A) \[6\,e\]
B) \[5\,e\]
C) \[4\,e\]
D) None of these
Correct Answer: A
Solution :
\[\sum\limits_{n=0}^{\infty }{\frac{{{n}^{2}}+4}{n!}}\] \[=\sum\limits_{n=0}^{\infty }{\left( \frac{{{n}^{2}}}{n!}+\frac{4}{n!} \right)}\] \[=\sum\limits_{n=0}^{\infty }{\left( \frac{n}{n(n-1)!}+\frac{4}{n!} \right)}\] \[=\sum\limits_{n=0}^{\infty }{\left( \frac{n}{(n-1)!}+\frac{4}{n!} \right)}\] \[=\sum\limits_{n=0}^{\infty }{\left\{ \frac{(n-1)}{(n-1)}+\frac{1}{(n-1)!}+\frac{4}{n!} \right\}}\] \[=\sum\limits_{n=0}^{\infty }{\left\{ \frac{1}{(n-2)!}+\frac{1}{(n-1)!}+\frac{4}{n!} \right\}}\] \[=\sum\limits_{n=2}^{\infty }{\frac{1}{(n-2)!}+}\sum\limits_{n=1}^{\infty }{\frac{1}{(n-1)!}}+4\sum\limits_{n=0}^{\infty }{\frac{1}{n!}}\] \[=\left\{ 1+\frac{1}{1!}+\frac{1}{2!}+....\infty \right\}\] \[+\left\{ 1+\frac{1}{1!}+\frac{1}{2!}+....\infty \right\}+4\left\{ 1+\frac{1}{1!}+\frac{1}{2!}+....\infty \right\}\] \[=e+e+4e=6e\]
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