diff --git a/chapter_5/exercises.md b/chapter_5/exercises.md
index e69de29..e9a0f29 100644
--- a/chapter_5/exercises.md
+++ b/chapter_5/exercises.md
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+**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 $k \geq 1$.
+
+2. $b_j = \dfrac{5 - j}{5 + j}$, for every integer $j \geq 1$.
+
+3. $c_i = \dfrac{(-1)^i}{3^i}$, for every integer $i \geq 0$.
+
+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
+   $n \geq 0$.
+
+6. $f_n = \left\lfloor \dfrac{n}{4} \right\rfloor \cdot 4$, for every integer
+   $n \geq 1$.
+
+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.
+
+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 $n \geq 1$.
+
+9 $h_n = n\lfloor \log_{2}n \rfloor$ for every integer $n \geq 1$.
+
+Find explicit formulas for sequences of the form $a_1, a_2, a_3, \dots$ with the
+initial terms given in 10-16.
+
+10. $-1, 1, -1, 1, -1, 1$
+
+11. $0, 1, -2, 3, -4, 5$
+
+12. $\dfrac{1}{4}, \dfrac{2}{9}, \dfrac{3}{16}, \dfrac{4}{25}, \dfrac{5}{36}, \dfrac{6}{49}$
+
+13. $1 - \dfrac{1}{2}, \dfrac{1}{2} - \dfrac{1}{3}, \dfrac{1}{3} - \dfrac{1}{4}, \dfrac{1}{4} - \dfrac{1}{5}, \dfrac{1}{5} - \dfrac{1}{6}, \dfrac{1}{6} - \dfrac{1}{7}$
+
+14. $\dfrac{1}{3}, \dfrac{4}{9}, \dfrac{9}{27}, \dfrac{16}{81}, \dfrac{25}{243}, \dfrac{36}{729}$
+
+15. $0, -\dfrac{1}{2}, \dfrac{2}{3}, -\dfrac{3}{4}, \dfrac{4}{5}, -\dfrac{5}{6}, \dfrac{6}{7}$
+
+16. $3, 6, 12, 24, 48, 96$
+
+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.
+
+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. $\sum_{i = 0}^{6}{a_i}$
+
+b. $\sum_{i = 0}^{0}{a_i}$
+
+c. $\sum_{j = 1}^{3}{a_{2j}}$
+
+d. $\prod_{k = 0}^{6}{a_k}$
+
+e. $\prod_{k = 2}^{2}{a_k}$
+
+Compute the summations and products in 19-28.
+
+19. $\sum_{k = 1}^{5}{(k + 1)}$
+
+20. $\prod_{k = 2}^{4}{k^2}$
+
+21. $\sum_{k = 1}^{3}{(k^2 + 1)}$
+
+22. $\prod_{j = 0}^{4}{(-1)^j}$
+
+23. $\sum_{i = 1}^{1}{i(i + 1)}$
+
+24. $\sum_{j = 0}^{0}{(j + 2) \cdot 2^j}$
+
+25. $\prod_{k = 2}^{2}{\left(1 - \dfrac{1}{k}\right)}$
+
+26. $\sum_{k = -1}^{1}{(k^2 + 3)}$
+
+27. $\sum_{n = 1}^{6}{\left(\dfrac{1}{n} - \dfrac{1}{n + 1}\right)}$
+
+28. $\prod_{i = 2}^{5}{\dfrac{i(i + 2)}{(i - 1) \cdot (i + 1)}}$
+
+Write the summations in 29-32 in expanded form.
+
+29. $\sum_{i = 1}^{n}{(-2)^i}$
+
+30. $\sum_{j = 1}^{n}{j(j + 1)}$
+
+31. $\sum_{k = 0}^{n + 1}{\dfrac{1}{k!}}$
+
+32. $\sum_{i = 1}^{k + 1}{i(i!)}$
+
+Evaluate the summations and products in 33-36 for the indicated values of the
+variable.
+
+33. $\dfrac{1}{1^2} + \dfrac{1}{2^2} + \dfrac{1}{3^2} + \dots + \dfrac{1}{n^2}; n = 1$
+
+34. $1(1!) + 2(2!) + 3(3!) + \dots + m(m!); m = 2$
+
+35. $\left(\dfrac{1}{1 + 1}\right)\left(\dfrac{2}{2 + 1}\right)\left(\dfrac{3}{3 + 1}\right) \dots \left(\dfrac{k}{k + 1}\right); k = 3$
+
+36. $\left(\dfrac{1 \cdot 2}{3 \cdot 4}\right)\left(\dfrac{4 \cdot 5}{6 \cdot 7}\right)\left(\dfrac{6 \cdot 7}{8 \cdot 9}\right) \dots \left(\dfrac{m \cdot (m + 1)}{(m + 2) \cdot (m + 3)}\right); m = 1$
+
+Write each of 37-39 as a single summation.
+
+37. $\sum_{i = 1}^{k}{i^3 + (k + 1)^3}$
+
+38. $\sum_{k = 1}^{m}{\dfrac{k}{k + 1} + \dfrac{m + 1}{m + 2}}$
+
+39. $\sum_{m = 0}^{n}{(m + 1)2^m + (n + 2)2^{n + 1}}$
+
+Rewrite 40-42 by separating off the final term.
+
+40. $\sum_{i = 1}^{k + 1}{i(i!)}$
+
+41. $\sum_{k = 1}^{m + 1}{k^2}$
+
+42. $\sum_{m = 1}^{n + 1}{m(m + 1)}$
+
+Write each of 43-52 using summation or product notation.
+
+43. $1^2 - 2^2 + 3^2 - 4^2 + 5^2 - 6^2 + 7^2$
+
+44. $(1^3 - 1) - (2^3 - 1) + (3^3 - 1) - (4^3 - 1) + (5^3 - 1)$
+
+45. $(2^2 - 1) \cdot (3^2 - 1) \cdot (4^2 - 1)$
+
+46. $\dfrac{2}{3 \cdot 4} - \dfrac{3}{4 \cdot 5} + \dfrac{4}{5 \cdot 6} - \dfrac{5}{6 \cdot 7} + \dfrac{6}{7 \cdot 8}$
+
+47. $1 - r + r^2 - r^3 + r^4 - r^5$
+
+48. $(1 - t) \cdot (1 - t^2) \cdot (1 - t^3) \cdot (1 - t^4)$
+
+49. $1^3 + 2^3 + 3^3 + \dots + n^3$
+
+50. $\dfrac{1}{2!} + \dfrac{2}{3!} + \dfrac{3}{4!} + \dots + \dfrac{n}{(n + 1)!}$
+
+51. $n + (n - 1) + (n - 2) + \dots + 1$
+
+52. $n + \dfrac{n - 1}{2!} + \dfrac{n - 2}{3!} + \dfrac{n - 3}{4!} + \dots + \dfrac{1}{n!}$
+
+Transform each of 53 and 54 by making the change of variable $i = k + 1$.
+
+53. $\sum_{k = 0}^{5}{k(k - 1)}$
+
+54. $\prod_{k = 1}^{n}{\dfrac{k}{k^2 + 4}}$
+
+Transform each of 55-58 by making the change of variable $j = i - 1$.
+
+55. $\sum_{i = 1}^{n + 1}{\dfrac{(i - 1)^2}{i \cdot n}}$
+
+56. $\sum_{i = 3}^{n}{\dfrac{i}{i + n - 1}}$
+
+57. $\sum_{i = 1}^{n - 1}{\dfrac{i}{(n - i)^2}}$
+
+58. $\prod_{i = n}^{2n}{\dfrac{n - i + 1}{n + i}}$
+
+Write each of 59-61 as a single summation or product.
+
+59. $3 \cdot \sum_{k = 1}^{n}{(2k - 3)} + \sum_{k = 1}^{n}{(4 - 5k)}$
+
+60. $2 \cdot \sum_{k = 1}^{n}{(3k^2 + 4)} + 5 \cdot \sum_{k = 1}^{n}{(2k^2 - 1)}$
+
+61. $\left(\prod_{k = 1}^{n}{\dfrac{k}{k + 1}}\right) \cdot \left(\prod_{k = 1}^{n}{\dfrac{k + 1}{k + 2}}\right)$
+
+Compute each of 62-76. Assume the values of the variables are restricted so that
+the expressions are defined.
+
+62. $\dfrac{4!}{3!}$
+
+63. $\dfrac{6!}{8!}$
+
+64. $\dfrac{4!}{0!}$
+
+65. $\dfrac{n!}{(n - 1)!}$
+
+66. $\dfrac{(n - 1)!}{(n + 1)!}$
+
+67. $\dfrac{n!}{(n - 2)!}$
+
+68. $\dfrac{((n + 1)!)^2}{(n!)^2}$
+
+69. $\dfrac{n!}{(n - k)!}$
+
+70. $\dfrac{n!}{(n - k + 1)!}$
+
+71. $\dbinom{5}{3}$
+
+72. $\dbinom{7}{4}$
+
+73. $\dbinom{3}{0}$
+
+74. $\dbinom{5}{5}$
+
+75. $\dbinom{n}{n - 1}$
+
+76. $\dbinom{n + 1}{n - 1}$
+
+77.
+
+a. Prove that $n! + 2$ is divisible by $2$, for every integer $n \geq 2$.
+
+b. Prove that $n! + k$ is divisible by $k$, for every integer $n \geq 2$ and
+$k = 2, 3, \dots, n$.
+
+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.
+
+78. Prove that for all nonnegative integers $n$ and $r$ with
+
+$$ r + 1 \leq n, \binom{n}{r + 1} = \frac{n - r}{r + 1}\binom{n}{r} $$
+
+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 $p$.
+
+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.
+
+$\text{sum } := 0\\ \text{\textbf{for }} i := 0 \text{\textbf{ to \_\_\_\_}}\\ \ \ \text{sum } := \text{\_\_\_\_}\\ \text{\textbf{next }} i$
+
+b.
+
+$\text{sum } := 0\\ \text{\textbf{for }} j := 2 \text{\textbf{ to \_\_\_\_}}\\ \ \ \text{sum } := \text{\_\_\_\_}\\ \text{\textbf{next }} j$
+
+Use repeated division by $2$ to convert (by hand) the integers in 81-83 from
+base 10 to base 2.
+
+81. $90$
+
+82. $98$
+
+83. $205$
+
+Make a trace table to trace the action of Algorithm 5.1.1 on the input in 84-86.
+
+84. $23$
+
+85. $28$
+
+86. $44$
+
+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).
+
+Use the algorithm you developed for exercise 87 to convert the integers in 88-90
+to hexadecimal notation.
+
+88. $287$
+
+89. $693$
+
+90. $2,301$
+
+91. Write a formal version of the algorithm you developed for exercise 87.
diff --git a/chapter_5/notes.md b/chapter_5/notes.md
index e69de29..a9a1aa1 100644
--- a/chapter_5/notes.md
+++ b/chapter_5/notes.md
@@ -0,0 +1,120 @@
+Page 284
+
+**Definition**
+
+If $m$ and $n$ are integers and $m \leq n$, the symbol $\sum_{k=m}^{n}{a_k}$,
+read the **summation from $k$ equals $m$ to $n$ of $a$-sub-$k$**, is the sum of
+all the terms $a_m, a_{m + 1}, a_{m + 2}, \dots, a_n$. We say that
+$a_m + a_{m + 1} + a_{m + 2} + \dots + a_n$ is the **expanded form** of the sum,
+and we write
+
+$$ \sum_{k=m}^{n}{a_k} = a_m + a_{m + 1} + a_{m + 2} + \dots + a_n $$
+
+We call $k$ the **index** of the summation, $m$ the **lower limit** of the
+summation, and $n$ the **upper limit** of the summation.
+
+---
+
+Page 287
+
+**Definition**
+
+If $m$ and $n$ are integers and $m \leq n$, the symbol $\prod_{k = m}^{n}{a_k}$
+read the **product from $k$ equals $m$ to $n$ of $a$-sub-$k$**, is the product
+of all the terms $a_m, a_{m + 1}, a_{m + 2}, \dots, a_n$.
+
+We write
+
+$$ \prod_{k = m}^{n}{a_k} = a_m \cdot a_{m + 1} \cdot a_{m + 1} \dots a_n $$
+
+---
+
+Page 288
+
+**Theorem 5.1.1**
+
+If $a_m, a_{m + 1}, a_{m + 1}, \dots$ and $b_m, b_{m + 1}, b_{m + 1}, \dots$ are
+sequences of real numbers and $c$ is any real number, then the following
+equations hold for any integer $n \geq m$:
+
+1. $\sum_{k = m}^{n}{a_k} + \sum_{k = m}^{n}{b_k} = \sum_{k = m}^{n}{(a_k + b_k)}$
+
+2. $c \cdot \sum_{k = m}^{n}{a_k} = \sum_{k = m}^{n}{c \cdot a_k} \quad \text{generalized distributive law}$
+
+3. $\left(\prod_{k = m}^{n}{a_k}\right) \cdot \left(\prod_{k = m}^{n}{b_k}\right) = \prod_{k = m}^{n}{(a_k \cdot b_k)}$
+
+---
+
+Page 291
+
+**Definition**
+
+For each positive integer $n$, the quantity **$n$ factorial** denoted $n!$, is
+defined to be the product of all the integers from $1$ to $n$:
+
+$$ n! = n \cdot (n - 1) \dots 3 \cdot 2 \cdot 1 $$
+
+**Zero factorial**, denoted $0!$, is defined to be $1$:
+
+$$ 0! = 1 $$
+
+---
+
+Page 292
+
+**Definition**
+
+Let $n$ and $r$ be integers with $0 \leq r \leq n$. The symbol
+
+$$ \binom{n}{r} $$
+
+is read "**$n$ choose $r$**" and represents the number of subsets of size $r$
+that can be chosen from a set with $n$ elements.
+
+---
+
+Page 292
+
+**Formula for Computing $\dbinom{n}{r}$**
+
+For all integers $n$ and $r$ with $0 \leq r \leq n$,
+
+$$ \binom{n}{r} = \frac{n!}{r!(n - r)!} $$
+
+---
+
+Page 295
+
+**Algorithm 5.1.1 Decimal to Binary Conversion Using Repeated Division by $2$**
+
+_[In Algorithm 5.1.1 the input is a nonnegative integer $a$. The aim of the
+algorithm is to produce a sequence of binary digits $r[0], r[1], r[2], \dots
+r[k] so that the binary representation of $n$ is_
+
+$$ \left(r[k]r[k - 1] \dots r[2]r[1]r[0]\right)_2 $$
+
+_That is,_
+
+$$ a = 2^k \cdot r[k] + 2^{k - 1} \cdot r[k - 1] + \dots + 2^3 \cdot r[2] + 2^1 \cdot r[1] + 2^0 \cdot r[0] $$
+
+_.]_
+
+**Input:** $a$ _[a nonegative integer]_
+
+**Algorithm Body:**
+
+$q := a, i := 0$
+
+_[Repeatedly perform the integer division of $q$ by $2$ 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 $q \neq 0$.]_
+
+$\text{\textbf{while }}(i = 0 \text{ or } q \new 0)\\ \ \ r[i] := q \mod 2\\ \ \ q := q \text{ div } 2\\ \ \ \text{[r[i] and q can be obtained by calling the division algorithm.]}\\ \ \ i := i + 1\\ \text{\textbf{end while}}$
+
+_[After execution of this step, the values of $r[0], r[1], \dots, r[i - 1]$ are
+all $0$'s and $1$'s, and
+$a = \left(r[i - 1]r[i - 2] \dots r[2]r[1]r[0]\right)_2$.]_
+
+**Output:** $r[0], r[1], r[2], \dots, r[i - 1]$ _[a sequence of integers]_
diff --git a/chapter_5/test_yourself.md b/chapter_5/test_yourself.md
index e69de29..68d23a0 100644
--- a/chapter_5/test_yourself.md
+++ b/chapter_5/test_yourself.md
@@ -0,0 +1,18 @@
+**Test Yourself**
+
+Page 296
+
+1. The notation $\sum_{k = m}^{n}{a_k}$ is read "_____."
+
+2. The expanded form of $\sum_{k = m}^{n}{a_k}$ is _____.
+
+3. The value of $a_1 + a_2 + a_3 + \dots + a_n$ when $n = 2$ is "_____."
+
+4. The notation $\prod_{k = m}^{n}{a_k}$ is read "_____."
+
+5. If $n$ is a positive integer, then $n! =$ _____.
+
+6. $\sum_{k = m}^{n}{a_k} + c\sum_{k = m}^{n}{b_k} =$ _____.
+
+7. $\left(\prod_{k = m}^{n}{a_k}\right)\left(\prod_{k = m}^{n}{b_k}\right) =$
+   _____.