Question

Prove : $1^2+\left(1^2+2^2\right)+\left(1^2+2^2+3^2\right)+\dots$ upto $n$ terms $=\frac{n{\left(n+1\right)}^2\left(n+2\right)}{12}$

Answer 1

Given series is $1^2+(1^2+2^2)+(1^2+2^2+3^2)+........$

We must find the $n^{th}$ term $t_n$ of the series

Clearly $n^{th}$ term of the series is,

$1^2+2^2+3^2+............n^2=\sum n^2$

We know that $\sum n^2=\frac{n(n+1)(2n+1)}{6}$

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

We know that sum of n terms of any series $S_n=\sum t_n$

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

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

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

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

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

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

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