Sequence And Series Question 52
Question: $ 1+( \frac{1}{2}+\frac{1}{3} ),\frac{1}{4}+( \frac{1}{4}+\frac{1}{5} ),\frac{1}{4^{2}}+( \frac{1}{6}+\frac{1}{7} ),\frac{1}{4^{3}}+….\infty = $
Options:
A) $ {\log_{e}}(2\sqrt{3}) $
B) $ 2{\log_{e}}2 $
C) $ {\log_{e}}2 $
D) $ {\log_{e}}( \frac{2}{\sqrt{3}} ) $
Show Answer
Answer:
Correct Answer: A
Solution:
$ S={ 1+\frac{{{( \frac{1}{2} )}^{2}}}{2}+\frac{{{( \frac{1}{2} )}^{4}}}{4}+….. } $ $ +2{ \frac{1}{2}+\frac{{{( \frac{1}{2} )}^{3}}}{3}+\frac{{{( \frac{1}{2} )}^{5}}}{5\ }+….. }-1 $ $ =1-\frac{1}{2}{\log_{e}}( 1+\frac{1}{2} )( 1-\frac{1}{2} )+{\log_{e}}( \frac{1+\frac{1}{2}}{1-\frac{1}{2}} )-1 $ $ =-\frac{1}{2}{\log_{e}}\frac{3}{4}+{\log_{e}}3={\log_{e}}2\sqrt{3} $ .
BETA
