# First Principle of Finite Induction

Author

Ernest Michael Nelson

License

Creative Commons CC BY 4.0

Abstract

Mathematical Induction paper

Share your thoughts on the Overleaf Template Gallery!

Author

Ernest Michael Nelson

License

Creative Commons CC BY 4.0

Abstract

Mathematical Induction paper

```
\documentclass[article]{article}
\begin{document}
%%%%%%%%% This is the title page %%%%%%%%%%%%%%%%
\title{First Principle of Finite Induction}
\author{Ernest Michael Nelson}
\date{\today}
\maketitle
%%%%%%%%%%% End of title page %%%%%%%%%%%%%%%%%%%
%%%%%%%%%%% beginning of Introduction section %%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Introduction}
In this paper, we will be doing a step by step solution to a common induction problem.Also it is show and given in many books in the study of number theory and reasoning and proofs.
%%%%%%%%%%%%%%%%%%%%%%%%% End of Introduction Section %%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%% Formula %%%%%%%%%%%%%%%%%%%%%%%%%
\section{Formula}
We are given the formula
$$\sum_{n=1}^{\infty}n^2=\frac{n(2n+1)(n+1)}{6}$$ with n element of the natural numbers.
\section{Proof:}
Now we begin by running off a few terms to help see the pattern emerge.
$$\sum_{n=1}^{\infty}1^2+2^2+3^3+4^2+5^2+{\cdots}+n^2=\frac{n(2n+1)(n+1)}{6}$$
Let's assume that n=1
$$1=\frac{1(2(1)+1)(1+1)}{6}=1$$
$$1=\frac{(2+1)(1+1)}{6}=1$$
$$1=\frac{6}{6}=1$$
%%%%%%%%%%%%%%%%%%%%%End of formula %%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%% Induction Hypothesis %%%%%%%%%%%%%%%%
Next we assume that n=k and that k is an element of of the natural numbers.An we will use this equation in the Induction hypothesis that we denote as equation 1.
$$\sum_{k=1}^{\infty}1^2+2^2+3^3+4^+5^2+{\cdots}+k^2=\frac{k(2k+1)(k+1)}{6}$$
To obtain the next term, we add k square with k+1 square to both sides of the equation.
$$\sum_{k=1}^{\infty}1^2+2^2+3^2+4^2+5^2+{\cdots}+k^2+(k+1)^2 =\frac{(k+1)(2(k+1)+1)((k+1)+1)}{6}\\$$
Now we simplify the formula to aide in seeing the connection.
$$\sum_{k=1}^{\infty}1^2+2^+3^2+4^2+5^2+{\cdots}+k^2+(k+1)^2 =\frac{(k+1)(2k+3)(k+2)}{6}\\$$
From the induction hypotheses, we use equation 1 on the left hand side (LHS) and substitute it in for the induction process.
$$\frac{k(2k+1)(k+1)}{6}+(k+1)^2=\frac{(k+1)(2k+3)(k+2)}{6}$$
Now we factor out a (k+1) on the (LHS) of the equation.
$$(k+1)(\frac{k(2k+1)}{6}]+(k+1))=\frac{(k+1)(2k+3)(k+2)}{6}$$
Next we combine the left hand side (LHS) equation by finding the greatest conman factor (GCF).
$$(k+1)(\frac{k(2k+1)}{6}+\frac{6(k+1)}{6})=\frac{(k+1)(2k+3)(k+2)}{6}$$
$$(k+1)(\frac{k(2k+1)+6(k+1)}{6})=\frac{(k+1)(2k+3)(k+2)}{6}$$
The next step is that we factor the (LHS) of the equation.
$$(k+1)(\frac{2k^2+k+6k+6}{6})=\frac{(k+1)(2k+3)(k+2)}{6}$$
Then we combine like terms on the (LHS).
$$(k+1)(\frac{2k^2+7k+6}{6})=\frac{(k+1)(2k+3)(k+2)}{6}$$
Now we factor the (LHS) numerator.
$$(k+1)(\frac{(2k+3)(k+2)}{6})=\frac{(k+1)(2k+3)(k+2)}{6}$$
Final we factor the (k+1) back in the (LHS) of the equation.
$$\frac{(k+1)(2k+3)(k+1)}{6} =\frac{(k+1)(2k+3)(k+2)}{6}$$
%%%%%%%%%%% End of Induction hypothesis %%%%%%%%%%%%%%%%%%%
Thus we achieved what we desired.\\
%%%%%%%%%%%% References %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
Reference:
David M.Burton, Elements of Number Theory,page 3
%%%%%%%%%%%% End of Reference %%%%%%%%%%%%%%%%%%%%%%%%%%%%
\end{document}
```