A) \[\frac{e}{4}\]
B) \[\frac{e}{2}\]
C) \[8e\]
D) \[\frac{e({{e}^{2}}-1)}{2({{e}^{2}}+1)}\]
Correct Answer: B
Solution :
\[\frac{1+\frac{1}{2!}+\frac{2}{3!}+\frac{{{2}^{2}}}{4!}+\frac{{{2}^{3}}}{5!}+....}{1+\frac{1}{2!}+\frac{1}{4!}+\frac{1}{6!}+....}\] \[=\frac{{{2}^{\frac{1}{2}}}\left[ {{2}^{2}}+\frac{{{2}^{2}}}{2!}+\frac{{{2}^{3}}}{3!}+\frac{{{2}^{4}}}{4!}+\frac{{{2}^{5}}}{5!}+.... \right]}{\left( \frac{e+{{e}^{-1}}}{2} \right)}\] \[=\frac{\left[ 4+\frac{{{2}^{2}}}{2!}+\frac{{{2}^{3}}}{3!}+\frac{{{2}^{4}}}{4!}+\frac{{{2}^{5}}}{5!}+.... \right]}{4\left( \frac{e+{{e}^{-1}}}{2} \right)}\] \[=\frac{\left[ 1+1+2+\frac{{{2}^{2}}}{2!}+\frac{{{2}^{3}}}{3!}+\frac{{{2}^{4}}}{4!}+.... \right]}{2({{e}^{2}}+1)/e}\] \[=\frac{(1+{{e}^{2}}).e}{2({{e}^{2}}+1)}=\frac{e}{2}\]
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