A) \[\frac{{{e}^{2}}-1}{{{e}^{2}}+1}\]
B) \[\frac{{{e}^{2}}+1}{{{e}^{2}}-1}\]
C) \[\frac{e+1}{e-1}\]
D) \[\frac{1-{{e}^{2}}}{e+{{e}^{2}}}\]
Correct Answer: B
Solution :
We know that, \[{{e}^{x}}=1+\frac{x}{1\,!}+\frac{{{x}^{2}}}{2\,!}+\frac{{{x}^{3}}}{3\,!}+....\] Put \[x=1\] and \[-1,\] \[e=1+\frac{1}{1\,!}+\frac{1}{2\,!}+\frac{1}{3\,!}+\frac{1}{4\,!}+....\] ?.. (i) \[{{e}^{-1}}=1-\frac{1}{1\,!}+\frac{1}{2\,!}-\frac{1}{3\,!}+\frac{1}{4\,!}-....\] ?.. (ii) Adding Eqs. (i) and (i), we get \[e+\frac{1}{e}=2\left[ 1+\frac{1}{2\,!}+\frac{1}{4!}+... \right]\] ?. (iii) Subtracting Eq. (ii) from Eq. (i), we get \[e-\frac{1}{e}=2\left[ \frac{1}{1\,!}+\frac{1}{3\,!}+\frac{1}{5\,!}+.... \right]\] ....(iv) On dividing Eq. (iii) by Eq. (iv), we get \[\frac{e+\frac{1}{e}}{e-\frac{1}{e}}=\frac{1+\frac{1}{2!}+\frac{1}{4!}+....}{\frac{1}{1\,!}+\frac{1}{3\,!}+\frac{1}{5!}+....}\] \[\Rightarrow \] \[\frac{1+\frac{1}{2!}+\frac{1}{4!}+....}{\frac{1}{1\,!}+\frac{1}{3!}+\frac{1}{5!}+....}=\frac{{{e}^{2}}+1}{{{e}^{2}}-1}\]
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