A) \[\sqrt{e}\]
B) \[\sqrt{e}+1\]
C) \[\sqrt{e}-1\]
D) \[e-1\]
Correct Answer: C
Solution :
[c] \[{{T}_{n}}=\frac{1.3.5.......(2n-1)}{(2n)!}\] \[=\frac{1.2.3.4....(2n-1).2n}{(2n)!2.\,4.\,6.....2n}\] \[=\frac{(2n)!}{(2n)!{{2}^{n}}n!}=\frac{1}{{{2}^{n}}n!}\] Now putting \[n=1,2,3,.........\] we see that the sum of series \[S=\frac{1}{2}+\frac{{{(1/2)}^{2}}}{2!}+\frac{{{(1/2)}^{3}}}{3!}+....\] \[={{e}^{\frac{1}{2}}}-1=\sqrt{e}-1\]
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