Question

The sum of the series $1+\frac{2^2}{2!}+\frac{3^2}{3!}+\frac{4^2}{4!}+...$ is?

• A. $2e$
• B. $3e$
• C. $2e-1$
• D. None of these

Answer 1

The correct option is A $2e$

Let $S=1+\frac{2^2}{2!}+\frac{3^2}{3!}+\frac{4^2}{4!}+...$
Therefore, $T_n=\frac{n^2}{n!}=\frac{n}{(n-1)!}$
$\Rightarrow T_n=\frac{n-1+1}{(n-1)!}=\frac{1}{(n-2)!}+\frac{1}{(n-1)!}$
Thus $S_n=\sum T_n=\sum \frac{1}{(n-2)!}+\sum \frac{1}{(n-1)!}$
$=e+e$
$=2e$