Sequence And Series Question 51
Question: $ \frac{1}{1,.,2}-\frac{1}{2,.,3}+\frac{1}{3,.,4}-\frac{1}{4,.,5}+…..\infty = $
[Roorkee 1992; AIEEE 2003]
Options:
A) $ {\log_{e}}\frac{4}{e} $
B) $ {\log_{e}}\frac{e}{4} $
C) $ {\log_{e}}4 $
D) $ {\log_{e}}2 $
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Answer:
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-( \frac{1}{2.3} )-( \frac{1}{4.5} )-( \frac{1}{6.7} )-……. $ …..(ii) (when $ x=-1 $ in $ {\log_{e}}(1-x) $ ) By adding (i) and (ii), we get $ 2{\log_{e}}2=1+( \frac{1}{1.2}-\frac{1}{2.3} )+( \frac{1}{3.4}-\frac{1}{4.5} )+…… $
$ \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}}( \frac{4}{e} ) $ .
BETA
