A) \[e\]
B) \[3\,e\]
C) \[e/2\]
D) \[3e/2\]
Correct Answer: D
Solution :
\[{{T}_{n}}=\frac{\Sigma n}{n\,!}=\frac{n(n+1)}{2\,(n)\,!}\] \[=\frac{1}{2}\left[ \frac{(n+1)}{(n-1)!} \right]=\frac{1}{2}\left[ \frac{n-1}{(n-1)!}+\frac{2}{(n-1)!} \right]\] \[=\frac{1}{2}\left[ \frac{1}{(n-2)!}+\frac{2}{(n-1)!} \right]=\frac{(e+2e)}{2}=\frac{3e}{2}\].
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