Chapter 11.7, Problem 1CYP

To determine

To prove

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

Explanation of Solution

Given:

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

Concept Used:

An identity can be proved using mathematical induction in two steps:

1. First prove the identity is true for $n=1$ .
2. Prove that if it is true for $n=k$ then it is also true for $n=k+1$ .

Calculation:

In order to prove

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

Use mathematical induction.

First, let’s proof that the identity is true for $n=1$ .

The left hand side of given identity is:

  $1^2=1$

And right hand side is:

  $\begin{array}{c}\frac{1\left(n+1\right)\left(2\left(n\right)+1\right)}{6}=\frac{1\left(1+1\right)\left(2\left(1\right)+1\right)}{6}\left(\text{Putting }n=1\right)\\ =\frac{2\times 3}{6}\\ =1\\ =\text{Left side}\end{array}$

Thus, the given identity is true for $n=1$ .

Now, let it is true for $n=k;\text{ for some }k>1$ . Thus,

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

Now, let’s show that it is also true for $n=k+1$ , i.e. prove

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

Take left side of above and simplify as

  $\begin{array}{c}{1}^2+2^2+3^2+\mathrm{........}+k^2+{\left(k+1\right)}^2=\frac{k\left(k+1\right)\left(2k+1\right)}{6}+{\left(k+1\right)}^2\\ \because 1^2+2^2+3^2+\mathrm{........}+k^2=\frac{k\left(k+1\right)\left(2k+1\right)}{6}\\ \Rightarrow 1^2+2^2+3^2+\mathrm{........}+k^2+{\left(k+1\right)}^2=\frac{k\left(k+1\right)\left(2k+1\right)+6{\left(k+1\right)}^2}{6}\\ =\frac{\left(k+1\right)\left(k\left(2k+1\right)+6\left(k+1\right)\right)}{6}\\ =\frac{\left(k+1\right)\left(2k^2+k+6k+6\right)}{6}\\ =\frac{\left(k+1\right)\left(2k^2+7k+6\right)}{6}\\ =\frac{\left(k+1\right)\left(2k^2+4k+3k+6\right)}{6}\\ =\frac{\left(k+1\right)\left(2k\left(k+2\right)+3\left(k+2\right)\right)}{6}\\ =\frac{\left(k+1\right)\left(2k+3\right)\left(k+2\right)}{6}\\ =\frac{\left(k+1\right)\left(k+1+1\right)\left(2\left(k+1\right)+1\right)}{6}\\ =\text{Right side}\end{array}$

Thus, the identity is also true for $n=k+1$ , when it is true for $n=k$ .

Thus by mathematical induction it is proved that

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