Question

The sum of the series $1+(1+2)+(1+2+3)+.......$upto $n$ terms will be

• A. $n^2-2n+6$
• B. $\frac{n(n+1)(2n-1)}{6}$
• C. $n^2-2n-6$
• D. $\frac{n(n+1)(n+2)}{6}$

Answer 1

The correct option is D

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

Explanation for correct options

Given: $1+(1+2)+(1+2+3)+.......$

Let $S=1+(1+2)+(1+2+3)+.......$

$S=1+(1+2)+(1+2+3)+..............+(1+2+3+4+..........+n)$

The last term$=(1+2+3+.........+n)$

$$\begin{gathered} =\sum _1^{n}n \\ =\frac{n(n+1)}{2} \end{gathered}$$

$$\begin{gathered} \therefore S=\sum _1^n\frac{n(n+1)}{2} \\ \Rightarrow S=\sum _1^n\frac{n^2}{2}+\sum _1^n\frac{n}{2} \\ \Rightarrow S=\frac{1}{2}\left(\frac{n(n+1)(2n+1)}{6}\right)+\frac{1}{2}\left(\frac{n(n+1)}{2}\right) \\ \Rightarrow S=\frac{1}{4}\left(n\right)(n+1)\left(\frac{2n+1}{3}+1\right) \\ \Rightarrow S=\frac{1}{4}\left(n\right)(n+1)\left(\frac{2n+1+3}{3}\right) \\ \Rightarrow S=\frac{1}{4}\left(n\right)(n+1)\left(\frac{2n+4}{3}\right) \\ \Rightarrow S=\frac{1}{6}\left(n\right)(n+1)(n+2) \end{gathered}$$

Hence, $\mathrm{Option}\left(\mathrm{D}\right)$ is correct.