Question

Find the sum to n terms

$1^2+(1^2+2^2)+(1^2+2^2+3^2)+\dots$

Answer 1

The given series is $1^2+(1^2+2^2)+(1^2+2^2+3^2)+\dots \dots$ to n terms.

Let $a_n$ be the nth term of the given series and $S_n$ be the sum of n terms of the given series.

$\therefore a_n=(1^2+2^2+3^2+\dots \dots +n^2)$

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

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

$\therefore S_n={\sum }_{k=1}^n{a}_k$

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

$=\frac{1}{6}[{2.1}^3+{3.1}^2+1]+\frac{1}{6}[{2.2}^3+{3.2}^2+2]+\dots \dots +\frac{1}{6}[2n^3+3n^2+n]$

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

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

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

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

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

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

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