A) \[{{\log }_{e}}\frac{4}{e}\]
B) \[{{\log }_{e}}\frac{e}{4}\]
C) \[{{\log }_{e}}4\]
D) \[{{\log }_{e}}2\]
Correct Answer: A
Solution :
We know that, \[{{\log }_{e}}2=\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+.....\] .....(i) (when \[x=1\] in \[{{\log }_{e}}(1+x)\]) Also, \[{{\log }_{e}}2=1-\left( \frac{1}{2.3} \right)-\left( \frac{1}{4.5} \right)-\left( \frac{1}{6.7} \right)-.......\] .....(ii) (when \[x=-1\] in \[{{\log }_{e}}(1-x)\]) By adding (i) and (ii), we get \[2{{\log }_{e}}2=1+\left( \frac{1}{1.2}-\frac{1}{2.3} \right)+\left( \frac{1}{3.4}-\frac{1}{4.5} \right)+......\] \[\Rightarrow 2{{\log }_{2}}2-1=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{3.4}-\frac{1}{4.5}+........\] \[\Rightarrow \frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{3.4}-\frac{1}{4.5}+........\] \[={{\log }_{e}}4-{{\log }_{e}}e={{\log }_{e}}\left( \frac{4}{e} \right)\].
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