Question

$\sum _{n=0}^{\infty }\left(\frac{n^2+4}{n!}\right)$ is equal to

• A. $6e$
• B. $5e$
• C. $4e$
• D. None of these

Answer 1

The correct option is A

$6e$

Explanation for the correct answer:

Simplifying the given expression:

$$\begin{gathered} \Rightarrow \sum _{n=0}^{\infty }\left(\frac{n^2+4}{n!}\right) \\ \Rightarrow \sum _{n=0}^{\infty }\left(\frac{n^2}{n!}\right)+\sum _{n=0}^{\infty }\left(\frac{4}{n!}\right) \\ \Rightarrow \sum _{n=0}^{\infty }\left[\frac{1}{(n-2)!}+\frac{1}{(n-1)!}\right]+\sum _{n=0}^{\infty }\left(\frac{4}{n!}\right) \\ \Rightarrow e+e+4e\left[\because e=1+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\dots .\right] \\ \Rightarrow 6e \end{gathered}$$

Thus, $\sum _{n=0}^{\infty }\left(\frac{n^2+4}{n!}\right)=6e$

Therefore, the correct answer is option (A).