A) \[e\]
B) \[2\,e\]
C) \[3\,e\]
D) None of these
Correct Answer: C
Solution :
\[S=\frac{2}{1\ !}+\frac{2+4}{2\ !}+\frac{2+4+6}{3\ !}+......+\frac{\frac{n}{2}(2+2n)}{n\ !}+.....\infty \] Here \[{{T}_{n}}=\frac{n(n+1)}{n\ !}=\frac{n-1+2}{(n-1)\ !}=\frac{1}{(n-2)\ !}+\frac{2}{(n-1)\ !}\] \[\Rightarrow S=\sum\limits_{n=1}^{\infty }{{{T}_{n}}}=\sum\limits_{n=1}^{\infty }{\frac{1}{(n-2)\ !}+2\sum\limits_{n=1}^{\infty }{\frac{1}{(n-1)\ !}}}\] \[=e+2e=3e\].
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