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%%%%%%%%%%%%%%%%%%%%%% Title %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\title{Simple Mathematical Induction}
\author{Ernest Michael Nelson}
\date{\today}
\maketitle
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%%%%%%%%%%%%%%%%%%%%% Introduction %%%%%%%%%%%%%%%%%%%%%%%%%
\section{Introduction}
With this paper we will establish formula by the use of mathematical induction. An it will be a step by step way to solve. An hopefully it will help out the reader
%%%%%%%%%%%%%%%%%%%% End of Introduction %%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%% Formula %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Formula}
We are given the formula
$$\sum_{n=1}^{\infty}i=\frac{n(n+1)}{2}, \forall n \geq1$$
And n is element of the positive integers.
%%%%%%%%%%%%%%%%%%%% End of Formula %%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%% Proof %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Proof}
We began by running off a few tern to aid us in seeing a pattern emerge in the formula.
$$\sum_{n=1}^{\infty} 1+2+3+4+5+\cdot+n=\frac{n(n+1)}{2}$$
%%%%%%%%%%%%%%%%%%%% Initial Step %%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Initial Step}
Let's assume that n=1,
$$1=\frac{1(1+1)}{2}=1$$
$$1 = \frac{2}{2}=1$$
%%%%%%%%%%%%%%%%%%% End of Initial Step %%%%%%%%%%%%%%%%%%%%%%
Next we assume that n=k and k is an element of the positive integers. And we denote it as equation 1.
\begin{equation}\label{1}
\sum_{k=1}^{\infty} 1+2+3+4+5+\cdots+k=\frac{k(k+1)}{2}
\end{equation}
$$\sum_{k=1}^{\infty} 1+2+3+4+5+\cdots+k=\frac{k(k+1)}{2}$$
Now we do k+1 step and use equation 1 to help us solve the proof.
$$\sum_{k=1}^{\infty} 1+2+3+4+5+\cdots+k+k+1=\frac{(k+1)((k+1)+1)}{2}$$
To aid us in seeing the proof we will simplify the right hand side (RHS) of the equation.
$$\sum_{k=1}^{\infty} 1+2+3+4+5+\cdots+k+k+1=\frac{(k+1)(k+2)}{2}$$
From \eqref{1} we will use in the Induction hypothesis t
o prove the formula.
$$\frac{k(k+1)}{2}+ k+1 =\frac{(k+1)(k+2)}{2}$$
The next step we do is find the greatest common factor (GCF) on the left hand side (LHS).
$$\frac{k(k+1)}{2}+\frac{2(k+1)}{2}=\frac{(k+1)(k+2)}{2}$$
Now we can combine the (LHS) of the equation since we have the same denominator.
$$\frac{k(k+1)+2(k+1)}{2}=\frac{(k+1)(k+2)}{2}$$
Next we factor the equation on the (LHS).
$$\frac{k^2+k+2k+2}{2}=\frac{(k+1)(k+2)}{2}$$
Combine like terms on the (LHS).
$$\frac{k^2+3k+2}{2}=\frac{(k+1)(k+2)}{2}$$
Final we factor the equation on the (LHS) and we will be done.
$$\frac{(k+1)(k+2)}{2}=\frac{(k+1)(k+2)}{2}$$
Thus we achieved what we desired>
%%%%%%%%%%%%%%%%%%%% End of Proof %%%%%%%%%%%%%%%%%%%%%%%%%%%
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