Question

Prove by mathematical induction,
$1^2+2^2+3^2+....+n^2=\frac{n(n+1)(2n+1)}{6}$

Answer 1

$P\left(n\right):$

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

$P\left(1\right):$

$1^2=\frac{1(1+1)\left(2\right(1)+1)}{6}$

$1=\frac{6}{6}=1$

$\therefore LHS=RHS$

Assume P(k) is true

$P\left(k\right):$

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

P(k+1) is given by,

$P(k+1):$

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

$\Rightarrow (k+1)\frac{k(2k+1)+6(k+1)}{6}=\frac{(k+1)(k+2)(2k+3)}{6}$

$\Rightarrow (k+1)\frac{2k^2+7k+6}{6}=\frac{(k+1)(k+2)(2k+3)}{6}$

$\Rightarrow \frac{(k+1)(k+2)(2k+3)}{6}=\frac{(k+1)(k+2)(2k+3)}{6}$

True for P(k+1)

Hence by Principle of mathematical induction

$1^2+2^2+3^2+........+n^2=\frac{n(n+1)(2n+1)}{6}$ is true for $\forall n\in N$