Question

$2+4+7+11+16+......n\mathrm{terms}=$

• A. $\frac{1}{6}\left(n^2+3n+8\right)$
• B. $\frac{n}{6}\left(n^2+3n+8\right)$
• C. $\frac{1}{6}\left(n^2-3n+8\right)$
• D. $\frac{n}{6}\left(n^2-3n+8\right)$

Answer 1

The correct option is B

$\frac{n}{6}\left(n^2+3n+8\right)$

Explanation for the correct option:

Step-1: Finding $n^{th}$term of the given series.

Let the given series, $\mathrm{S}=2+4+7+11+16+......n\mathrm{terms}$

$\Rightarrow \mathrm{S}=\left(1+\frac{1^2+1}{2}\right)+\left(1+\frac{2^2+2}{2}\right)+\left(1+\frac{3^2+3}{2}\right)+\left(1+\frac{4^2+4}{2}\right)+16+......\left(1+\frac{n^2+n}{2}\right)\mathrm{terms}$

$\Rightarrow a_n=\left(1+\frac{n^2+n}{2}\right)$

Step-2: Finding sum of $n^{th}$term of the given series.

$\Rightarrow {\mathrm{S}}_{\mathrm{n}}=\sum _{\mathrm{n}-1}^{\infty }{\mathrm{a}}_{\mathrm{n}}$

$\Rightarrow {\mathrm{S}}_{\mathrm{n}}=\sum _{\mathrm{n}-1}^{\infty }\left(1+\frac{{\mathrm{n}}^2+\mathrm{n}}{2}\right)$

$\Rightarrow {\mathrm{S}}_{\mathrm{n}}=\sum _{\mathrm{n}-1}^{\infty }1+\sum _{\mathrm{n}-1}^{\infty }\frac{n^2}{2}+\sum _{\mathrm{n}-1}^{\infty }\frac{n}{2}$

Step-3: Apply formula $\sum _{\mathrm{n}-1}^{\infty }1=n,\sum _{\mathrm{n}-1}^{\infty }{n}^2=\frac{1}{2}\left(\frac{n\left(n+1\right)\left(2n+1\right)}{6}\right),\sum _{\mathrm{n}-1}^{\infty }n=\left(\frac{n\left(n+1\right)}{2}\right)$

$$\begin{gathered} \Rightarrow {\mathrm{S}}_{\mathrm{n}}=n+\frac{1}{2}\left(\frac{n\left(n+1\right)\left(2n+1\right)}{6}\right)+\frac{1}{2}\left(\frac{n\left(n+1\right)}{2}\right) \\ \Rightarrow {\mathrm{S}}_{\mathrm{n}}=n\left(\frac{12+\left(n+1\right)\left(2n+1\right)+\left(3n+3\right)}{12}\right) \\ \Rightarrow {\mathrm{S}}_{\mathrm{n}}=n\left(\frac{12+\left(n+1\right)\left(2n+4\right)}{12}\right) \\ \Rightarrow {\mathrm{S}}_{\mathrm{n}}=n\left(\frac{6+\left(n+1\right)\left(n+2\right)}{6}\right) \\ \Rightarrow {\mathrm{S}}_{\mathrm{n}}=n\left(\frac{n^2+3n+2+6}{6}\right) \\ \Rightarrow {\mathrm{S}}_{\mathrm{n}}=\frac{n}{6}\left(n^2+3n+8\right) \end{gathered}$$

Hence, correct answer is option $\left(B\right)$