🚧 Fin 5.1

chapter_5/exercises.md

@@ -6,220 +6,725 @@ 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 $m \geq 0$.
 
-5. $e_n = \left\lfloor \dfrac{n}{4} \right\rfloor \cdot 2$, for every integer
+$$ d_0 = 1 + \left(\dfrac{1}{2}\right)^0 = 1 $$
+
+$$ d_1 = 1 + \left(\dfrac{1}{2}\right)^1 = \frac{1}{2} $$
+
+$$ d_2 = 1 + \left(\dfrac{1}{2}\right)^2 = \frac{1}{4} $$
+
+$$ d_3 = 1 + \left(\dfrac{1}{2}\right)^3 = \frac{1}{8} $$
+
+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:
 
@@ -232,38 +737,103 @@ 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. $693$
 
-90. $2,301$
+$$ 693_{10} = 1BF_{16} $$
+
+91. $2,301$
+
+$$ 2301_{10} = 8FD_{16} $$
 
 91. Write a formal version of the algorithm you developed for exercise 87.
+
+Already done.

chapter_5/notes.md

@@ -111,7 +111,7 @@ $r[0], r[1], r[2], \dots r[k]$. 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 $q \neq 0$.]_
 
-$\text{\textbf{while }}(i = 0 \text{ or } q \new 0)\\ \ \ r[i] := q \mod 2\\ \ \ q := q \text{ div } 2\\ \ \ \text{[r[i] and q can be obtained by calling the division algorithm.]}\\ \ \ i := i + 1\\ \text{\textbf{end while}}$
+$\text{\textbf{while }}(i = 0 \text{ or } q \neq 0)\\ \ \ r[i] := q \mod 2\\ \ \ q := q \text{ div } 2\\ \ \ \text{[r[i] and q can be obtained by calling the division algorithm.]}\\ \ \ i := i + 1\\ \text{\textbf{end while}}$
 
 _[After execution of this step, the values of $r[0], r[1], \dots, r[i - 1]$ are
 all $0$'s and $1$'s, and

chapter_5/test_yourself.md

@@ -4,15 +4,29 @@ Page 296
 
 1. The notation $\sum_{k = m}^{n}{a_k}$ is read "_____."
 
+The summation from $k$ equals $m$ to $n$ of $a$ sub $k$.
+
 2. The expanded form of $\sum_{k = m}^{n}{a_k}$ is _____.
 
+$$ a_m + a_{m + 1} + a_{m + 2} + \dots + a_n $$
+
 3. The value of $a_1 + a_2 + a_3 + \dots + a_n$ when $n = 2$ is "_____."
 
+$$ a_1 + a_2 $$
+
 4. The notation $\prod_{k = m}^{n}{a_k}$ is read "_____."
 
+The product from $k$ equals $m$ to $n$ of $a$ sub $k$.
+
 5. If $n$ is a positive integer, then $n! =$ _____.
 
+$$ n \cdot (n - 1) \dots \cdot 3 \cdot 2 \cdot 1 $$
+
 6. $\sum_{k = m}^{n}{a_k} + c\sum_{k = m}^{n}{b_k} =$ _____.
 
+$$ \sum_{k = m}^{n}{a_k + cb_k} $$
+
 7. $\left(\prod_{k = m}^{n}{a_k}\right)\left(\prod_{k = m}^{n}{b_k}\right) =$
    _____.
+
+$$ \prod_{k = m}^{n}{a_kb_k} $$