Question

If $n^{th}$ term of a series be $3+n\left(n-1\right)$, then the sum of $n$ terms of the series is:

• A. $\frac{n^2+n}{3}$
• B. $\frac{n^3+8n}{3}$
• C. $\frac{n^2+8n}{5}$
• D. $\frac{n^2-8n}{3}$

Answer 1

The correct option is B

$\frac{n^3+8n}{3}$

Explanation for the correct option

Given $n^{th}$ term of a series is $t_n=$$3+n\left(n-1\right)=3+n^2-n$

Sum of $n$ terms of the series is $S=\underset{}{\overset{}{\sum t_n}}$

$\begin{array}{cccc}\Rightarrow & S& =& \sum _{}^{}\left(3+n^2-n\right)\\ & & =& \sum _{}^{}3+\sum _{}^{}{n}^2-\sum _{}^{}n\\ & & =& 3n+\frac{n\left(n+1\right)\left(2n+1\right)}{6}-\frac{n\left(n+1\right)}{2}\\ & & =& n\left[\frac{18+\left(2n^2+3n+1\right)-\left(3n+3\right)}{6}\right]\\ & & =& n\left[\frac{2n^2+16}{6}\right]\\ & & =& \frac{2n\left(n^2+8\right)}{6}\\ \Rightarrow & S& =& \frac{n^3+8n}{3}\end{array}$

Hence the correct option is option(B) i.e. $\frac{n^3+8n}{3}$