Question

The sum of the series 1.2.3 + 2.3.4 + 3.4.5 + .......to n terms is

Answer 1

$T_{\pi }$ = n(n+1)(n+2) = n( $n^2$ + 3n + 2) = $n^3$ + 3 $n^2$ + 2n

∴ $S_{\pi }$ = $\sum$( $n^3$) + $\sum$( ${3n}^2$) + $\sum$(2n)

$S_{\pi }$ =${\left[\frac{n(n+1)}{2}\right]}^2$ + $\frac{3.n(n+1)(2n+1)}{6}$ + $\frac{2.n(n+1)}{2}$

$S_{\pi }$ = $\frac{1}{4}$n(n+1)(n+2)(n+3).