# Find the sum to $n$ terms of the series $1^2+(1^2+2^2)+(1^2+2^2+3^2)+........$

$\begin{array}{1 1}\large\frac{n(n+1)(n+2)^2}{12}\\\large\frac{n^2(n+1)(n+2)}{12} \\\large\frac{n(n+1)^2(n+2)}{12} \\ \large\frac{n(n+1)^2(n+2)}{24}\end{array}$

Toolbox:
• $1^2+2^2+3^2+.......=\sum \:n^2=\large\frac{n(n+1)(2n+1)}{6}$
• Sum of $n$ terms of any series $S_n=\sum t_n$
• $\sum n^3=\large\frac{n^2(n+1)^2}{4}$
• $\sum n=\large\frac{n(n+1)}{2}$
Given series is $1^2+(1^2+2^2)+(1^2+2^2+3^2)+........$
We have to 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 $\sum n^2=\large\frac{n(n+1)(2n+1)}{6}$
$\Rightarrow\:t_n=\sum n^2=\large\frac{n(n+1)(2n+1)}{6}=\large\frac{1}{6}$$(2n^3+3n^2+n) We know that sum of n terms of any series S_n=\sum t_n \Rightarrow\:S_n=\sum t_n=\sum \large\frac{1}{6}$$(2n^3+3n^2+n)$
$\qquad=\large\frac{1}{6}$$\big[\sum2n^3+\sum 3n^2+\sum n\big] \qquad=\large\frac{1}{6}$$\bigg[2\times\large\frac{n^2(n+1)^2}{4}$$+3\times \large\frac{n(n+1)(2n+1)}{6}$$+\large\frac{n(n+1)}{2}\bigg]$
$\qquad=\large\frac{1}{6}$$\bigg[\large\frac{n(n+1)}{2}$$\big(n(n+1)+(2n+1)+1\big)\bigg]$
$\qquad=\large\frac{n(n+1)}{12}$$\big[n^2+3n+2\big] \qquad=\large\frac{n(n+1)}{12}$$\big[(n+1)(n+2)\big]$
$\qquad=\large\frac{n(n+1)^2(n+2)}{12}$