Chapter 12.5, Problem 6E

To determine

To Prove: $1^2+2^2+3^2+\mathrm{.......}+n^2=\frac{n\left(n+1\right)\left(2n+1\right)}{6}$ is True for all natural numbers $n$ .

Explanation of Solution

Given information:

Use Mathematical induction to prove the above mentioned formula.

Concept used:

1.By using Mathematical induction technique we need to prove that this formula holds True for all natural numbers. The first step is to prove True for $n=1$ .

2.Second step by using induction hypothesis of mathematical induction we assume for $n=k$ given formula is True.

3.Third step is we need to prove that above formula is True for $n=k+1$ by using formula is True for $n=k$ .

Proof:

First we need to prove the above formula is True for $n=1$ .

Given formula is

  $1^2+2^2+3^2+\mathrm{.......}+n^2=\frac{n\left(n+1\right)\left(2n+1\right)}{6}$

Put $n=1$

We get

Left hand side = $1^2=1$ .

Right hand side = $\frac{1\left(1+1\right)\left(2\left(1\right)+1\right)}{6}=\frac{1\left(2\right)\left(3\right)}{6}=\frac{6}{6}=1$

So Left hand side = Right hand side

It True for $n=1$ .

Let us assume it is True for $n=k$

So

  $1^2+2^2+3^2+\mathrm{.......}+k^2=\frac{k\left(k+1\right)\left(2k+1\right)}{6}$ is True

We need to prove it is True for $n=k+1$

Left hand side = $1^2+2^2+3^2+\mathrm{.......}+k^2+{\left(k+1\right)}^2$

  $\Rightarrow$ Left hand side = $\frac{k\left(k+1\right)\left(2k+1\right)}{6}+{\left(k+1\right)}^2$

Taking $\left(k+1\right)$ common from both terms

We get

  $\Rightarrow$ Left hand side = $\left(k+1\right)\left[\frac{k\left(2k+1\right)}{6}+\left(k+1\right)\right]$

  $\Rightarrow$ Left hand side = $\left(k+1\right)\left[\frac{k\left(2k+1\right)+6\left(k+1\right)}{6}\right]$

  $\Rightarrow$ Left hand side = $\left(k+1\right)\left[\frac{2k^2+k+6k+6}{6}\right]$

  $\Rightarrow$ Left hand side = $\left(k+1\right)\left[\frac{2k^2+7k+6}{6}\right]$

Right hand side = $\frac{\left(k+1\right)\left(k+1+1\right)\left(2\left(k+1\right)+1\right)}{6}$

  $\Rightarrow$ Right hand side = $\left(k+1\right)\left[\frac{\left(k+2\right)\left(2k+2+1\right)}{6}\right]$

  $\Rightarrow$ Right hand side = $\left(k+1\right)\left[\frac{\left(k+2\right)\left(2k+3\right)}{6}\right]$

  $\Rightarrow$ Right hand side = $\left(k+1\right)\left[\frac{2k^2+3k+4k+6}{6}\right]$

  $\Rightarrow$ Right hand side = $\left(k+1\right)\left[\frac{2k^2+7k+6}{6}\right]$

Left hand side = Right hand side

And hence the proof by Mathematical induction.