Question

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

Answer 1

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

We have to prove for $n=k+1$

i.e. To Prove $1^2+2^2+...+k^2+(k+1{)}^2=\frac{(k+1)(k+2)(2k+3)}{6}\dots$

But We Know, $1^2+2^2+...+k^2=\frac{k(k+1)(2k+1)}{6}$

Therfore, We have to prove $\frac{k(k+1)(2k+1)}{6}+(k+1{)}^2=\frac{(k+1)(k+2)(2k+3)}{6}\dots$

Divide by (k+1) and multiply by 6

$k(2k+1)+6(k+1)=(k+2)(2k+3)$

$2k^2+k+6k+6=2k^2+3k+4k+6$

$2k^2+7k+6=2k^2+7k+6$