Question

Find the sum to nn terms of the series:$1\times 2+2\times 3+3\times 4+4\times 5+\dots$

Answer 1

$a_n=n(n+1)=n^2+n\therefore S_n=\sum n^2+n=\left(\frac{n(n+1)(2n+1)}{6}\right)+\left(\frac{n(n+1)}{2}\right)=\left(\frac{n}{2}\right)\left[\right(\frac{(n+1)(2n+1)+3(n+1)}{3}\left)\right]=\left(\frac{n}{2}\right)\left(\frac{2n^2+n+2n+1+3n+3}{3}\right)=\left(\frac{n(2n^2+6n+4)}{6}\right)=\left(\frac{n(n^2+3n+2)}{3}\right)=\left(\frac{n(n+1)(n+2)}{3}\right)$