A) \[{{e}^{3/2}}\]
B) \[{{e}^{3}}\]
C) \[{{e}^{-3/2}}\]
D) \[{{e}^{-3}}\]
Correct Answer: B
Solution :
We have,\[S=\sum\limits_{n=2}^{\infty }{^{n}{{C}_{2}}}\frac{{{3}^{n-2}}}{n!}\] \[=\sum\limits_{n=2}^{\infty }{\frac{n!}{(n-2)!2n!}\cdot {{3}^{n-2}}}\] \[=\frac{1}{2}\sum\limits_{n=2}^{\infty }{\frac{{{3}^{n-2}}}{(n-2)!}=\frac{1}{2}{{e}^{3}}}\] \[\Rightarrow \] \[2S={{e}^{3}}\]You need to login to perform this action.
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