Question

Find the sum to n terms of the series whose $n^{th}$ terms is given by-

n(n+1)

Answer 1

Given :$a_n=n(n+1)$

$S_n=\sum _{n=1}^{n}n(n+1)$

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

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

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

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

$=\frac{n(n+1)}{2}\left(\frac{2n+1+3}{3}\right)$

$=\frac{n(n+1)}{2}\times \frac{2n+4}{3}$

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

$\therefore S_n=\frac{n(n+1)(n+2)}{3}$