Question

The sum of n terms of the series whose $n^{th}$ term is n(n+1) is equal to

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

Answer 1

The correct option is A

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

$T_{\pi }$ = $n^2$ + n ⇒ $S_{\pi }$ = $\sum T_{\pi }$ = $\sum$ $n^2$ + $\sum$n

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

= $\frac{n(n+1)}{6}$ {2n + 1 + 3} = $\frac{n(n+1)(n+2)}{3}$