6629 lines
161 KiB
Markdown
6629 lines
161 KiB
Markdown
**Exercise Set 5.1**
|
|
|
|
Page 296
|
|
|
|
Write the first four terms of the sequences defined by the formulas 1-6.
|
|
|
|
1. $a_k = \dfrac{k}{10 + k}$, for every integer $k \geq 1$.
|
|
|
|
$$ a_1 = \frac{1}{10 + 1} = \frac{1}{11} $$
|
|
|
|
$$ a_2 = \frac{2}{10 + 2} = \frac{2}{12} $$
|
|
|
|
$$ a_3 = \frac{3}{10 + 3} = \frac{3}{13} $$
|
|
|
|
$$ a_4 = \frac{4}{10 + 4} = \frac{4}{14} $$
|
|
|
|
2. $b_j = \dfrac{5 - j}{5 + j}$, for every integer $j \geq 1$.
|
|
|
|
$$ b_1 = \dfrac{5 - 1}{5 + 1} = \frac{4}{6} $$
|
|
|
|
$$ b_2 = \dfrac{5 - 2}{5 + 2} = \frac{3}{7} $$
|
|
|
|
$$ b_3 = \dfrac{5 - 3}{5 + 3} = \frac{2}{8} $$
|
|
|
|
$$ b_4 = \dfrac{5 - 4}{5 + 4} = \frac{1}{9} $$
|
|
|
|
3. $c_i = \dfrac{(-1)^i}{3^i}$, for every integer $i \geq 0$.
|
|
|
|
$$ c_0 = \dfrac{(-1)^0}{3^0} = \frac{1}{1} $$
|
|
|
|
$$ c_1 = \dfrac{(-1)^1}{3^1} = \frac{-1}{3} $$
|
|
|
|
$$ c_2 = \dfrac{(-1)^2}{3^2} = \frac{1}{9} $$
|
|
|
|
$$ c_3 = \dfrac{(-1)^3}{3^3} = \frac{-1}{27} $$
|
|
|
|
4. $d_m = 1 + \left(\dfrac{1}{2}\right)^m$ for every integer $m \geq 0$.
|
|
|
|
$$ d_0 = 1 + \left(\dfrac{1}{2}\right)^0 = 1 $$
|
|
|
|
$$ d_1 = 1 + \left(\dfrac{1}{2}\right)^1 = \frac{1}{2} $$
|
|
|
|
$$ d_2 = 1 + \left(\dfrac{1}{2}\right)^2 = \frac{1}{4} $$
|
|
|
|
$$ d_3 = 1 + \left(\dfrac{1}{2}\right)^3 = \frac{1}{8} $$
|
|
|
|
5. $e_n = \left\lfloor \dfrac{n}{2} \right\rfloor \cdot 2$, for every integer
|
|
$n \geq 0$.
|
|
|
|
$$ e_0 = \left\lfloor \dfrac{0}{2} \right\rfloor \cdot 2 = 0 $$
|
|
|
|
$$ e_1 = \left\lfloor \dfrac{1}{2} \right\rfloor \cdot 2 = 0 $$
|
|
|
|
$$ e_2 = \left\lfloor \dfrac{2}{2} \right\rfloor \cdot 2 = 2 $$
|
|
|
|
$$ e_3 = \left\lfloor \dfrac{3}{2} \right\rfloor \cdot 2 = 2 $$
|
|
|
|
6. $f_n = \left\lfloor \dfrac{n}{4} \right\rfloor \cdot 4$, for every integer
|
|
$n \geq 1$.
|
|
|
|
$$ f_1 = \left\lfloor \dfrac{1}{4} \right\rfloor \cdot 4 = 0 $$
|
|
|
|
$$ f_2 = \left\lfloor \dfrac{2}{4} \right\rfloor \cdot 4 = 0 $$
|
|
|
|
$$ f_3 = \left\lfloor \dfrac{3}{4} \right\rfloor \cdot 4 = 0 $$
|
|
|
|
$$ f_4 = \left\lfloor \dfrac{4}{4} \right\rfloor \cdot 4 = 4 $$
|
|
|
|
7. Let $a_k = 2k + 1$ and $b_k = (k - 1)^3 + k + 2$ for every integer
|
|
$k \geq 0$. Show that the first three terms of these sequences are identical
|
|
but that their fourth terms differ.
|
|
|
|
$$ a_0 = 2(0) + 1 = 1 $$
|
|
|
|
$$ a_1 = 2(1) + 1 = 3 $$
|
|
|
|
$$ a_2 = 2(2) + 1 = 5 $$
|
|
|
|
$$ a_3 = 2(3) + 1 = 7 $$
|
|
|
|
$$ b_0 = (0 - 1)^3 + 0 + 2 = 1 $$
|
|
|
|
$$ b_1 = (1 - 1)^3 + 1 + 2 = 3 $$
|
|
|
|
$$ b_2 = (2 - 1)^3 + 2 + 2 = 5 $$
|
|
|
|
$$ b_3 = (3 - 1)^3 + 3 + 2 = 13 $$
|
|
|
|
Compute the first fifteen terms of each of the sequences in 8 and 9, and
|
|
describe the general behavior of these sequences in words. (A definition of
|
|
logarithm is given in Section 7.1.)
|
|
|
|
8. $g_n = \lfloor \log_{2}n \rfloor$ for every integer $n \geq 1$.
|
|
|
|
$$ g_1 = \lfloor \log_{2}(1) \rfloor = 0 $$
|
|
|
|
$$ g_2 = \lfloor \log_{2}(2) \rfloor = 1 $$
|
|
|
|
$$ g_3 = \lfloor \log_{2}(3) \rfloor = 1 $$
|
|
|
|
$$ g_4 = \lfloor \log_{2}(4) \rfloor = 2 $$
|
|
|
|
$$ g_5 = \lfloor \log_{2}(5) \rfloor = 2 $$
|
|
|
|
$$ g_6 = \lfloor \log_{2}(6) \rfloor = 2 $$
|
|
|
|
$$ g_7 = \lfloor \log_{2}(7) \rfloor = 2 $$
|
|
|
|
$$ g_8 = \lfloor \log_{2}(8) \rfloor = 3 $$
|
|
|
|
$$ g_9 = \lfloor \log_{2}(9) \rfloor = 3 $$
|
|
|
|
$$ g_{10} = \lfloor \log_{2}(10) \rfloor = 3 $$
|
|
|
|
$$ g_{11} = \lfloor \log_{2}(11) \rfloor = 3 $$
|
|
|
|
$$ g_{12} = \lfloor \log_{2}(12) \rfloor = 3 $$
|
|
|
|
$$ g_{13} = \lfloor \log_{2}(13) \rfloor = 3 $$
|
|
|
|
$$ g_{14} = \lfloor \log_{2}(14) \rfloor = 3 $$
|
|
|
|
$$ g_{15} = \lfloor \log_{2}(15) \rfloor = 3 $$
|
|
|
|
The general behavior of this sequence is that it increments in binary
|
|
increments, as in it increments every 1, then 2, then 4, then 8 iterations of
|
|
the index $n$.
|
|
|
|
9 $h_n = n\lfloor \log_{2}n \rfloor$ for every integer $n \geq 1$.
|
|
|
|
$$ h_1 = (1)\lfloor \log_{2}(1) \rfloor = 0 $$
|
|
|
|
$$ h_2 = (2)\lfloor \log_{2}(2) \rfloor = 2 $$
|
|
|
|
$$ h_3 = (3)\lfloor \log_{2}(3) \rfloor = 3 $$
|
|
|
|
$$ h_4 = (4)\lfloor \log_{2}(4) \rfloor = 8 $$
|
|
|
|
$$ h_5 = (5)\lfloor \log_{2}(5) \rfloor = 10 $$
|
|
|
|
$$ h_6 = (6)\lfloor \log_{2}(6) \rfloor = 12 $$
|
|
|
|
$$ h_7 = (7)\lfloor \log_{2}(7) \rfloor = 14 $$
|
|
|
|
$$ h_8 = (8)\lfloor \log_{2}(8) \rfloor = 24 $$
|
|
|
|
$$ h_9 = (9)\lfloor \log_{2}(9) \rfloor = 27 $$
|
|
|
|
$$ h_{10} = (10)\lfloor \log_{2}(10) \rfloor = 30 $$
|
|
|
|
$$ h_{11} = (11)\lfloor \log_{2}(11) \rfloor = 33 $$
|
|
|
|
$$ h_{12} = (12)\lfloor \log_{2}(12) \rfloor = 36 $$
|
|
|
|
$$ h_{13} = (13)\lfloor \log_{2}(13) \rfloor = 39 $$
|
|
|
|
$$ h_{14} = (14)\lfloor \log_{2}(14) \rfloor = 42 $$
|
|
|
|
$$ h_{15} = (15)\lfloor \log_{2}(15) \rfloor = 45 $$
|
|
|
|
The sequence finds the minimal (floor) power of $log_{2}n$ and then multiplies
|
|
it by $n$, which is why there are sudden "jumps" when the floor calculates a
|
|
jump to the next power of $2$. For example, at $n = 7$ to $n = 8$, there is a
|
|
noticeable jump because $\lfloor \log_{2}7 \rfloor$ is $2$, and then
|
|
$\lfloor \log_{2}8 \rfloor$ is $3$.
|
|
|
|
Find explicit formulas for sequences of the form $a_1, a_2, a_3, \dots$ with the
|
|
initial terms given in 10-16.
|
|
|
|
10. $-1, 1, -1, 1, -1, 1$
|
|
|
|
$a_n = (-1)^n$ where $n$ is an integer such that $n \geq 1$.
|
|
|
|
11. $0, 1, -2, 3, -4, 5$
|
|
|
|
$a_n = (n - 1)(-1)^{n}$ where $n$ is an integer such that $n \geq 1$.
|
|
|
|
12. $\dfrac{1}{4}, \dfrac{2}{9}, \dfrac{3}{16}, \dfrac{4}{25}, \dfrac{5}{36}, \dfrac{6}{49}$
|
|
|
|
$a_n = \dfrac{n}{(n + 1)^2}$ where $n$ is an integer such that $n \geq 1$.
|
|
|
|
13. $1 - \dfrac{1}{2}, \dfrac{1}{2} - \dfrac{1}{3}, \dfrac{1}{3} - \dfrac{1}{4}, \dfrac{1}{4} - \dfrac{1}{5}, \dfrac{1}{5} - \dfrac{1}{6}, \dfrac{1}{6} - \dfrac{1}{7}$
|
|
|
|
$a_n = \dfrac{1}{n} - \dfrac{1}{n + 1}$ where $n$ is an integer such that
|
|
$n \geq 1$.
|
|
|
|
14. $\dfrac{1}{3}, \dfrac{4}{9}, \dfrac{9}{27}, \dfrac{16}{81}, \dfrac{25}{243}, \dfrac{36}{729}$
|
|
|
|
$a_n = \dfrac{n^2}{3^n}$ where $n$ is an integer such that $n \geq 1$.
|
|
|
|
15. $0, -\dfrac{1}{2}, \dfrac{2}{3}, -\dfrac{3}{4}, \dfrac{4}{5}, -\dfrac{5}{6}, \dfrac{6}{7}$
|
|
|
|
$a_n = \dfrac{(n - 1)(-1)^{n + 1}}{n}$ where $n$ is an integer such that
|
|
$n \geq 1$.
|
|
|
|
16. $3, 6, 12, 24, 48, 96$
|
|
|
|
$a_n = 3 \cdot 2^{n - 1}$ where $n$ is an integer such that $n \geq 1$.
|
|
|
|
17. Consider the sequence defined by $a_n = \dfrac{2n + (-1)^n - 1}{4}$ for
|
|
every integer $n \geq 0$. Find an alternative explicit formula for $a_n$
|
|
that uses the floor notation.
|
|
|
|
Omitted.
|
|
|
|
18. Let
|
|
$a_0 = 2, a_1 = 3, a_2 = -2, a_3 = 1, a_4 = 0, a_5 = -1, \text{ and } a_6 = -2$.
|
|
Compute each of the summations and products below.
|
|
|
|
a. $\sum_{i = 0}^{6}{a_i}$
|
|
|
|
$$ \sum_{i = 0}^{6}{a_i} = 2 + 3 + (-2) + 1 + 0 + (-1) + (-2) = 1 $$
|
|
|
|
b. $\sum_{i = 0}^{0}{a_i}$
|
|
|
|
$$ \sum_{i = 0}^{0}{a_i} = 2 $$
|
|
|
|
c. $\sum_{j = 1}^{3}{a_{2j}}$
|
|
|
|
$$ \sum_{j = 1}^{3}{a_{2j}} = (-2) + 0 + (-2) = -4 $$
|
|
|
|
d. $\prod_{k = 0}^{6}{a_k}$
|
|
|
|
$$ \prod_{k = 0}^{6}{a_k} = 2 \cdot 3 \cdot (-2) \cdot 1 \cdot 0 \cdot (-1) \cdot (-2) = 0 $$
|
|
|
|
e. $\prod_{k = 2}^{2}{a_k}$
|
|
|
|
$$ \prod_{k = 2}^{2}{a_k} = -2 $$
|
|
|
|
Compute the summations and products in 19-28.
|
|
|
|
19. $\sum_{k = 1}^{5}{(k + 1)}$
|
|
|
|
$$ \sum_{k = 1}^{5}{(k + 1)} = (1 + 1) + (2 + 1) + (3 + 1) + (4 + 1) + (5 + 1) = 20 $$
|
|
|
|
20. $\prod_{k = 2}^{4}{k^2}$
|
|
|
|
$$ \prod_{k = 2}^{4}{k^2} = (2)^2 \cdot (3)^2 \cdot (4)^2 = 576 $$
|
|
|
|
21. $\sum_{k = 1}^{3}{(k^2 + 1)}$
|
|
|
|
$$ \sum_{k = 1}^{3}{(k^2 + 1)} = ((1)^2 + 1) + ((2)^2 + 1) + ((3)^2 + 1) = 17 $$
|
|
|
|
22. $\prod_{j = 0}^{4}{(-1)^j}$
|
|
|
|
$$ \prod_{j = 0}^{4}{(-1)^j} = (-1)^{(0)} \cdot (-1)^{(1)} \cdot (-1)^{(2)} \cdot (-1)^{(3)} \cdot (-1)^{(4)} = 1 $$
|
|
|
|
23. $\sum_{i = 1}^{1}{i(i + 1)}$
|
|
|
|
$$ \sum_{i = 1}^{1}{i(i + 1)} = (1)((1) + 1) = 2 $$
|
|
|
|
24. $\sum_{j = 0}^{0}{(j + 2) \cdot 2^j}$
|
|
|
|
$$ \sum_{j = 0}^{0}{(j + 2) \cdot 2^j} = (0 + 2) \cdot 2^0 = 2 $$
|
|
|
|
25. $\prod_{k = 2}^{2}{\left(1 - \dfrac{1}{k}\right)}$
|
|
|
|
$$ \prod_{k = 2}^{2}{\left(1 - \dfrac{1}{k}\right)} = \left(1 - \dfrac{1}{2}\right) = \frac{1}{2} $$
|
|
|
|
26. $\sum_{k = -1}^{1}{(k^2 + 3)}$
|
|
|
|
$$ \sum_{k = -1}^{1}{(k^2 + 3)} = ((-1)^2 + 3) + ((0)^2 + 3) + ((1)^2 + 3) = 11 $$
|
|
|
|
27. $\sum_{n = 1}^{6}{\left(\dfrac{1}{n} - \dfrac{1}{n + 1}\right)}$
|
|
|
|
$$ \sum_{n = 1}^{6}{\left(\dfrac{1}{n} - \dfrac{1}{n + 1}\right)} = \left(\dfrac{1}{(1)} - \dfrac{1}{(1) + 1}\right) + \left(\dfrac{1}{(2)} - \dfrac{1}{(2) + 1}\right) + \left(\dfrac{1}{(3)} - \dfrac{1}{(3) + 1}\right) + \left(\dfrac{1}{(4)} - \dfrac{1}{(4) + 1}\right) + \left(\dfrac{1}{(5)} - \dfrac{1}{(5) + 1}\right) + \left(\dfrac{1}{(6)} - \dfrac{1}{(6) + 1}\right) = \frac{6}{7} $$
|
|
|
|
28. $\prod_{i = 2}^{5}{\dfrac{i(i + 2)}{(i - 1) \cdot (i + 1)}}$
|
|
|
|
$$ \prod_{i = 2}^{5}{\dfrac{i(i + 2)}{(i - 1) \cdot (i + 1)}} = \dfrac{(2)((2) + 2)}{((2) - 1) \cdot ((2) + 1)} + \dfrac{(3)((3) + 2)}{((3) - 1) \cdot ((3) + 1)} + \dfrac{(4)((4) + 2)}{((4) - 1) \cdot ((4) + 1)} + \dfrac{(5)((5) + 2)}{((5) - 1) \cdot ((5) + 1)} = \frac{35}{3} $$
|
|
|
|
Write the summations in 29-32 in expanded form.
|
|
|
|
29. $\sum_{i = 1}^{n}{(-2)^i}$
|
|
|
|
$$ \sum_{i = 1}^{n}{(-2)^i} = (-2)^1 + (-2)^2 + (-2)^3 + \dots + (-2)^{n} $$
|
|
|
|
30. $\sum_{j = 1}^{n}{j(j + 1)}$
|
|
|
|
$$ \sum_{j = 1}^{n}{j(j + 1)} = ((1)((1) + 1)) + ((2)((2) + 1)) + ((3)((3) + 1)) + \dots + ((n)((n) + 1)) = 2 + 6 + 12 + \dots n(n + 1) $$
|
|
|
|
31. $\sum_{k = 0}^{n + 1}{\dfrac{1}{k!}}$
|
|
|
|
$$ \sum_{k = 0}^{n + 1}{\dfrac{1}{k!}} = \dfrac{1}{0!} + \dfrac{1}{1!} + \dfrac{1}{2!} + \dots + \dfrac{1}{(n + 1)!} $$
|
|
|
|
32. $\sum_{i = 1}^{k + 1}{i(i!)}$
|
|
|
|
$$ \sum_{i = 1}^{k + 1}{i(i!)} = 1(1!) + 2(2!) + 3(3!) + \dots + (k + 1)((k + 1)!) $$
|
|
|
|
Evaluate the summations and products in 33-36 for the indicated values of the
|
|
variable.
|
|
|
|
33. $\dfrac{1}{1^2} + \dfrac{1}{2^2} + \dfrac{1}{3^2} + \dots + \dfrac{1}{n^2}; n = 1$
|
|
|
|
$$ \frac{1}{1^2} = 1 $$
|
|
|
|
34. $1(1!) + 2(2!) + 3(3!) + \dots + m(m!); m = 2$
|
|
|
|
$$ 1(1!) + 2(2!) = 5 $$
|
|
|
|
35. $\left(\dfrac{1}{1 + 1}\right)\left(\dfrac{2}{2 + 1}\right)\left(\dfrac{3}{3 + 1}\right) \dots \left(\dfrac{k}{k + 1}\right); k = 3$
|
|
|
|
$$ \left(\dfrac{1}{1 + 1}\right)\left(\dfrac{2}{2 + 1}\right)\left(\dfrac{3}{3 + 1}\right) = \frac{1}{4} $$
|
|
|
|
36. $\left(\dfrac{1 \cdot 2}{3 \cdot 4}\right)\left(\dfrac{4 \cdot 5}{6 \cdot 7}\right)\left(\dfrac{6 \cdot 7}{8 \cdot 9}\right) \dots \left(\dfrac{m \cdot (m + 1)}{(m + 2) \cdot (m + 3)}\right); m = 1$
|
|
|
|
$$ \left(\frac{1 \cdot 2}{3 \cdot 4}\right) = \frac{2}{12} = \frac{1}{6} $$
|
|
|
|
Write each of 37-39 as a single summation.
|
|
|
|
37. $\sum_{i = 1}^{k}{i^3 + (k + 1)^3}$
|
|
|
|
$$ \sum_{i = 1}^{k}{i^3 + (k + 1)^3} = \sum_{i = 1}^{k + 1}{i^3} $$
|
|
|
|
38. $\sum_{k = 1}^{m}{\dfrac{k}{k + 1} + \dfrac{m + 1}{m + 2}}$
|
|
|
|
$$ \sum_{k = 1}^{m}{\dfrac{k}{k + 1} + \dfrac{m + 1}{m + 2}} = \sum_{k = 1}^{m + 1}{\dfrac{k}{k + 1}} $$
|
|
|
|
39. $\sum_{m = 0}^{n}{(m + 1)2^m + (n + 2)2^{n + 1}}$
|
|
|
|
$$ \sum_{m = 0}^{n}{(m + 1)2^m + (n + 2)2^{n + 1}} = \sum_{m = 0}^{n + 1}{(m + 1)2^m} $$
|
|
|
|
Rewrite 40-42 by separating off the final term.
|
|
|
|
40. $\sum_{i = 1}^{k + 1}{i(i!)}$
|
|
|
|
$$ \sum_{i = 1}^{k + 1}{i(i!)} = \sum_{i = 1}^{k}{i(i!) + (k + 1)(k + 1)!} $$
|
|
|
|
41. $\sum_{k = 1}^{m + 1}{k^2}$
|
|
|
|
$$ \sum_{k = 1}^{m + 1}{k^2} = \sum_{k = 1}^{m}{k^2 + (m + 1)^2} $$
|
|
|
|
42. $\sum_{m = 1}^{n + 1}{m(m + 1)}$
|
|
|
|
$$ \sum_{m = 1}^{n + 1}{m(m + 1)} = \sum_{m = 1}^{n}{m(m + 1) + (n + 1)(n + 2)} $$
|
|
|
|
Write each of 43-52 using summation or product notation.
|
|
|
|
43. $1^2 - 2^2 + 3^2 - 4^2 + 5^2 - 6^2 + 7^2$
|
|
|
|
$$ 1^2 - 2^2 + 3^2 - 4^2 + 5^2 - 6^2 + 7^2 $$
|
|
|
|
$$ \sum_{k = 1}^{7}{(-1)^{k + 1}k^2} $$
|
|
|
|
44. $(1^3 - 1) - (2^3 - 1) + (3^3 - 1) - (4^3 - 1) + (5^3 - 1)$
|
|
|
|
$$ \sum_{k = 1}^{5}{(-1)^{k + 1}(k^3 - 1)} $$
|
|
|
|
45. $(2^2 - 1) \cdot (3^2 - 1) \cdot (4^2 - 1)$
|
|
|
|
$$ \prod_{k = 2}^{4}{(k^2 - 1)} $$
|
|
|
|
46. $\dfrac{2}{3 \cdot 4} - \dfrac{3}{4 \cdot 5} + \dfrac{4}{5 \cdot 6} - \dfrac{5}{6 \cdot 7} + \dfrac{6}{7 \cdot 8}$
|
|
|
|
$$ \sum_{k=2}^{6}{\dfrac{(-1)^k \cdot k}{(k + 1) \cdot (k + 2)}} $$
|
|
|
|
47. $1 - r + r^2 - r^3 + r^4 - r^5$
|
|
|
|
$$ \sum_{k = 0}^{5}{(-1)^kr^{k + 1}} $$
|
|
|
|
48. $(1 - t) \cdot (1 - t^2) \cdot (1 - t^3) \cdot (1 - t^4)$
|
|
|
|
$$ \prod_{k = 1}^{4}{(1 - t^k)} $$
|
|
|
|
49. $1^3 + 2^3 + 3^3 + \dots + n^3$
|
|
|
|
$$ \sum_{k}^{n}{k^3} $$
|
|
|
|
50. $\dfrac{1}{2!} + \dfrac{2}{3!} + \dfrac{3}{4!} + \dots + \dfrac{n}{(n + 1)!}$
|
|
|
|
$$ \sum_{k = 1}^{n}{\frac{k}{(k + 1)!}} $$
|
|
|
|
51. $n + (n - 1) + (n - 2) + \dots + 1$
|
|
|
|
$$ \sum_{k = 0}^{n - 1}{(n - k)} $$
|
|
|
|
52. $n + \dfrac{n - 1}{2!} + \dfrac{n - 2}{3!} + \dfrac{n - 3}{4!} + \dots + \dfrac{1}{n!}$
|
|
|
|
$$ \sum_{k = 0}^{n - 1}{\frac{n - k}{(k + 1)!}} $$
|
|
|
|
Transform each of 53 and 54 by making the change of variable $i = k + 1$.
|
|
|
|
$$ i = k + 1 $$
|
|
|
|
$$ i - 1 = k $$
|
|
|
|
53. $\sum_{k = 0}^{5}{k(k - 1)}$
|
|
|
|
$$ \sum_{k = 0}^{5}{k(k - 1)} = \sum_{i = 1}^{6}{(i - 1)(i - 2)} $$
|
|
|
|
54. $\prod_{k = 1}^{n}{\dfrac{k}{k^2 + 4}}$
|
|
|
|
$$ \prod_{k = 1}^{n}{\dfrac{k}{k^2 + 4}} = \prod_{i = 2}^{n + 1}{\frac{i - 1}{(i - 1)^2 + 4}} $$
|
|
|
|
Transform each of 55-58 by making the change of variable $j = i - 1$.
|
|
|
|
$$ j = i - 1 $$
|
|
|
|
$$ i = j + 1 $$
|
|
|
|
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.
|
|
|
|
59. $3 \cdot \sum_{k = 1}^{n}{(2k - 3)} + \sum_{k = 1}^{n}{(4 - 5k)}$
|
|
|
|
$$ \sum_{k = 1}^{n}{3(2k - 3) + (4 - 5k)} $$
|
|
|
|
$$ \sum_{k = 1}^{n}{(6k - 9 + 4 - 5k)} $$
|
|
|
|
$$ \sum_{k = 1}^{n}{(k - 5)} $$
|
|
|
|
60. $2 \cdot \sum_{k = 1}^{n}{(3k^2 + 4)} + 5 \cdot \sum_{k = 1}^{n}{(2k^2 - 1)}$
|
|
|
|
$$ \sum_{k = 1}^{n}{2(3k^2 + 4) + 5(2k^2 - 1)} $$
|
|
|
|
$$ \sum_{k = 1}^{n}{(6k^2 + 8 + 10k^2 - 5)} $$
|
|
|
|
$$ \sum_{k = 1}^{n}{(16k^2 + 3)} $$
|
|
|
|
61. $\left(\prod_{k = 1}^{n}{\dfrac{k}{k + 1}}\right) \cdot \left(\prod_{k = 1}^{n}{\dfrac{k + 1}{k + 2}}\right)$
|
|
|
|
$$ \prod_{k = 1}^{n}{\left(\frac{k}{k + 1}\right)\left(\frac{k + 1}{k + 2}\right)} $$
|
|
|
|
$$ \prod_{k = 1}^{n}{\left(\frac{k(k + 1)}{(k + 1)(k + 2)}\right)} $$
|
|
|
|
$$ \prod_{k = 1}^{n}{\left(\frac{k\cancel{(k + 1)}}{\cancel{(k + 1)}(k + 2)}\right)} $$
|
|
|
|
$$ \prod_{k = 1}^{n}{\left(\frac{k}{k + 2}\right)} $$
|
|
|
|
Compute each of 62-76. Assume the values of the variables are restricted so that
|
|
the expressions are defined.
|
|
|
|
62. $\dfrac{4!}{3!}$
|
|
|
|
$$ \frac{4!}{3!} = \frac{4 \cdot 3 \cdot 2 \cdot 1}{3 \cdot 2 \cdot 1} = 4 $$
|
|
|
|
63. $\dfrac{6!}{8!}$
|
|
|
|
$$ \frac{6!}{8!} = \frac{6!}{8 \cdot 7 \cdot 6!} = \frac{1}{56} $$
|
|
|
|
64. $\dfrac{4!}{0!}$
|
|
|
|
$$ \frac{4!}{0!} = \frac{4 \cdot 3 \cdot 2 \cdot 1}{1} = 24 $$
|
|
|
|
65. $\dfrac{n!}{(n - 1)!}$
|
|
|
|
$$ \frac{n!}{(n - 1)!} = \frac{n(n - 1)!}{(n - 1)!} = n $$
|
|
|
|
66. $\dfrac{(n - 1)!}{(n + 1)!}$
|
|
|
|
$$ \frac{(n - 1)!}{(n + 1)!} = \frac{(n - 1)!}{(n + 1)(n)(n - 1)!} = \frac{1}{n(n + 1)} $$
|
|
|
|
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.
|