A) \[2\,e\]
B) \[3\,e\]
C) \[4\,e\]
D) \[5\,e\]
Correct Answer: D
Solution :
\[S=\frac{{{1}^{3}}}{1\,!}+\frac{{{2}^{3}}}{2\ !}+\frac{{{3}^{3}}}{3\ !}+........+\frac{{{n}^{3}}}{n\ !}+......\] Here \[{{T}_{n}}=\frac{{{n}^{3}}}{n\ !}=\frac{{{n}^{2}}}{(n-1)\ !}=\frac{{{n}^{2}}-1}{(n-1)\ !}+\frac{1}{(n-1)\ !}\] \[=\frac{n+1}{(n-2)\ !}+\frac{1}{(n-1)\ !}=\frac{n}{(n-2)\ !}+\frac{1}{(n-2)\ !}+\frac{1}{(n-1)\ !}\] \[=\frac{n-2}{(n-2)\ !}+\frac{2}{(n-2)\ !}+\frac{1}{(n-2)\ !}+\frac{1}{(n-1)\ !}\] \[=\frac{1}{(n-3)\ !}+\frac{3}{(n-2)\ !}+\frac{1}{(n-1)\ !}\] Hence, sum = \[\sum\limits_{n=1}^{\infty }{\frac{1}{(n-3)\,!}}+3\sum\limits_{n=1}^{\infty }{\frac{1}{(n-2)\,!}}+\sum\limits_{n=1}^{\infty }{\frac{1}{(n-1)\,!}}\] = \[e+3e+e=5e\].
You need to login to perform this action.
You will be redirected in
3 sec
