Question

Sum of the series $\frac{1^2}{1!}+\frac{2^2}{2!}+\frac{3^2}{3!}+\frac{4^2}{4!}+.....$(infinite terms) is?

• A. e
• B. $2e$
• C. $e^2$
• D. $\infty$

Answer 1

Answer: (B)

$2e$

$\begin{array}{cc}& S=\frac{1^2}{1!}+\frac{2^2}{2!}+\frac{3^2}{3!}+\frac{4^2}{4b}+\\ & \text{(infinite terems)}is\\ & S=\sum _{n=1}^{\infty }{n}^2\\ & n^2=a_0+a_{1}n+a_{2}n(n-1)\\ & n=00=a_0\\ & n=11=0+a,\Rightarrow a_1=1\\ & n=24=0+2+2a_2\Rightarrow a_2=1\\ & S=\sum _{n=1}^{\infty }(a_0+a_{1}n+a_{2}n(n-1\left)\right)\end{array}$

$S={\sum }_{n=1}^{\infty }\left(\frac{0+n+n(n-1)}{n_0}\right)$ $S={\sum }_{n=1}^{\infty }\frac{n}{n_b}+{\sum }_{n=1}^{\infty }\frac{n(n-1)}{n_0^1}$

$S=e+e=2e$

$e=1+\frac{1+\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+\cdots }{\Rightarrow }{\sum }_{n=1}^{\infty }\frac{1}{(n-1)!}$ $e={\sum }_{n=2(n-2)\}}=1+1+\frac{1}{2!}+\frac{1}{3!}\cdots \cdot \cdots$

Option $B$ is correct