Question

The sum of the series $1^2$.2 + $2^2$.3 + $3^2$.4 + ........ to n terms is

• A. $\frac{n^3{(n+1)}^3(2n+1)}{24}$
• B. $\frac{n(n+1)(3n^2+7n+2)}{12}$
• C. $\frac{n(n+1)}{6}$[n(n+1) + (2n + 1)]
• D. $\frac{n(n+1)}{12}$[6n(n+1) + 2(2n + 1)]

Answer 1

The correct option is B

$\frac{n(n+1)(3n^2+7n+2)}{12}$

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

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

= $\frac{n(n+1)}{2}$[$\frac{n(n+1)}{2}$ + $\frac{2n+1}{3}$] = $\frac{n(n+1)(3n^2+7n+2)}{12}$