Question

Prove by method of induction, for all $n\in$
$1^2+2^2+3^2+...+n^2=\frac{n(n+1)(2n+1)}{6}$.

Answer 1

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

For $n=1$,

$L.H.S=1^2=1$

$R.H.S=\frac{1\left(1+1\right)\left(2\times 1+1\right)}{6}=\frac{1\times 2\times 3}{6}=1$

Hence,

$L.H.S=R.H.S$

$\therefore P\left(n\right)$ is true for $n=1$.