A) \[2-{{\log }_{e}}2\]
B) \[2+{{\log }_{e}}2\]
C) \[{{\log }_{e}}4\]
D) None of these
Correct Answer: A
Solution :
\[\frac{1}{1}.\frac{1}{2}+\frac{3}{2}.\frac{1}{{{2}^{2}}}+\frac{5}{3}.\frac{1}{{{2}^{3}}}+\frac{7}{4}.\frac{1}{{{2}^{4}}}+.........\] \[=\left( 2-\frac{1}{1} \right)\frac{1}{2}+\left( 2-\frac{1}{2} \right)\frac{1}{{{2}^{2}}}+\left( 2-\frac{1}{3} \right)\frac{1}{{{2}^{3}}}+\left( 2-\frac{1}{4} \right)\frac{1}{{{2}^{4}}}+..\] \[=2\left\{ \frac{1}{2}+\frac{1}{{{2}^{2}}}+\frac{1}{{{2}^{3}}}+........ \right\}-\left\{ \frac{\frac{1}{2}}{1}+\frac{\frac{1}{{{2}^{2}}}}{2}+\frac{\frac{1}{{{2}^{3}}}}{3}+....... \right\}\] \[=\frac{1}{1-\frac{1}{2}}-\left\{ -{{\log }_{e}}\left( 1-\frac{1}{2} \right) \right\}=2-{{\log }_{e}}2\].
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