Question

$\sum _{n=1}^{\infty }\left(\frac{2n^2+n+1}{n!}\right)$ is equal to

• A. $2e-1$
• B. $2e+1$
• C. $6e-1$
• D. $6e+1$

Answer 1

The correct option is C

$6e-1$

Explanation for the correct answer:

Simplifying the given expression:

Let,

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

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

Therefore, the correct answer is option (C).