Question

$1\times 2+2\times 3+3\times 4+4\times 5+...$

Answer 1

$n^{th}$ term of given series is, $a_n=n(n+1)$
$\therefore S_n={\sum }_{k=1}^n{a}_k={\sum }_{k=1}^{n}k(k-1)={\sum }_{k=1}^n{k}^2+{\sum }_{k=1}^{n}k=\frac{n(n+1)(2n+1)}{6}+\frac{n(n+1)}{2}=\frac{n(n+1)}{2}(\frac{2n+1}{3}+1)=\frac{n(n+1)}{2}\left(\frac{2n+4}{3}\right)=\frac{n(n+1)(n+2)}{3}$