# First Principle of Finite Induction

Författare

Ernest Michael Nelson

Last Updated

för 9 år sedan

Licens

Creative Commons CC BY 4.0

Sammanfattning

Mathematical Induction paper

Författare

Ernest Michael Nelson

Last Updated

för 9 år sedan

Licens

Creative Commons CC BY 4.0

Sammanfattning

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}
```

We only use cookies for essential purposes and to improve your experience on our site. You can find out more in our cookie policy.