In this section are completed proofs using induction. As time progresses I will slowly begin to improve the English explanation included in each proof. Below are the proof statements.

I. Bernoulli’s Inequality: If \( x > -1 \) then \( \forall n \in \mathbb{N} \) it is the case that \( \left( 1 + x \right)^n \ge 1 + nx \)

II. Prove that \( n^3 + 5n \) is divisible by \( 6 \ \forall n \in \mathbb{N} \)

III. Prove \( 1^2 – 2^2 +3^2 + \ldots + \left(-1 \right)^{n+1}n^2 = \left(-1 \right)^{n+1} \frac{n \left(n + 1 \right)}2 \ \forall n \in \mathbb{N} \)

IV. Prove using induction $$ E_n \left( x \right) = 1 + x + \frac{ x^2 }{2} + \frac{ x^3 }{3!}+ \cdots + \frac{ x^n }{n!} $$

I. Bernoulli’s Inequality:

If \( x > -1 \) then \( \forall n \in \mathbb{N} \) it is the case that \( \left( 1 + x \right)^n \ge 1 + nx \) \( \left( 1 + x \right)^1 \ge 1 + \left(1 \right) x \) By the inductive hypothesis \( \forall k \in \mathbb{N} \) it is the case that \( \left( 1 + x \right)^k \ge 1 + kx \). Since \( \left( 1 + x \right) > 0 \) it is true that \( \left( 1 + x \right)^k \left( 1 + x \right) \ge \left( 1 + kx \right) \left( 1 + x \right)\) $$ \left( 1 + x \right)^{k + 1} \ge 1 + x + kx + kx^2 = 1 + \left(k + 1 \right)x + kx^2 \ge 1 + \left(k + 1 \right)x $$. Therefore, by mathematical induction Bernoulli’s equation is correct.

II. Prove that \( n^3 + 5n \) is divisible by \( 6 \ \forall n \in \mathbb{N} \)

By definition \( 6 \vert \left(n^3 + 5n \right) \) if and only if there exists some \( m \in \mathbb{Z} \) such that \( n^3 + 5n = 6m \).
\( \left(1 \right)^3 + 5\left(1\right) = 6m \)
By the inductive hypothesis it is true that \( k^3 + 5k = 6m \) for all \( k \in \mathbb{N} \) and for some \( m \in \mathbb{Z} \)
\( k^3 + 5k +3k \left( 3k + 1 \right) + 6= 6m + 3k \left( 3k + 1 \right) + 6 \)
\( \left(k + 1 \right)^3 + 5 \left(k + 1 \right)= 6m + 3k \left( 3k + 1 \right) + 6 \)
If the quantity \( 3k \left( 3k + 1 \right) \) contains a factor of six then the proof is complete.
Since \( k \in \mathbb{Z}\) and since all integers are either even or odd then
i. if \( k \) is even then by definition \( k = 2p \) which implies \( 3k \left(3k + 1 \right) = 6p \left(6p + 1 \right)=6\left(6p + p \right) \)
ii. if \( k \) is odd then by definition \( k = 2p +1 \) which implies \( 3k \left(3k + 1 \right) = \left( 6p +3 \right) \left(6p + 4 \right)=6 \left( 6p^2 +7p + 2 \right) \)

III. Prove \( 1^2 – 2^2 +3^2 + \ldots + \left(-1 \right)^{n+1}n^2 = \left(-1 \right)^{n+1} \frac{n \left(n + 1 \right)}2 \ \forall n \in \mathbb{N} \)

\( \left(-1 \right)^2 1^2 = \left(-1 \right)^2 \frac{1 \left( 1 + 1 \right)}2 \).
By the inductive hypothesis \( \forall k \in \mathbb{N} \) it is the case that \( 1^2 – 2^2 +3^2 + \ldots + \left(-1 \right)^{k+1}k^2 = \left(-1 \right)^{k+1} \frac{k \left(k + 1 \right)}2 \). \( \sum_{l=1}^k \left(-1 \right)^{l+1}l^2 + \left(-1 \right)^{k+2} \left(k+1 \right)^2 = \left(-1 \right)^{k+1} \frac{k \left(k + 1 \right)}2 + \left(-1 \right)^{k+2} \left(k+1 \right)^2 \).
Simplifying the left-hand side and creating a common denominator on the right yields \( \sum_{l=1}^{k+1} \left(-1 \right)^{l+1}l^2 = \left(-1 \right)^{k+1} \frac{k \left(k + 1 \right)}2 + \frac{2 \left(-1 \right)^{k+2} \left(k+1 \right)^2}2 \).
\( \sum_{l=1}^{k+1} \left(-1 \right)^{l+1}l^2 = \left(-1 \right)^{k+1} \left(k+1 \right) \frac{k – 2 \left(k+1 \right)}2 = \left(-1 \right)^{k+1} \left(k+1 \right) \frac{\left(-1 \right) \left(k+2 \right)}2 = \left(-1 \right)^{k+2} \frac{\left(k + 1 \right) \left(k+2 \right)}2 \)

IV. Let \(E_n\) be a sequence of continuous functions defined as $$ E_1 \left( x \right) = 1 + x $$ $$ E_{ n + 1 } = 1 + \int_0^x E_n \left( x \right) \ dt $$ Prove using induction $$ E_n \left( x \right) = 1 + x + \frac{ x^2 }{2} + \frac{ x^3 }{3!}+ \cdots + \frac{ x^n }{n!} $$ \( \forall n \in \mathbb{N} \) and \( \forall x \in \mathbb{R} \)

First show that \( E_1 \) is true.$$ E_1 = 1 + x $$ $$E_2 = 1 + \int_0^x E_1 \left( x \right) \ dt = 1 + x + \frac{ x^2 }{2} $$
Since \(E_1\) is true then by the induction hypothesis $$ E_k \left( x \right) = 1 + \int_0^x E_{k – 1} \left( x \right) \ dt = 1 + x + \frac{ x^2 }{2} + \frac{ x^3 }{3!}+ \cdots + \frac{ x^k }{k!} $$ $$E_k + \frac{ x^{ k + 1 }}{ \left( k + 1 \right)! } = 1 + x + \frac{ x^2 }{2} + \frac{ x^3 }{3!}+ \cdots + \frac{ x^k }{k!} + \frac{ x^{ k + 1 }}{ \left( k + 1 \right)! } = E_{ k + 1 } $$
therefore by mathematical induction
$$ E_n \left( x \right) = 1 + x + \frac{ x^2 }{2} + \frac{ x^3 }{3!}+ \cdots + \frac{ x^n }{n!} $$