Question

The series $9+\frac{16}{2!}+\frac{27}{2!}+\frac{42}{4!}+....+\infty$is equal to

• A. $11e-4$
• B. $11e-6$
• C. $10e+5$
• D. $3e+4$

Answer 1

The correct option is B

$11e-6$

Explanation for the correct options:

Step1. Given series:

Given, $y=9+\frac{16}{2!}+\frac{27}{2!}+\frac{42}{4!}+....+\infty$

$\Rightarrow$ $y=2+1+6+\frac{2\left(4\right)+2+6}{2!}+\frac{2\left(9\right)+3+6}{3!}+\frac{2\left(16\right)+4+6}{4!}+.......$

$\Rightarrow$ $y=\sum _{n=1}^{\infty }{T}_n$

Where,$T_n=\frac{(2n^2+n+6)}{n!}$

Where, $T_n$ is the $n$th term, and $S_n$is the sum of the series.

Step2. Find the sum:

$S_n=\sum _{n=1}^{\infty }\left[\frac{2n^2}{n!}+\frac{n}{n!}+\frac{6}{n!}\right]$

$=\sum _{n=1}^{\infty }\left[\frac{2n}{\left(n-1\right)!}+\frac{1}{\left(n-1\right)!}+\frac{6}{n!}\right]$

$=\sum _{n=1}^{\infty }\left[\frac{2\left(n+1-1\right)}{\left(n-1\right)!}+\frac{1}{\left(n-1\right)!}+\frac{6}{n!}\right]$

$=\sum _{n=1}^{\infty }\left[\frac{2\left(n-1\right)}{\left(n-1\right)!}+\frac{2\left(1\right)}{\left(n-1\right)!}+\frac{1}{\left(n-1\right)!}+\frac{6}{n!}\right]$

$=\sum _{n=1}^{\infty }\left[\frac{2\left(n-1\right)}{\left(n-1\right)\left(n-2\right)!}+\frac{3}{\left(n-1\right)!}+\frac{6}{n!}\right]$

$=\sum _{n=1}^{\infty }\frac{2}{\left(n-2\right)!}+\sum _{n=1}^{\infty }\frac{3}{\left(n-1\right)!}+\sum _{n=1}^{\infty }\frac{6}{n!}$

$=2\left(1+\frac{1}{1!}+\frac{1}{2!}+...+\infty \right)+3\left(1+\frac{1}{1!}+\frac{1}{2!}+...+\infty \right)+6\left(\frac{1}{1!}+\frac{1}{2!}+...+\infty \right)$

$=2e+3e+6(e-1)$ $\left[\mathbf{\because }{\mathit{e}}^{\mathbf{x}}\mathbf{=}\mathbf{}\mathbf{1}\mathbf{+}\frac{\mathbf{x}}{\mathbf{1}\mathbf{!}}\mathbf{+}\frac{{\mathbf{x}}^{\mathbf{2}}}{\mathbf{2}\mathbf{!}}\mathbf{+}\frac{{\mathbf{x}}^{\mathbf{3}}}{\mathbf{3}\mathbf{!}}\mathbf{+}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{\infty }\right]$

$=\mathbf{11}\mathit{e}\mathbf{-}\mathbf{6}$

Hence, the correct option is (B).