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Mathematics Question on Sequence and series

The value of ∑n=0∞n2+4n !\sum\limits_{n=0}^{\infty }{\frac{{{n}^{2}}+4}{n\,!}} is equal to

A

6 e6\,e

B

5 e5\,e

C

4 e4\,e

D

None of theseNone \,of\, these

Answer

6 e6\,e

Explanation

Solution

∑n=0∞n2+4n!\sum\limits_{n=0}^{\infty }{\frac{{{n}^{2}}+4}{n!}}
=∑n=0∞(n2n!+4n!)=\sum\limits_{n=0}^{\infty }{\left( \frac{{{n}^{2}}}{n!}+\frac{4}{n!} \right)}
=∑n=0∞(nn(n−1)!+4n!)=\sum\limits_{n=0}^{\infty }{\left( \frac{n}{n(n-1)!}+\frac{4}{n!} \right)}
=∑n=0∞(n(n−1)!+4n!)=\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\\}}
=∑n=2∞1(n−2)!+∑n=1∞1(n−1)!+4∑n=0∞1n!=\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=e+e+4e=6e

So, the correct answer is (A): 6 e6\ e