A) \[6\,e\]
B) \[7\,e\]
C) \[8\,e\]
D) \[9\,e\]
Correct Answer: B
Solution :
\[S=\frac{{{1}^{2}}.2}{1!}+\frac{{{2}^{2}}.3}{2!}+\frac{{{3}^{2}}.4}{3!}+....\] Here \[{{T}_{n}}=\frac{{{n}^{2}}.(n+1)}{n!}=\frac{n(n+1)}{(n-1)!}\] \[=\frac{(n-1)(n-2)+4n-2}{(n-1)!}=\frac{1}{(n-3)!}+\frac{4(n-1)+2}{(n-1)!}\] \[\frac{1}{(n-3)!}\] \[\Rightarrow \,\,\,S=\sum{{{T}_{n}}=e+4e+2e=7e}\].You need to login to perform this action.
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