Question

Find the sum to $n$ terms of the series
${1.2}^2+{2.3}^2+{3.4}^2+...$

Answer 1

$1,2^2+2,3^2+3,4^2+$….

$Tn\Rightarrow n{(n+1)}^2$ (General Term of this series)

$=n(n^2+1+2n)$

$Tn=n^3+2n^2+n$

$Sn=\sum Tn=\sum n^3+\sum 2n^2+\sum n$

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

$\Rightarrow \frac{n(n+1)}{2}[\frac{n(n+1)}{2}\frac{2(2n+1)}{3}+1]$

$Sn\Rightarrow \frac{n(n+1)}{2}\left[\frac{3\left(n(n+1)\right)+4(2n+1)+6}{6}\right)$