Question

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

Answer 1

$\left(1^2\right)+(1^2+2^2).....+(1^2+2^2.........n^2)$

We know that $\sum _{i=1}^{n}i=\frac{n(n+1)}{2}$

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

$\sum _{i=1}^n{i}^3=\frac{n^2(n+1{)}^2}{4}$

$\left(1^2\right)+(1^2+2^2)..............(1^2+2^2............n^2)$

$=\sum _{i=1}^n(1^2+2^2.........i^2)$

$=\sum _{i=1}^n\sum _{i=1}^i{i}^2$

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

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

$=\frac{1}{3}\left[n^2\frac{(n+1{)}^2}{4}\right]+\frac{1}{2}\left[\frac{n(n+1)(2n+1)}{6}\right]+\frac{1}{6}\left[\frac{n(n+1)}{2}\right]$

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

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