Sequence And Series Question 62
Question: The sum of $ \frac{1}{2}+\frac{1}{3}.\frac{1}{2^{3}}+\frac{1}{5}.\frac{1}{2^{5}}+…..\infty $ is
[MP PET 1991]
Options:
A) $ {\log_{e}}\sqrt{\frac{3}{2}} $
B) $ {\log_{e}}\sqrt{3} $
C) $ {\log_{e}}\sqrt{\frac{1}{2}} $
D) $ {\log_{e}}3 $
Show Answer
Answer:
Correct Answer: B
Solution:
Sum of $ \frac{1}{2}+\frac{1}{3}.\frac{1}{{2^{.3}}}+\frac{1}{5}.\frac{1}{2^{5}}+….\infty $ = $ \frac{1}{2}[ 1+\frac{1}{3}.\frac{1}{2^{2}}+\frac{1}{5}.\frac{1}{2^{4}}+….\infty ] $ $ =\frac{1}{2}.{\log_{e}}( \frac{1+\frac{1}{2}}{1-\frac{1}{2}} ) $ = $ \frac{1}{2}.{\log_{e}}( \frac{3/2}{1/2} )=\frac{1}{2}.{\log_{e}}(3)={\log_{e}}\sqrt{3} $ .
BETA
