Question

Prove that $1^2+2^2+3^2+....+n^2=\frac{n(n+1)(2n+1)}{6}$

Answer 1

Prove that $1^2+2^2+3^2+....+n^2=\frac{n(n+1)(2n+1)}{6}$

Principle of Mathematical Induction

Step 1: Check for $n=1$

Let

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

For $n=1$
L.H.S$=1^2=1$
And R.H.S

$$\begin{gathered} =\frac{\left(1\right)(1+1)(2\times 1+1)}{6}=\frac{1\times 2\times 3}{6} \\ =1 \end{gathered}$$

L.H.S = R.H.S

$P\left(n\right)$ is true for $n=1$
Step 2: Assumed for $n=k$ identity is true

Assume that $P\left(k\right)$ is true
$1^2+2^2+3^2+...+k^2=\frac{k(k+1)(2k+1)}{6}$

Step 3: Check for $n=k+1$
we will prove $P(k+1)$ is true
$\begin{array}{rcl}{1}^2+2^2+3^2+....+(k+1{)}^2& =& \frac{k(k+1)(2k+1)}{6}+(k+1{)}^2\\ & =& \frac{k(k+1)(2k+1)+6(k+1{)}^2}{6}\\ & =& \frac{(k+1)\left(k\right(2k+1)+6(k+1\left)\right)}{6}\\ & =& \frac{(k+1)(2k2+k+6k+1)}{6}\\ & =& \frac{(k+1)(2k2+7k+1)}{6}\\ & =& \frac{(k+1)(k+2)(2k+3)}{6}\end{array}$
Thus $1^2+2^2+3^2+...+(k+1{)}^2=\frac{(k+1)(k+2)(2k+3)}{6}$
$\therefore$$P(k+1)$ is true when $P\left(k\right)$ is true
$\therefore$ By principle of mathematical induction, $P\left(n\right)$ true for $n$ where $n$ is a natural number.

Hence, the given expression is proved.