A) \[{{e}^{2}}\]
B) \[{{\log }_{e}}2+1\]
C) \[{{\log }_{e}}3-2\]
D) \[1-{{\log }_{e}}2\]
Correct Answer: D
Solution :
We have \[{{\log }_{4}}2-{{\log }_{8}}2-{{\log }_{16}}2{{\log }_{32}}2+...\] \[=\frac{1}{{{\log }_{2}}4}-\frac{1}{{{\log }_{2}}8}+\frac{1}{{{\log }_{2}}16}-\frac{1}{{{\log }_{2}}32}\] \[\Rightarrow \]\[\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}+...\] \[=1-\left( 1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}... \right)\] \[=1-{{\log }_{e}}2\]
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