A) e
B) \[2\,e\]
C) e/2
D) e/3
Correct Answer: C
Solution :
\[S=\frac{1}{1\ !}+\frac{2}{3\ !}+\frac{3}{5\ !}+\frac{4}{7\ !}+.......+\frac{n}{(2n-1)\ !}+.....\] Here \[{{T}_{n}}=\frac{1}{2}.\frac{2n}{(2n-1)\ !}=\frac{1}{2}\frac{(2n-1)+1}{(2n-1)\ !}\] \[=\frac{1}{2}\left\{ \frac{1}{(2n-2)\ !}+\frac{1}{(2n-1)\ !} \right\}\] \[\Rightarrow S=\sum{{{T}_{n}}=\frac{1}{2}\left\{ \frac{e+{{e}^{-1}}}{2}+\frac{e-{{e}^{-1}}}{2} \right\}=\frac{e}{2}}\]. Trick: The sum of this series upto 4 terms is 1.359 ...... and this is value of e/2 approximately.
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