A) \[13\,e\]
B) \[15\,e\]
C) \[9\,e+1\]
D) \[5\,e\]
Correct Answer: A
Solution :
\[1.5+\frac{2.6}{1!}+\frac{3.7}{2!}+\frac{4.8}{3!}+....\] \[{{T}_{n}}=\frac{n(n+4)}{(n-1)!}=\frac{(n-1)(n+4)}{(n-1)!}+\frac{(n+4)}{(n-1)!}\] \[=\frac{n+4}{(n-2)!}+\frac{1}{(n-2)!}+\frac{5}{(n-1)!}\] \[=\frac{1}{(n-3)!}+\frac{7}{(n-2)!}+\frac{5}{(n-1)!}\] \[{{S}_{\infty }}=\sum\limits_{n=1}^{\infty }{\frac{1}{(n-3)!}+7\sum\limits_{n=1}^{\infty }{\frac{1}{(n-2)!}+5\sum\limits_{n=1}^{\infty }{\frac{1}{(n-1)!}}}}\] \[=e+7e+5e=13e\].
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