A) \[{{e}^{2}}\]
B) \[{{e}^{2}}-1\]
C) \[{{e}^{3/2}}\]
D) None of these
Correct Answer: B
Solution :
\[\frac{1+\frac{{{2}^{2}}}{2\ !}+\frac{{{2}^{4}}}{3\ !}+\frac{{{2}^{6}}}{4\ !}+....\infty }{1+\frac{1}{2\ !}+\frac{2}{3\ !}+\frac{{{2}^{2}}}{4\ !}+.....\infty }\] \[=\frac{\frac{1}{{{2}^{2}}}\left\{ \frac{{{2}^{2}}}{1\ !}+\frac{{{({{2}^{2}})}^{2}}}{2\ !}+\frac{{{({{2}^{2}})}^{3}}}{3\ !}+...... \right\}}{\frac{1}{{{2}^{2}}}\left\{ 2+2+\frac{{{2}^{2}}}{2\ !}+\frac{{{2}^{3}}}{3\ !}+...... \right\}}=\frac{{{e}^{({{2}^{2}})}}-1}{1+{{e}^{2}}}={{e}^{2}}-1\].
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