Sequence And Series Question 43
Question: $ {\log_{e}},[{{(1+x)}^{1+x}}{{(1-x)}^{1-x}}],= $
Options:
A) $ \frac{x^{2}}{2}+\frac{x^{4}}{4}+\frac{x^{6}}{6}+….\infty $
B) $ \frac{x^{2}}{1.2}+\frac{x^{4}}{3.4}+\frac{x^{6}}{5.6}+….\infty $
C) $ 2[ \frac{x^{2}}{1.2}+\frac{x^{4}}{3.4}+\frac{x^{6}}{5.6}+..\infty ] $
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
$ {\log_{e}}{{{(1+x)}^{1+x}}{{(1-x)}^{1-x}}} $ $ =(1+x){\log_{e}}(1+x)+(1-x){\log_{e}}(1-x) $ $ =(1+x){ x-\frac{x^{2}}{2}+\frac{x^{3}}{3}-\frac{x^{4}}{4}+…… } $ $ +(1-x){ -x-\frac{x^{2}}{2}-\frac{x^{3}}{3}-\frac{x^{4}}{4}-……. } $ $ =2{ -\frac{x^{2}}{2}-\frac{x^{4}}{4}-\frac{x^{6}}{6}-….. }+2{ x^{2}+\frac{x^{4}}{3}+\frac{x^{6}}{5}+…… } $ $ =2[ x^{2}( 1-\frac{1}{2} )+x^{4}( \frac{1}{3}-\frac{1}{4} )+x^{6}( \frac{1}{5}-\frac{1}{6} )+…… ] $ $ =2[ \frac{x^{2}}{1.2}+\frac{x^{4}}{3.4}+\frac{x^{6}}{5.6}+……. ] $ .
BETA
