Chapter 11.4, Problem 3E

To determine

To prove:

Given statement is true using principle of mathematical induction.

Explanation of Solution

Give information:

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

Formula Used:

Principle of mathematical Induction.

Proof:

It is given that,

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

By using principle of mathematical induction,

Put $n=1$ ,

  $1^2+2^2+3^2+\mathrm{...}+n^2=\frac{1\left(1+1\right)\left(2.1+1\right)}{6}=\frac{1\left(2\right)\left(3\right)}{6}=\frac{6}{6}=1=1^2$

And $1^2+2^2+3^2+\mathrm{...}+n^2$ is restricted to only the term $1^2$ when $n=1$ .

Both sides of equation $P\left(1\right)$ are equal to $1^2=1$ .

Let, $P\left(k\right)$ be true, thus $1^2+2^2+3^2+\mathrm{...}+k^2=\frac{k\left(k+1\right)\left(2k+1\right)}{6}$

Need to $P\left(k+1\right)$ be true, thus need to show that

  $\begin{array}{l}{1}^2+2^2+3^2+\mathrm{...}+{\left(k+1\right)}^2=\frac{\left(k+1\right)\left(\left(k+1\right)+1\right)\left(2\left(k+1\right)+1\right)}{6}\\ 1^2+2^2+3^2+\mathrm{...}+{\left(k+1\right)}^2=\left[1^2+2^2+\mathrm{...}+k^2\right]+{\left(k+1\right)}^2\\ 1^2+2^2+3^2+\mathrm{...}+{\left(k+1\right)}^2=\frac{k\left(k+1\right)\left(2k+1\right)}{6}+\frac{6{\left(k+1\right)}^2}{6}\\ 1^2+2^2+3^2+\mathrm{...}+{\left(k+1\right)}^2=\frac{\left(k+1\right)\left[\left(2k+1\right)+6\left(k+1\right)\right]}{6}\\ 1^2+2^2+3^2+\mathrm{...}+{\left(k+1\right)}^2=\frac{\left(k+1\right)\left[2k^2+k+6k+6\right]}{6}\\ 1^2+2^2+3^2+\mathrm{...}+{\left(k+1\right)}^2=\frac{\left(k+1\right)\left[2k^2+7k+6\right]}{6}\\ =\frac{\left(k+1\right)\left(k+2\right)\left(2k+3\right)}{6}\\ =\frac{\left(k+1\right)\left(k+1+1\right)\left(2k+2+1\right)}{6}\\ =\frac{\left(k+1\right)\left(\left(k+1\right)+1\right)\left(2\left(k+1\right)+1\right)}{6}\end{array}$

Therefore, $P\left(k+1\right)$ is true.

Therefore, given statement is true.