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chapter_5/exercises.md
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@ -837,3 +837,347 @@ $$ 2301_{10} = 8FD_{16} $$
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
$\left(1 - \dfrac{1}{2}\right)\left(1 - \dfrac{1}{3}\right)\left(1 - \dfrac{1}{4}\right)\left(1 - \dfrac{1}{5}\right)\left(1 - \dfrac{1}{6}\right) = \dfrac{1}{6}$.
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
$\left(1 - \dfrac{1}{2}\right)\left(1 - \dfrac{1}{3}\right)\left(1 - \dfrac{1}{4}\right)\left(1 - \dfrac{1}{5}\right)\left(1 - \dfrac{1}{6}\right)\left(1 - \dfrac{1}{7}\right) = \dfrac{1}{7}$.
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?
b. Write $P(k)$.
c. Write $P(k + 1)$.
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?
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?
b. Write $P(k)$.
c. Write $P(k + 1)$.
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?
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 $P(1)$. Is $P(1)$ true?
b. Write $P(k)$.
c. Write $P(k + 1)$.
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?
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.
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 $n \geq 1$,
$$ 2 + 4 + 6 + \dots + 2n = n^2 + n $$
7. For every integer $n \geq 1$,
$$ 1 + 6 + 11 + 16 + \dots + (5n - 4) = \frac{n(5n - 3)}{2} $$
8. For every integer $n \geq 0$,
$$ 1 + 2 + 2^2 + \dots + 2^n = 2^{n + 1} + 1 $$
9. For every integer $n \geq 3$,
$$ 4^3 + 4^4 + 4^5 + \dots + 4^n = \frac{4(4^n - 16)}{3} $$
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
$n \geq 1$.
11. $1^3 + 2^3 + \dots + n^3 = \left[\dfrac{n(n + 1)}{2}\right]^2$, for every
integer $n \geq 1$.
12. $\dfrac{1}{1 \cdot 2} + \dfrac{1}{2 \cdot 3} + \dots + \dfrac{1}{n(n + 1)} = \dfrac{n}{n + 1}$,
for every integer $n \geq 1$.
13. $\sum_{i = 1}^{n - 1}{i(i + 1)} = \dfrac{n(n - 1)(n + 1)}{3}$, for every
integer $n \geq 2$.
14. $\sum_{i = 1}^{n + 1}{i \cdot 2^i} = n \cdot 2^{n + 2} + 2$, for every
integer $n \geq 0$.
15. $\sum_{i = 1}^{n}{i(i!)} = (n + 1)! - 1$, for every integer $n \geq 1$.
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 $n \geq 2$.
17. $\prod_{i = 0}^{n}{\left(\dfrac{1}{2i + 1} \cdot \dfrac{1}{2i + 2}\right)} = \dfrac{1}{(2n + 2)!}$,
for every integer $n \geq 0$.
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.
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 $n \geq 1$,
$\dfrac{d(x^n)}{dx} = nx^{n - 1}$.
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. $4 + 8 + 12 + 16 + \dots + 200$
21. $5 + 10 + 15 + 20 + \dots + 300$
22.
a. $3 + 4 + 5+ 6 + \dots + 1000$
b. $3 + 4 + 5 + 6 + \dots + m$
23.
a. $7 + 8 + 9 + 10 + \dots + 600$
b. $7 + 8 + 9 + 10 + \dots + k$
24. $1 + 2 + 3 + \dots + (k - 1)$, where $k$ is any integer with $k \geq 2$.
25.
a. $1 + 2 + 2^2 + \dots + 2^{25}$
b. $2 + 2^2 + 2^3 + \dots + 2^{26}$
c. $2 + 2^2 + 2^3 + \dots + 2^n$
26. $3 + 3^2 + 3^3 + \dots + 3^n$, where $n$ is any integer with $n \geq 1$.
27. $5^3 + 5^4 + 5^5 + \dots + 5^k$, where $k$ is any integer with $k \geq 3$.
28. $1 + \dfrac{1}{2} + \dfrac{1}{2^2} + \dots + \dfrac{1}{2^n}$, where $n$ is
any positive integer.
29. $1 - 2 + 2^2 - 2^3 + \dots + (-1)^n2^n$, where $n$ is any positive integer.
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.
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.
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.
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.
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.)
b. Assuming that each generation represents 25 years, how long is 40
generations?
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?
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
$(k + 1)^2 = \dfrac{(k + 1)((k + 1) + 1)(2(k + 1) + 1)}{6}$."
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."
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."
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.
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 $p$.

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chapter_5/notes.md
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@ -118,3 +118,203 @@ 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]_
---
Page 300
**Principle of Mathematical Induction**
Let $P(n)$ be a property that is defined for integers $n$, and let $a$ be a
fixed integer. Suppose the following two statements are true:
1. $P(a)$ is true.
2. For every integer $k \geq a$, if $P(k)$ is true then $P(k + 1)$ is true.
Then the statement
$$ \text{for every integer } n \geq a, P(a) $$
is true.
---
Page 301
**Method of Proof by Mathematical Induction**
Consider the statement of the form, "For every integer $n \geq a$, a property
$P(n)$ is true." To prove such a statement, perform the following two steps:
**Step 1 (basis step):**
Show that $P(a)$ is true.
**Step 2 (inductive step):**
Show that for every integer $k \geq a$, if $P(k)$ is true then $P(k + 1)$ is
true. To perform this step,
**suppose** that $P(k)$ is true, where $k$ is any particular but arbitrarily
chosen integer with $k \geq a$. _[This supposition is called the **inductive
hypothesis**.]_
Then **show** that $P(k + 1)$ is true.
---
Page 303
**Theorem 5.2.1 Sum of the First $n$ Integers**
For every integer $n \geq 1$,
$$ 1 + 2 + \dots + n = \frac{n(n + 1)}{2} $$
**Proof (by mathematical induction):**
Let the property $P(n)$ be the equation
$$ 1 + 2 + 3 + \dots + n = \frac{n(n + 1)}{2} $$
_Show that $P(1)$ is true:_
To establish $P(1)$, we must show that
$$ 1 = \frac{1(1 + 1)}{2} $$
But the left-hand side of this equation is $1$ and the right-hand side is
$$ \frac{1(1 + 1)}{2} = \frac{2}{2} = 1 $$
also. Hence $P(1)$ is true.
_Show that for every integer $k \geq 1$, if $P(k)$ is true then $P(k + 1)$ is
also true:_
_[Suppose that $P(k)$ is true for a particular but arbitrarily chosen integer
$k \geq 1$. That is:]_
Suppose that $k$ is any integer with $k \geq 1$ such that
$$ 1 + 2 + 3 + \dots + k = \frac{k(k + 1)}{2} $$
_[We must show that $P(k + 1)$ is true. That is:]_ We must show that
$$ 1 + 2 + 3 + \dots + (k + 1) = \frac{(k + 1)[(k + 1) + 1]}{2} $$
or, equivalently, that
$$ 1 + 2 + 3 + \dots + (k + 1) = \frac{(k + 1)(k + 2)}{2} $$
_[We will show that the left-hand side and the right-hand side of $P(k + 1)$ are
equal to the same quantity and thus are equal to each other.]_
The left-hand side of $P(k + 1)$ is
$$ 1 + 2 + 3 + \dots + (k + 1) $$
$$ = 1 + 2 + 3 + \dots + k + (k + 1) $$
$$ = \frac{k(k + 1)}{2} + (k + 1) $$
$$ = \frac{k(k + 1)}{2} + \frac{2(k + 1)}{2} $$
$$ = \frac{k^2 + k}{2} + \frac{2k + 2}{2} $$
$$ = \frac{k^2 + 3k + 2}{2} $$
And the right-hand side of $P(k + 1)$ is
$$ \frac{(k + 1)(k + 2)}{2} = \frac{k^2 + 3k + 2}{2} $$
Thus the two sides of $P(k + 1)$ are equal to the same quantity and so they are
equal to each other. Therefore, the equation $P(k + 1)$ is true _[as was to be
shown]_.
_[Since we have proved both the basis step and the inductive step, we conclude
that the theorem is true.]_
---
Page 304
**Definition**
If a sum with a variable number of terms is shown to equal an expression that
does not contain an ellipsis or a summation symbol, we say that the sum is
written **in closed form.**
---
Page 306
**Theorem 5.2.2 Sum of a Geometric Sequence**
For any real number $r$ except $1$, and any integer $n \geq 0$,
$$ \sum_{i = 0}^{n}{r^i} = \frac{r^{n + 1} - 1}{r - 1} $$
**Proof (by mathematical induction):**
Suppose $r$ is a particular but arbitrarily chosen real number that is not equal
to $1$, and let the property $P(n)$ be the equation
$$ \sum_{i = 0}^{n}{r^i} = \frac{r^{n + 1} - 1}{r - 1} $$
We must show that $P(n)$ is true for every integer $n \geq 0$. We do this by
mathematical induction on $n$.
_Show that $P(0)$ is true:_
To establish $P(0)$, we must show that
$$ \sum_{i = 0}^{0}{r^i} = \frac{r^{0 + 1} - 1}{r - 1} $$
The left-hand side of this equation is $r^0 = 1$ and the right-hand side is
$$ \frac{r^{0 + 1} - 1}{r - 1} = \frac{r - 1}{r - 1} = 1 $$
also because $r^1 = r$ and, since $r \neq 1$, $r - 1 \neq 0$. Hence $P(0)$ is
true.
_Show that for every integer $k \geq 0$, if $P(k)$ is true then $P(k + 1)$ is
also true:_
_[Suppose that $P(k)$ is true for a particular but arbitrarily chosen integer
$k \geq 0$. That is:]_
Let $k$ be any integer with $k \geq 0$, and suppose that
$$ \sum_{i = 0}^{k}{r^j} = \frac{r^{k + 1} - 1}{r - 1} $$
_[We must show that $P(k + 1)$ is true. That is:]_ We must show that
$$ \sum_{i = 0}^{k + 1}{r^j} = \frac{r^{(k + 1) + 1} - 1}{r - 1} $$
or, equivalently, that
$$ \sum_{i = 0}^{k + 1}{r^j} = \frac{r^{k + 2} - 1}{r - 1} $$
_[We will show that the left-hand side of $P(k + 1)$ equals the right-hand
side.]_
The left-hand side of $P(k + 1)$ is
$$ \sum_{i = 0}^{k + 1}{r^j} = \sum_{i = 0}^{k}{r^i + r^{k + 1}} $$
$$ = \frac{r^{k + 1} - 1}{r - 1} + r^{k + 1} $$
$$ = \frac{r^{k + 1} - 1}{r - 1} + \frac{r^{k + 1}(r - 1)}{r - 1} $$
$$ = \frac{(r^{k + 1} - 1) + r^{k + 1}(r - 1)}{r - 1} $$
$$ = \frac{r^{k + 1} - 1 + r^{k + 2} - r^{k + 1}}{r - 1} $$
$$ = \frac{r^{k + 2} - 1}{r - 1} $$
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 theorem is true.]_

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chapter_5/test_yourself.md
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@ -30,3 +30,22 @@ $$ \sum_{k = m}^{n}{a_k + cb_k} $$
_____.
$$ \prod_{k = m}^{n}{a_kb_k} $$
---
**Test Yourself**
Page 309
1. Mathematical induction is a method for proving that a property defined for
integers $n$ is true for all values of $n$ that are _____.
2. Let $P(n)$ be a property defined for integers $n$ and consider constructing a
proof by mathematical induction for the statement "P(n) is true for all
$n \geq a$."
a. In the basis step one must show _____.
b. In the inductive step one supposes that _____ for a particular but
arbitrarily chosen value of an integer $k \geq a$. This supposition is called
the _____. One then has to show that _____.