)(1) + (-D)(\sqrt{3}) = 3 - \left[D \cdot \left(1 - \sqrt{3}\right)\right] $$
|
|
|
|
$$ (1 + \sqrt{3}) + (-D)(1 + \sqrt{3}) = 3 - \left[D \cdot \left(1 - \sqrt{3}\right)\right] $$
|
|
|
|
$$ (1 + \sqrt{3}) + D(-1 - \sqrt{3}) = 3 - \left[D \cdot \left(1 - \sqrt{3}\right)\right] $$
|
|
|
|
$$ D(-1 - \sqrt{3}) + D \cdot \left(1 - \sqrt{3}\right) = 3 - (1 - \sqrt{3}) $$
|
|
|
|
$$ D\left[(-1 - \sqrt{3}) + \left(1 - \sqrt{3}\right)\right] = 3 - 1 + \sqrt{3} $$
|
|
|
|
$$ D\left[-1 - \sqrt{3} + 1 - \sqrt{3}\right] = 2 + \sqrt{3} $$
|
|
|
|
$$ D\left[-\sqrt{3} - \sqrt{3}\right] = 2 + \sqrt{3} $$
|
|
|
|
$$ D(-2\sqrt{3}) = 2 + \sqrt{3} $$
|
|
|
|
$$ D = \frac{2 + \sqrt{3}}{-2\sqrt{3}} $$
|
|
|
|
$$ D = \frac{2\sqrt{3} + 3}{-2 \cdot 3} $$
|
|
|
|
$$ D = \frac{2\sqrt{3} + 3}{-6} $$
|
|
|
|
$$ C = 1 - \left(\frac{2\sqrt{3} + 3}{-6}\right) $$
|
|
|
|
$$ C = \frac{-6}{-6} - \left(\frac{2\sqrt{3} + 3}{-6}\right) $$
|
|
|
|
$$ C = \frac{-6 - (2\sqrt{3} + 3)}{-6} $$
|
|
|
|
$$ C = \frac{-6 - 2\sqrt{3} - 3}{-6} $$
|
|
|
|
$$ C = \frac{-9 - 2\sqrt{3}}{-6} $$
|
|
|
|
$$ C = \frac{9 + 2\sqrt{3}}{6} $$
|
|
|
|
$$ s_n = \frac{9 + 2\sqrt{3}}{6} \cdot \left(1 + \sqrt{3}\right)^n + \frac{2\sqrt{3} + 3}{-6} \cdot \left(1 - \sqrt{3}\right)^n $$
|
|
|
|
17. Find an explicit formula for the sequence of exercise 39 in Section 5.6.
|
|
|
|
The recurrence relation found in exercise 39 is:
|
|
|
|
$$ c_k = c_{k - 1} + c_{k - 2} $$
|
|
|
|
$$ t^k = t^{k - 1} + t^{k - 2} $$
|
|
|
|
$$ t^2 = t^{2 - 1} + t^{2 - 2} $$
|
|
|
|
$$ t^2 = t^1 + t^0 $$
|
|
|
|
$$ t^2 = t + 1 $$
|
|
|
|
$$ t^2 - t - 1 = 0 $$
|
|
|
|
$$ t = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(1)(-1)}}{2(1)} $$
|
|
|
|
$$ t = \frac{1 \pm \sqrt{5}}{2} $$
|
|
|
|
$$ c_n = Cr^n + Ds^n $$
|
|
|
|
$$ c_n = C \cdot \left(\frac{1 + \sqrt{5}}{2}\right)^n + D \cdot \left(\frac{1 - \sqrt{5}}{2}\right)^n $$
|
|
|
|
$$ c_1 = 1 = C \cdot \left(\frac{1 + \sqrt{5}}{2}\right)^1 + D \cdot \left(\frac{1 - \sqrt{5}}{2}\right)^1 $$
|
|
|
|
$$ c_2 = 2 = C \cdot \left(\frac{1 + \sqrt{5}}{2}\right)^2 + D \cdot \left(\frac{1 - \sqrt{5}}{2}\right)^2 $$
|
|
|
|
Let $\phi = \dfrac{1 + \sqrt{5}}{2}$ and $\psi = \dfrac{1 - \sqrt{5}}{2}$. Then:
|
|
|
|
$$ c_1 = 1 = C \cdot \phi + D \cdot \psi $$
|
|
|
|
$$ c_2 = 2 = C \cdot \phi^2 + D \cdot \psi^2 $$
|
|
|
|
$$ C \cdot \phi + D \cdot \psi = 1 $$
|
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|
|
$$ C\phi = 1 - D\psi $$
|
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|
|
$$ C = \frac{1 - D\psi}{\phi} $$
|
|
|
|
$$ C \cdot \phi^2 + D \cdot \psi^2 = 2 $$
|
|
|
|
$$ \frac{1 - D\psi}{\phi} \cdot \phi^2 + D \cdot \psi^2 = 2 $$
|
|
|
|
$$ (1 - D\psi) \cdot \phi + D \cdot \psi^2 = 2 $$
|
|
|
|
$$ \phi - D\psi\phi + D\psi^2 = 2 $$
|
|
|
|
$$ -D\psi\phi + D\psi^2 = 2 - \phi $$
|
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|
$$ D(-\psi\phi + \psi^2) = 2 - \phi $$
|
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|
$$ D = \frac{2 - \phi}{-\psi\phi + \psi^2} $$
|
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|
|
$$ D = \frac{1}{\psi}\frac{2 - \phi}{-\phi + \psi} $$
|
|
|
|
Now back-substitution yields:
|
|
|
|
$$ D = \frac{1}{\dfrac{1 - \sqrt{5}}{2}} \cdot \frac{2 - \dfrac{1 + \sqrt{5}}{2}}{-\dfrac{1 + \sqrt{5}}{2} + \left(\dfrac{1 - \sqrt{5}}{2}\right)} $$
|
|
|
|
$$ D = \frac{2}{1 - \sqrt{5}} \cdot \frac{\dfrac{4 - (1 + \sqrt{5})}{2}}{\dfrac{-(1 + \sqrt{5}) + (1 - \sqrt{5})}{2}} $$
|
|
|
|
$$ D = \frac{2}{1 - \sqrt{5}} \cdot \frac{3 - \sqrt{5}}{-1 - \sqrt{5} + 1 - \sqrt{5}} $$
|
|
|
|
$$ D = \frac{2}{1 - \sqrt{5}} \cdot \frac{3 - \sqrt{5}}{-2\sqrt{5}} $$
|
|
|
|
$$ D = \frac{2(3 - \sqrt{5})}{(1 - \sqrt{5})(-2\sqrt{5})} $$
|
|
|
|
$$ D = \frac{6 - 2\sqrt{5}}{-2\sqrt{5} + (-2\sqrt{5})(-\sqrt{5})} $$
|
|
|
|
$$ D = \frac{6 - 2\sqrt{5}}{-2\sqrt{5} + (2 \cdot 5)} $$
|
|
|
|
$$ D = \frac{6 - 2\sqrt{5}}{-2\sqrt{5} + 10} $$
|
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|
|
$$ D = \frac{3 - \sqrt{5}}{-\sqrt{5} + 10} $$
|
|
|
|
$$ D = \frac{3 - \sqrt{5}}{10 - \sqrt{5}} $$
|
|
|
|
$$ D = \frac{3 - \sqrt{5}}{10 - \sqrt{5}} \cdot \frac{10 + \sqrt{5}}{10 + \sqrt{5}} $$
|
|
|
|
$$ D = \frac{(3 - \sqrt{5})(10 + \sqrt{5})}{(10 - \sqrt{5})(10 + \sqrt{5})} $$
|
|
|
|
$$ D = \frac{(3)(10) + (-\sqrt{5})(10) + (3)(\sqrt{5}) + (-\sqrt{5})(\sqrt{5})}{(10)(10) + (-\sqrt{5})(10) + (10)(\sqrt{5}) + (-\sqrt{5})(\sqrt{5})} $$
|
|
|
|
$$ D = \frac{30 - 10\sqrt{5} + 3\sqrt{5} - 5}{100 - 10\sqrt{5} + 10\sqrt{5} - 5} $$
|
|
|
|
$$ D = \frac{25 - 7\sqrt{5}}{95} $$
|
|
|
|
$$ C = \frac{1 - \left(\dfrac{25 - 7\sqrt{5}}{95}\right)\left(\dfrac{1 - \sqrt{5}}{2}\right)}{\dfrac{1 + \sqrt{5}}{2}} $$
|
|
|
|
$$ C = \frac{1 - \left(\dfrac{(25 - 7\sqrt{5})(1 - \sqrt{5})}{190}\right)}{\dfrac{1 + \sqrt{5}}{2}} $$
|
|
|
|
$$ C = \frac{\dfrac{190 - (25 - 7\sqrt{5})(1 - \sqrt{5})}{190}}{\dfrac{1 + \sqrt{5}}{2}} $$
|
|
|
|
$$ C = \frac{\dfrac{190 - ((25)(1) + (-7\sqrt{5})(1) + (25)(-\sqrt{5}) + (-7\sqrt{5})(-\sqrt{5}))}{190}}{\dfrac{1 + \sqrt{5}}{2}} $$
|
|
|
|
$$ C = \frac{\dfrac{190 - (25 - 7\sqrt{5} - 25\sqrt{5} + 35)}{190}}{\dfrac{1 + \sqrt{5}}{2}} $$
|
|
|
|
$$ C = \frac{\dfrac{190 - (60 - 32\sqrt{5})}{190}}{\dfrac{1 + \sqrt{5}}{2}} $$
|
|
|
|
$$ C = \frac{\dfrac{190 - 60 + 32\sqrt{5}}{190}}{\dfrac{1 + \sqrt{5}}{2}} $$
|
|
|
|
$$ C = \frac{\dfrac{130 + 32\sqrt{5}}{190}}{\dfrac{1 + \sqrt{5}}{2}} $$
|
|
|
|
$$ C = \frac{130 + 32\sqrt{5}}{190} \cdot \frac{2}{1 + \sqrt{5}} $$
|
|
|
|
$$ C = \frac{(130 + 32\sqrt{5})(2)}{(190)(1 + \sqrt{5})} $$
|
|
|
|
$$ C = \frac{(130 + 32\sqrt{5})(2)}{2(95)(1 + \sqrt{5})} $$
|
|
|
|
$$ C = \frac{130 + 32\sqrt{5}}{95(1 + \sqrt{5})} $$
|
|
|
|
$$ C = \frac{130 + 32\sqrt{5}}{95 + 95\sqrt{5}} $$
|
|
|
|
$$ C = \frac{130 + 32\sqrt{5}}{95 + 95\sqrt{5}} \cdot \frac{1 - \sqrt{5}}{1 - \sqrt{5}} $$
|
|
|
|
$$ C = \frac{(130 + 32\sqrt{5})(1 - \sqrt{5})}{(95 + 95\sqrt{5})(1 - \sqrt{5})} $$
|
|
|
|
$$ C = \frac{(130)(1) + (32\sqrt{5})(1) + (130)(-\sqrt{5}) + (32\sqrt{5})(-\sqrt{5})}{(95)(1) + (95\sqrt{5})(1) + (95)(-\sqrt{5}) + (95\sqrt{5})(-\sqrt{5})} $$
|
|
|
|
$$ C = \frac{130 + 32\sqrt{5} - 130\sqrt{5} - 160}{95 + 95\sqrt{5} - 95\sqrt{5} - 475} $$
|
|
|
|
$$ C = \frac{98\sqrt{5} - 30}{-380} $$
|
|
|
|
$$ C = \frac{30 - 98\sqrt{5}}{380} $$
|
|
|
|
$$ C = \frac{15 - 49\sqrt{5}}{190} $$
|
|
|
|
$$ c_n = \frac{15 - 49\sqrt{5}}{190} \cdot \left(\frac{1 + \sqrt{5}}{2}\right)^n + \frac{25 - 7\sqrt{5}}{95} \cdot \left(\frac{1 - \sqrt{5}}{2}\right)^n $$
|
|
|
|
18. Suppose that the sequences $s_0, s_1, s_2, \dots$ and $t_0, t_1, t_2, \dots$
|
|
both satisfy the same second-order linear homogeneous recurrence relation
|
|
with constant coefficients:
|
|
|
|
$$ s_k = 5s_{k - 1} - 4s_{k - 2} \quad \text{ for each integer } k \geq 2 $$
|
|
|
|
$$ t_k = 5t_{k - 1} - 4t_{k - 2} \quad \text{ for each integer } k \geq 2 $$
|
|
|
|
Show that the sequence $2s_0 + 3t_0, 2s_1 + 3t_1, 2s_2 + 3t_2, \dots$ also
|
|
satisfies the same relation. In other words, show that
|
|
|
|
$$ 2s_k + 3t_k = 5(2s_{k - 1} + 3t_{k - 1}) - 4(2s_{k - 2} + 3t_{k - 2}) $$
|
|
|
|
for each integer $k \geq 2$. Do _not_ use Lemma 5.8.2.
|
|
|
|
**Proof:**
|
|
|
|
$$ 2s_k + 3t_k $$
|
|
|
|
By substitution of the given recurrence relations:
|
|
|
|
$$ 2s_k + 3t_k = 2(5s_{k - 1} - 4s_{k - 2}) + 3(5t_{k - 1} - 4t_{k - 2}) $$
|
|
|
|
$$ = 10s_{k - 1} - 8s_{k - 2} + 15t_{k - 1} - 12t_{k - 2} $$
|
|
|
|
$$ = 10s_{k - 1} + 15t_{k - 1} - 8s_{k - 2} - 12t_{k - 2} $$
|
|
|
|
$$ = 5(2s_{k - 1} + 3t_{k - 1}) - 4(2s_{k - 2} + 3t_{k - 2}) $$
|
|
|
|
This is what was to be shown.
|
|
|
|
Q.E.D.
|
|
|
|
19. Show that if $r, s, a_0$, and $a_1$ are numbers with $r \neq s$, then there
|
|
exist unique numbers $C$ and $D$ so that
|
|
|
|
$$ C + D = a_0 $$
|
|
|
|
$$ Cr + Ds = a_1 $$
|
|
|
|
**Proof:**
|
|
|
|
Suppose $r$, $s$, $a_0$, and $a_1$ are numbers with $r \neq s$.
|
|
|
|
Consider the system of equations:
|
|
|
|
$$ C + D = a_0 $$
|
|
|
|
$$ Cr + Ds = a_1 $$
|
|
|
|
By solving for $D$ and substituting:
|
|
|
|
$$ D = a_0 - C $$
|
|
|
|
$$ Cr + (a_0 - C)s = a_1 $$
|
|
|
|
$$ Cr + a_0s - Cs = a_1 $$
|
|
|
|
$$ Cr - Cs = a_1 - a_0s $$
|
|
|
|
$$ C(r - s) = a_1 - a_0s $$
|
|
|
|
Since $r \neq s$, both sides can be divided by $(r - s)$:
|
|
|
|
$$ C = \frac{a_1 - a_0s}{r - s} $$
|
|
|
|
Then:
|
|
|
|
$$ D = a_0 - \frac{a_1 - a_0s}{r - s} $$
|
|
|
|
$$ D = \frac{a_0(r - s) - (a_1 - a_0s)}{r - s} $$
|
|
|
|
$$ D = \frac{a_0r - a_0s - a_1 + a_0s}{r - s} $$
|
|
|
|
$$ D = \frac{a_0r - a_1}{r - s} $$
|
|
|
|
Since $r \neq s$, division by $r - s$ is valid, yielding a unique value for $C$,
|
|
which in turn yields a unique value for $D$.
|
|
|
|
Q.E.D.
|
|
|
|
20. Show that if $r$ is a nonzero real number, $k$ and $m$ are distinct
|
|
integers, and $a_k$ and $a_m$ are any real numbers, then there exist unique
|
|
real numbers $C$ and $D$ so that
|
|
|
|
$$ Cr^k + kDr^k = a_k $$
|
|
|
|
$$ Cr^m + mDr^m = a_m $$
|
|
|
|
**Proof:**
|
|
|
|
Suppose $r$ is a nonzero real number, $k$ and $m$ are distinct integers, and
|
|
$a_k$ and $a_m$ are any real numbers.
|
|
|
|
Consider the system of equations:
|
|
|
|
$$ Cr^k + kDr^k = a_k $$
|
|
|
|
$$ Cr^m + mDr^m = a_m $$
|
|
|
|
By solving for $D$ and substituting:
|
|
|
|
$$ kDr^k = a_k - Cr^k $$
|
|
|
|
Since $r$ is a nonzero real number, we can divide by $r^k$:
|
|
|
|
$$ D = \frac{a_k - Cr^k}{kr^k} $$
|
|
|
|
Then:
|
|
|
|
$$ Cr^m + m \cdot \frac{a_k - Cr^k}{kr^k} \cdot r^m = a_m $$
|
|
|
|
$$ Cr^m + \frac{mr^m(a_k - Cr^k)}{kr^k} = a_m $$
|
|
|
|
$$ \frac{Cr^m(kr^k) + mr^m(a_k - Cr^k)}{kr^k} = a_m $$
|
|
|
|
$$ \frac{Ckr^{k + m} + mr^ma_k - mr^mCr^k}{kr^k} = a_m $$
|
|
|
|
$$ \frac{Ckr^{k + m} - Cmr^mr^k + mr^ma_k}{kr^k} = a_m $$
|
|
|
|
$$ \frac{Ckr^{k + m} - Cmr^{k + m} + mr^ma_k}{kr^k} = a_m $$
|
|
|
|
$$ \frac{Cr^{k + m}(k - m) + mr^ma_k}{kr^k} = a_m $$
|
|
|
|
$$ Cr^{k + m}(k - m) + mr^ma_k = a_m(kr^k) $$
|
|
|
|
$$ Cr^{k + m}(k - m) = a_m(kr^k) - mr^ma_k $$
|
|
|
|
$$ C = \frac{a_m(kr^k) - mr^ma_k}{r^{k + m}(k - m)} $$
|
|
|
|
Now we solve for $D$:
|
|
|
|
$$ D = \frac{a_k - \left(\frac{a_m(kr^k) - mr^ma_k}{r^{k + m}(k - m)}\right)r^k}{kr^k} $$
|
|
|
|
$$ D = \frac{a_k - \left(\frac{(a_m(kr^k) - mr^ma_k)r^k}{r^{k + m}(k - m)}\right)}{kr^k} $$
|
|
|
|
$$ D = \frac{a_k - \left(\frac{a_m(kr^k)r^k - mr^ma_kr^k}{r^{k + m}(k - m)}\right)}{kr^k} $$
|
|
|
|
$$ D = \frac{a_k - \left(\frac{a_mkr^{2k} - mr^{k + m}a_k}{r^{k + m}(k - m)}\right)}{kr^k} $$
|
|
|
|
$$ D = \frac{\frac{a_k(r^{k + m}(k - m)) - (a_mkr^{2k} - mr^{k + m}a_k)}{r^{k + m}(k - m)}}{kr^k} $$
|
|
|
|
$$ D = \frac{r^k(a_kr^k(k - m) - a_mkr^k + mr^ma_k)}{r^{k + m}(k - m)kr^k} $$
|
|
|
|
$$ D = \frac{a_kr^k(k - m) - a_mkr^k + mr^ma_k}{r^{k + m}(k - m)k} $$
|
|
|
|
$$ D = \frac{r^k(a_k(k - m) - a_mk) + mr^ma_k}{r^{k + m}(k - m)k} $$
|
|
|
|
$$ D = \frac{r^k(a_k(k - m) - a_mk)}{r^{k + m}(k - m)k} + \frac{mr^ma_k}{r^{k + m}(k - m)k} $$
|
|
|
|
$$ D = \frac{r^k(a_k(k - m) - a_mk)}{r^k(r^m(k - m)k)} + \frac{mr^ma_k}{r^m(r^k(k - m)k)} $$
|
|
|
|
$$ D = \frac{a_k(k - m) - a_mk}{r^m(k - m)k} + \frac{ma_k}{r^k(k - m)k} $$
|
|
|
|
$$ D = \frac{(a_k(k - m) - a_mk)(r^k) + (ma_k)(r^m)}{r^{m + k}(k - m)k} $$
|
|
|
|
$$ D = \frac{a_k(k - m)r^k - a_mk(r^k) + ma_kr^m}{r^{m + k}(k - m)k} $$
|
|
|
|
$$ D = \frac{a_k[(k - m)r^k + mr^m] - a_mk(r^k)}{r^{m + k}(k - m)k} $$
|
|
|
|
Since $r \neq 0$ and $k$ and $m$ are distinct integers, division by
|
|
$r^{m + k}(k - m)$ is valid, yielding a unique value for $C$, which in turn
|
|
yields a unique value for $D$.
|
|
|
|
Q.E.D.
|
|
|
|
21. Prove Theorem 5.8.5 for the case where the values of $C$ and $D$ are
|
|
determined by $a_0$ and $a_1$.
|
|
|
|
**Theorem 5.8.5 Single-Root Theorem**
|
|
|
|
Suppose a sequence $a_0, a_1, a_2, \dots$ satisfies a recurrence relation
|
|
|
|
$$ a_k = Aa_{k - 1} + Ba_{k - 2} $$
|
|
|
|
for some real numbers $A$ and $B$ with $B \neq 0$ and for every integer
|
|
$k \geq 2$. If the characteristic equation $t^2 - At - B = 0$ has a single
|
|
(real) root $r$, then $a_0, a_1, a_2, \dots$ is given by the explicit formula
|
|
|
|
$$ a_n = Cr^n + Dnr^n $$
|
|
|
|
where $C$ and $D$ are the real numbers whose values are determined by the values
|
|
of $a_0$ and any other known value of the sequence.
|
|
|
|
**Proof (by strong mathematical induction):**
|
|
|
|
Suppose a sequence $a_0, a_1, a_2, \dots$ satisfies a recurrence relation
|
|
|
|
$$ a_k = Aa_{k - 1} + Ba_{k - 2} $$
|
|
|
|
for some real numbers $A$ and $B$ with $B \neq 0$ and for every integer
|
|
$k \geq 2$. Assume the characteristic equation $t^2 - At - B = 0$ has a single
|
|
(real) root $r$.
|
|
|
|
It follows then that the sequence $a_0, a_1, a_2, \dots$ is given by the
|
|
explicit formula
|
|
|
|
$$ a_n = Cr^n + Dnr^n $$
|
|
|
|
where $C$ and $D$ are the real numbers whose values are determined by the values
|
|
of $a_0$ and any other known value of the sequence.
|
|
|
|
Let $P(n)$ be the recurrence relation:
|
|
|
|
$$ a_n = Cr^n + Dnr^n $$
|
|
|
|
_Basis Step:_
|
|
|
|
Prove $P(0)$ and $P(1)$, that is:
|
|
|
|
$$ a_0 = Cr^0 + D(0)r^0 = C $$
|
|
|
|
$$ a_1 = Cr^1 + D(1)r^1 = Cr + Dr $$
|
|
|
|
Therefore $P(0)$ and $P(1)$ are true by the assumption that $C$ and $D$ are
|
|
determined by the values of $a_0$ and $a_1$.
|
|
|
|
_Inductive Step:_
|
|
|
|
Let $k$ be any integer where $k \geq 2$. Let $i$ be some integer such that
|
|
$0 \leq i \leq k$.
|
|
|
|
Suppose $P(i)$, that is:
|
|
|
|
$$ a_i = Cr^i + Dir^i $$
|
|
|
|
Prove $P(k + 1)$, that is:
|
|
|
|
$$ a_{k + 1} = Cr^{k + 1} + D(k + 1)r^{k + 1} $$
|
|
|
|
By the definition of a recurrence relation:
|
|
|
|
$$ a_{k + 1} = Aa_k + Ba_{k - 1} $$
|
|
|
|
By the inductive hypothesis:
|
|
|
|
$$ = A(Cr^k + Dkr^k) + B(Cr^{k - 1} + D(k - 1)r^{k - 1}) $$
|
|
|
|
By algebra:
|
|
|
|
$$ = ACr^k + ADkr^k + BCr^{k - 1} + BD(k - 1)r^{k - 1} $$
|
|
|
|
$$ = ACr^k + BCr^{k - 1} + ADkr^k + BD(k - 1)r^{k - 1} $$
|
|
|
|
$$ = C(Ar^k + Br^{k - 1}) + D(Akr^k + B(k - 1)r^{k - 1}) $$
|
|
|
|
By Lemma 5.8.4, we know that because $r$ satisfies the characteristic equation,
|
|
it follows that $1, r, r^2, \dots, r^n, \dots$ and
|
|
$0, r, 2r^2, \dots, nr^n, \dots$ also satisfy the same recurrence relation as
|
|
$a_k$.
|
|
|
|
Therefore:
|
|
|
|
$$ = Cr^{k + 1} + D(k + 1)r^{k + 1} $$
|
|
|
|
This is what was to be shown.
|
|
|
|
Q.E.D.
|
|
|
|
Exercises 22 and 23 are intended for students who are familiar with complex
|
|
numbers.
|
|
|
|
22. Find an explicit formula for a sequence $a_0, a_1, a_2, \dots$ that
|
|
satisfies
|
|
|
|
$$ a_k = 2a_{k - 1} - 2a_{k - 1} \quad \text{ for every integer } k \geq 2 $$
|
|
|
|
with initial conditions $a_0 = 1$ and $a_1 = 2$.
|
|
|
|
Omitted.
|
|
|
|
23. Find an explicit formula for a sequence $b_0, b_1, b_2, \dots$ that
|
|
satisfies
|
|
|
|
$$ b_k = 2b_{k - 1} - 5b_{k - 2} \quad \text{ for each integer } k \geq 2 $$
|
|
|
|
with initial conditions $b_0 = 1$ and $b_1 = 1$.
|
|
|
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Omitted.
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24. The numbers $\dfrac{1 + \sqrt{5}}{2}$ and $\dfrac{1 - \sqrt{5}}{2}$ that
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appear in the explicit formula for the Fibonacci sequence are related to a
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quantity called the _golden ratio_ in Greek mathematics. Consider a
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rectangle of length $\phi$ units and height $1$, where $\phi > 1$.
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See page 387 for picture.
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Divide the rectangle into a rectangle and a square as shown in the preceding
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diagram. The square is $1$ unit on each side, and the rectangle has sides of
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length $1$ and $\phi - 1$. The ancient Greeks considered the outer rectangle to
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be perfectly proportioned (saying that the lengths of its sides are in a _golden
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ratio_ to each other) if the ratio of the length to the width of the outer
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rectangle equals the ratio of the length to the width of the inner rectangle.
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That is, if the number $\phi$ satisfies the equation
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$$ \frac{\phi}{1} = \frac{1}{\phi - 1} $$
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a. Show that if $\phi$ satisfies the equation above, then it also satisfies the
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quadratic equation: $t^2 - t - 1 = 0$.
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Omitted.
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b. Find the two solutions of $t^2 - t - 1 = 0$ and call them $\phi_1$ and
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$\phi_2$.
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Omitted.
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c. Express the explicit formula for the Fibonacci sequence in terms of $\phi_1$
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and $\phi_2$.
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Omitted.
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