Question

Find the sum of first $n$ terms of the series $1^2+\left(1^2+2^2\right)+\left(1^2+2^2+3^2\right)+.....$

Answer 1

$1st$ term $=1^2$

$2nd$ term $=1^2+2^2$

$3rd$ term $=1^2+2^2+3^2$

$nth$ term $=1^2+2^2+3^2+......+n^2$

Here,

$\begin{array}{c}{a}_n=1^2+2^2+3^2+......+n^2\\ a_n=\frac{n\left(n+1\right)\left(2n+1\right)}{6}\\ =\frac{\left(n^2+n\right)\left(2n+1\right)}{6}\\ =\frac{\left(2n^3+n^2+2n^2+n\right)}{6}\\ =\frac{\left(2n^3+3n^2+n\right)}{6}\end{array}$