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Use mathematical induction to prove each of the following:

For each natural number $n$, $2 + 5 + 8 + \cdots + \left( {3n - 1} \right) = \dfrac{{n\left( {3n + 1} \right)}}{2}$.

Use mathematical induction to prove each of the following:

For each natural number $n$, $1 + 5 + 9 + \cdots + \left(4n - 3 \right) = n \left(2n - 1 \right)$.

Use mathematical induction to prove each of the following:

For each natural number $n$, $1^3 + 2^3 + 3^3 + \cdots + n^3 = \left[ {\dfrac{{n\left( {n + 1} \right)}}{2}} \right]^2$.

Use mathematical induction to prove each of the following:

For each natural number $n$, 3 divides $\left( {4^n - 1} \right)$.

Use mathematical induction to prove each of the following:

For each natural number $n$, 6 divides $\left( {n^3 - n} \right)$.

Use mathematical induction to prove that the sum of the cubes of any three consecutive natural numbers is a multiple of 9.

In calculus, it can be shown that

$$\begin{aligned}\int_{}^{} \sin^{\,2} x \, dx & = \frac{x}{2} - \frac{1}{2} \sin x \cos x + c \text{ and } \\ \int_{}{} \cos^{\,2} x \, dx & = \frac{x}{2} + \frac{1}{2} \sin x \cos x + c\end{aligned}$$

. Using integration by parts, it is also possible to prove that for each natural number $n$,

$$\begin{aligned}\int_{}{} \sin^{\,n} x \, dx & = -\frac{1}{n} \sin^{n-1}x \cos x + \frac{n-1}{n} \int{}{} \sin^{\,n-2}x \, dx \text{ and } \\ \int_{}{} \cos^{\,n} x \, dx & = \frac{1}{n} \cos^{n-1}x \sin x + \frac{n-1}{n} \int{}{} \cos^{\,n-2}x \, dx\end{aligned}$$

.

Determine the values of

$$\int_{0}^{\pi/2} \sin^{\,2} x \, dx \qquad \text{ and } \qquad \int_{0}^{\pi/2} \sin^{\,4} x \, dx$$

.

In calculus, it can be shown that

$$\begin{aligned}\int_{}^{} \sin^{\,2} x \, dx & = \frac{x}{2} - \frac{1}{2} \sin x \cos x + c \text{ and } \\ \int_{}{} \cos^{\,2} x \, dx & = \frac{x}{2} + \frac{1}{2} \sin x \cos x + c\end{aligned}$$

. Using integration by parts, it is also possible to prove that for each natural number $n$,

$$\begin{aligned}\int_{}{} \sin^{\,n} x \, dx & = -\frac{1}{n} \sin^{n-1}x \cos x + \frac{n-1}{n} \int{}{} \sin^{\,n-2}x \, dx \text{ and } \\ \int_{}{} \cos^{\,n} x \, dx & = \frac{1}{n} \cos^{n-1}x \sin x + \frac{n-1}{n} \int{}{} \cos^{\,n-2}x \, dx\end{aligned}$$

.

Use mathematical induction to prove that for each natural number $n$,

$$\begin{aligned}\int_{0}^{\pi/2} \sin^{\,2n}x \, dx & = \frac{1 \cdot 3 \cdot 5 \cdots (2n - 1)}{2 \cdot 4 \cdot 6 \cdots (2n)} \frac{\pi}{2} \text{ and } \\ \int_{0}^{\pi/2} \sin^{\,2n+1}x \, dx & = \frac{2 \cdot 4 \cdot 6 \cdots (2n)}{1 \cdot 3 \cdot 5 \cdots (2n + 1)}\end{aligned}$$

. These are known as the Wallis sine formulas.

In calculus, it can be shown that

$$\begin{aligned}\int_{}^{} \sin^{\,2} x \, dx & = \frac{x}{2} - \frac{1}{2} \sin x \cos x + c \text{ and } \\ \int_{}{} \cos^{\,2} x \, dx & = \frac{x}{2} + \frac{1}{2} \sin x \cos x + c\end{aligned}$$

. Using integration by parts, it is also possible to prove that for each natural number $n$,

$$\begin{aligned}\int_{}{} \sin^{\,n} x \, dx & = -\frac{1}{n} \sin^{n-1}x \cos x + \frac{n-1}{n} \int{}{} \sin^{\,n-2}x \, dx \text{ and } \\ \int_{}{} \cos^{\,n} x \, dx & = \frac{1}{n} \cos^{n-1}x \sin x + \frac{n-1}{n} \int{}{} \cos^{\,n-2}x \, dx\end{aligned}$$

.

Use mathematical induction to prove that

$$\begin{aligned}\int_{0}^{\pi/2} \cos^{\,2n}x \, dx & = \frac{1 \cdot 3 \cdot 5 \cdots (2n - 1)}{2 \cdot 4 \cdot 6 \cdots (2n)} \frac{\pi}{2} \text{ and } \\ \int_{0}^{\pi/2} \cos^{\,2n+1}x \, dx & = \frac{2 \cdot 4 \cdot 6 \cdots (2n)}{1 \cdot 3 \cdot 5 \cdots (2n+1)}\end{aligned}$$

. These are known as the Wallis cosine formulas.

Complete the following.

Why is it not possible to use mathematical induction to prove a proposition of the form

$$\left( {\forall x \in \mathbb{Q}} \right)\left( {P( x )} \right)$$

, where $P( x )$ is some predicate?

Complete the following.

Why is it not possible to use mathematical induction to prove a proposition of the form

For each real number

x

with

x \geq 1

,

P( x )

, where

P( x )

is some predicate?

Consider the following question:

For which natural numbers

n

do there exist nonnegative integers

x

and

y

such that

n = 3x + 5y

?

To help answer this question, we will let $\mathbb{Z}^* = \left\{ {x \in \mathbb{Z} \mid x \geq 0} \right\}$, and let $P \left( n \right)$ be There exist $x,y \in \mathbb{Z}^*$ such that $n = 3x + 5y$. Notice that $P(1)$ is false since if both $x$ and $y$ are zero, then $3x + 5y = 0$ and if either $x > 0$ or $y > 0$, then $3x + 5y \geq 3$. Also notice that $P(6)$ is true since $6 = 3 \cdot 2 + 5 \cdot 0$ and $P(8)$ is true since $8 = 3 \cdot 1 + 5 \cdot 1$.

We could continue trying to determine other values of $n$ for which $P(n)$ is true. However, let us see if we can use the work in part (2) to determine if $P(13)$ is true. Notice that $13 = 3 + 10$ and we know that $P(10)$ is true. We should be able to use this to prove that $P(13)$ is true. This is formalized in the next part.

Let $k \in \mathbb{N}$ with $k \geq 10$. Prove that if $P( 8 )$, $P ( 9 )$, $\ldots$, $P( k )$ are true, then $P( {k+1} )$ is true.

Hint: $k + 1 = 3 + \left( {k - 2} \right)$.

Use mathematical induction to prove each of the following:

For each natural number $n$ with $n \geq 6$, $2^n > \left( n + 1 \right)^2$.

Use mathematical induction to prove each of the following:

For each natural number $n$ with $n \geq 3$, $\left( {1 + \dfrac{1}{n}} \right)^n \lt n$.

For which natural numbers $n$ is $n! > 3^n$? Justify your conclusion.

Complete the following.

Verify that $\left( {1 - \dfrac{1}{4}} \right) = \dfrac{3} {4}$ and that $\left( {1 - \dfrac{1}{4}} \right)\left( {1 - \dfrac{1}{9}} \right) = \dfrac{4} {6}$.

Complete the following.

Verify that $\left( {1 - \dfrac{1}{4}} \right)\left( {1 - \dfrac{1}{9}} \right) \left( {1 - \dfrac{1}{{16}}} \right) = \dfrac{5}{8}$ and that $\left( {1 - \dfrac{1}{4}} \right)\left( {1 - \dfrac{1}{9}} \right) \left( {1 - \dfrac{1}{{16}}} \right)\left( {1 - \dfrac{1}{{25}}} \right) = \dfrac{6}{{10}}$.

Let $y = \ln x$.

Let $n$ be a natural number. Formulate a conjecture for a formula for $\dfrac{{d^n y}}{{dx^n }}$. Then use mathematical induction to prove your conjecture.

For which natural numbers $n$ do there exist nonnegative integers $x$ and $y$ such that $n = 4x + 5y$? Justify your conclusion.