Question

$1+\frac{2^3}{2!}+\frac{3^3}{3!}+\frac{4^3}{4!}+\dots$ equals

• A. $5e$
• B. $4e$
• C. $3e$
• D. $2e$

Answer 1

The correct option is A

$5e$

Explanation for the correct option:

Step 1. We can write given series $1+\frac{2^3}{2!}+\frac{3^3}{3!}+\frac{4^3}{4!}+\dots$ as:

$\sum _{n=1}^{\infty }\frac{n^3}{n!}$ ……(1)

Let $n^3=a_0+a_{1}n+a_{2}n(n-1)+a_{3}n(n-1)(n-2)$ ……(2)

where, $a_0=0,a_1=1,a_2=3,a_3=1$

Put the values of $a_0,a_1,a_2,a_3$ in equation (2)

$\Rightarrow n^3=n+3n(n-1)+n(n-1)(n-2)$

Step 2. Put the value of $n^3$ in equation (1)

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

$\underset{n=1}{\overset{\infty }{=\sum }}\frac{n}{n!}+3\sum _{n=1}^{\infty }\frac{n(n-1)}{(n-1)n!}+\sum _{n=1}^{\infty }\frac{n(n-1)(n-2)}{(n-1)(n-2)n!}$ $\left[\mathbf{\because }{\mathit{e}}^{\mathbf{x}}\mathbf{=}\mathbf{}\mathbf{1}\mathbf{}\mathbf{+}\mathbf{}\mathit{x}\mathbf{}\mathbf{+}\mathbf{}\frac{{\mathbf{x}}^{\mathbf{2}}}{\mathbf{2}\mathbf{!}}\mathbf{}\mathbf{+}\mathbf{}\frac{{\mathbf{x}}^{\mathbf{3}}}{\mathbf{3}\mathbf{!}}\mathbf{+}\mathbf{}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{}\mathbf{+}\mathbf{}\frac{{\mathbf{x}}^{\mathbf{n}}}{\mathbf{n}\mathbf{!}}\right]$ $$\begin{gathered} =e+3e+e \\ =\mathbf{5}\mathit{e} \end{gathered}$$

Hence, Option ‘A’ is Correct.