\documentclass[letterpaper, 11pt]{article}
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\title{Induction Proof}
\author{Dr. Presser}
\begin{document}
\maketitle
\section{Theorem}
For all integers $n \geq 1$,
$$
\sum_{i=1}^n{i} = \frac{n(n+1)}{2}
$$
\section{Proof (by mathematical induction)}
Let the property $P(n)$ be the equation $1 + 2 + \cdots + n = n(n+1)/2$.
\subsection{Basis}
Show that the property $P(n)$ is true for $n=1$.
We must show that $1 = \frac{1(1+1)}{2}$. The right hand side of the
equation is $\frac{1(1+1)}{2} = \frac{2}{2} = 1$, which is the same
as the left hand side. So the property is true for $n=1$.
\subsection{Induction}
Show that for all integers $n=k$, if $P(k)$ is true, then so is $P(k+1)$.
For the induction hypothesis, suppose $1+2+ \cdots + k = \frac{k(k+1)}{2}$,
for some integer $k \geq 1$. From this we must show that
$1+2+ \cdots + (k+1) = \frac{(k+1)(k+2)}{2}$.
The left hand side of the equation can be expanded to:
$ (1 + 2 + \cdots + k) + (k+1) $. Substituting using the induction hypothesis, this is:
$$
\frac{k(k+1)}{2} + (k+1)
$$
Finding a common denominator and simplifying, we have:
\begin{eqnarray}
\frac{k(k+1)}{2} + (k+1) & = & \frac{k(k+1)}{2} + \frac{(k+1) \cdot 2}{2} \\
& = & \frac{(k+1)(k+2)}{2}
\end{eqnarray}
which is what we were trying to show. QED.
\end{document}