A) \[e/2\]
B) \[e\]
C) \[2\,e\]
D) \[3e\]
Correct Answer: C
Solution :
\[{{T}_{n}}=\frac{1+3+5+.......+(2n-1)}{n\ !}=\frac{\frac{n}{2}[2\ .1+(n-1)2]}{n\ !}\] \[=\frac{n-1+1}{(n-1)\ !}=\frac{1}{(n-2)\ !}+\frac{1}{(n-1)\ !}\] \[\Rightarrow S=\sum\limits_{n=1}^{\infty }{{{T}_{n}}}=e+e=2e\].
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