Sequence And Series Question 30
Question: $ \frac{x-1}{(x+1)}+\frac{1}{2},.,\frac{x^{2}-1}{{{(x+1)}^{2}}}+\frac{1}{3},.,\frac{x^{3}-1}{{{(x+1)}^{3}}}+……\infty = $
Options:
A) $ {\log_{e}}x $
B) $ {\log_{e}}(1+x) $
C) $ {\log_{e}}(1-x) $
D) $ {\log_{e}}\frac{x}{1+x} $
Show Answer
Answer:
Correct Answer: A
Solution:
$ S={ \frac{x}{x+1}+\frac{{{( \frac{x}{x+1} )}^{2}}}{2}+\frac{{{( \frac{x}{x+1} )}^{3}}}{3}+………\infty } $ $ -{ \frac{1}{x+1}+\frac{{{( \frac{1}{x+1} )}^{2}}}{2}+\frac{{{( \frac{1}{x+1} )}^{3}}}{3}+……..\infty } $ $ =-{\log_{e}}( 1-\frac{x}{x+1} )-{ -{\log_{e}}( 1-\frac{1}{x+1} ) } $ $ =-{\log_{e}}\frac{1}{x+1}+{\log_{e}}\frac{x}{x+1}={\log_{e}}x $ . Trick: Put $ x=2 $ and check..
BETA
