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

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

Explanation for the correct option:

Find the required sum.

Given: $n^{th}$ term is $n(n+1)$

$n^{th}$ term, $t_n=n(n+1)$

$=n^2+n$

Sum, $S_n=\sum t^n$

$$\begin{gathered} =\sum (n^2+n) \\ =\sum n^2+\sum n \\ =\frac{n(n+1)(2n+1)}{6}+\frac{n(n+1)}{2} \\ =\frac{n(n+1)(2n+1)+3n(n+1)}{6} \\ =\frac{n(n+1)}{6}(2n+1+3) \\ =\frac{n(n+1)(2n+4)}{6} \\ =\frac{2n(n+1)(n+2)}{6} \\ =\frac{n(n+1)(n+2)}{3} \end{gathered}$$

Therefore, the sum of $n$ terms of the series whose $n^{th}$ term is $n(n+1)$ is equal to $\frac{n(n+1)(n+2)}{3}$.

Hence, option (A) is the correct answer.