Question

Find the sum to n terms of the series.
$2+6+12+20+$_________.

Answer 1

Let $S=2+6+12+20+....+T_{n-1}+T_n$

$S=2+6+12+....+T_{n-2}+T_{n-1}+T_n$

On subtracting we get

$0=2+4+6+8+....+T_n-T_{n-1}-T_n$

So, $T_n=2+4+6+8+....$

$T_n=\frac{n}{2}[4+2n-2]=\frac{n}{2}[2n+2]=n[2+1]$

$S_n=\sum T_n=\sum n(n+1)=\sum (n^2+n)=\sum n^2+\sum n$

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

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