### 1 Name: Ernst : 2018-10-31 17:55 [Del]

\documentclass[letterpaper]{article}
\newtheorem{thm}{Theorem}
\newproof{proof}{Proof}
\begin{document}
\title{Sample Proof Using Mathematical Induction}
\author{Brandon Grasley}
\date{2014-01-21}

\titlepage
\begin{thm}
\\For any $n \in \mathbb{N}$,
$\sum_{i=1}^{n}i=\frac{n\left ( n+1 \right )}{2}$
\end{thm}
\begin{proof}
\\Base case $n=1$: If $n=1$, the left side is 1 and the right side is $\frac{1\left ( 2\right )}{2}=1$.
So, the theorem holds when $n=1$.
Inductive hypothesis: Suppose the theorem holds for all values of $n$ up to some $k$, $k \geq 1$.
Inductive step: Let $n=k+1$. Then our left side is
\begin{align}
$\sum_{i=1}^{k+1}i&=\left (k+1\right )+\sum_{i=1}^{k}i\\ &=\left (k+1\right )+\frac{k\left ( k+1 \right )}{2}$\text{, by our inductive hypothesis}\\
$&=\frac{2\left (k+1 \right )}{2}+\frac{k\left (k+1 \right )}{2}\\ &=\frac{2\left (k+1 \right )+k\left (k+1 \right )}{2}\\ &=\frac{\left (k+1 \right )\left (k+2\right )}{2}$
\end{align}
which is our right side. So, the theorem holds for $n=k+1$.
By the principle of mathematical induction, the theorem holds for all $n \in \mathbb{N}$.
\end{proof}
\end{document}

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