Question

Find the sum of $1^2+(1^2+2^2)+(1^2+2^2+3^2)+.......$

Answer 1

First term$=1^2$

Second term$=1^2+2^2$

Third term$=1^2+2^2+3^2$...

$n^{th}$ term$=1^2+2^2+3^2+...+n^2$

Here $a_n=1^2+2^2+3^2+...+n^2$

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

$=\frac{2n^3+n^2+2n^2+n}{6}$

$=\frac{2n^3+3n^2+n}{6}$

Now, the sum of $n$ terms is

$S_n={\sum }_{n=1}^n{a}_n$

$={\sum }_{n=1}^n\frac{2n^3+3n^2+n}{6}$

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

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

$=\frac{n(n+1)}{12}[n(n+1)+(2n+1)+1]$

$=\frac{n(n+1)}{12}[n^2+n+2n+1+1]$

$=\frac{n(n+1)}{12}[n^2+3n+2]$

$=\frac{n(n+1)}{12}[n^2+2n+n+2]$

$=\frac{n(n+1)}{12}\left[(n+1)(n+2)\right]$

$=\frac{n{(n+1)}^2(n+2)}{12}$

Thus, the required sum is $\frac{n{(n+1)}^2(n+2)}{12}$