A) \[\frac{e-1}{\sqrt{e}}\]
B) \[\frac{e+1}{\sqrt{e}}\]
C) \[\frac{e-1}{2\sqrt{e}}\]
D) \[\frac{e+1}{2\sqrt{e}}\]
Correct Answer: D
Solution :
[d] We know that \[{{e}^{x}}=1+\frac{x}{1!}+\frac{{{x}^{2}}}{2!}+\frac{{{x}^{3}}}{3!}+\frac{{{x}^{4}}}{4!}+...\infty \] ....(i) \[\therefore {{e}^{-x}}=1-\frac{x}{1!}+\frac{{{x}^{2}}}{2!}-\frac{{{x}^{3}}}{3!}+\frac{{{x}^{4}}}{4}-...\infty \] ...(ii) On adding equations (i) and (ii), we get \[\therefore {{e}^{x}}+{{e}^{-x}}=2x+\frac{2{{x}^{2}}}{2!}+\frac{2{{x}^{4}}}{4!}+...\infty \] \[\therefore \frac{{{e}^{x}}+{{e}^{-x}}}{2}=1+\frac{{{x}^{2}}}{2!}+\frac{{{x}^{4}}}{4}+\frac{{{x}^{6}}}{6!}+...\] Putting x=1/2, we get \[\frac{e+1}{2\sqrt{e}}=1+\frac{1}{4.2!}+\frac{1}{16.4!}+\frac{1}{16.4!}+\frac{1}{64.6!}+\]
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