A) \[n+\frac{1}{n}\]
B) \[\frac{n'}{2}+\frac{1}{2n}\]
C) \[{{\log }_{e}}\frac{1}{1-{{({{\log }_{e}}n)}^{2}}}\]
D) \[\frac{1}{2}{{\log }_{e}}\,\frac{1}{1-{{({{\log }_{e}}\,n)}^{2}}}\]
Correct Answer: B
Solution :
We know that, \[1+\frac{{{x}^{2}}}{2!}+\frac{{{x}^{4}}}{4!2}+....=\frac{{{e}^{x}}+{{e}^{-x}}}{{}}\] \[\therefore \] \[a+\frac{{{({{\log }_{e}}n)}^{2}}}{2!}+\frac{{{({{\log }_{e}}n)}^{4}}}{4!}+....\] \[=\frac{{{e}^{{{\log }_{e}}n}}+{{e}^{-{{\log }_{e}}n}}}{2}\] \[=\frac{n+\frac{1}{n}}{2}\] \[=\frac{n}{2}+\frac{1}{2n}\]
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