Sequence And Series Question 58
Question: $ {\log_{e}}\frac{4}{5}+\frac{1}{4}-\frac{1}{2}{{( \frac{1}{4} )}^{2}}+\frac{1}{3},{{( \frac{1}{4} )}^{3}}+….. $
Options:
A) $ 2{\log_{e}}\frac{4}{5} $
B) $ {\log_{e}}\frac{5}{4} $
C) 1
D) 0
Show Answer
Answer:
Correct Answer: D
Solution:
$ {\log_{e}}\frac{4}{5}+\frac{1}{4}-\frac{1}{2}{{( \frac{1}{4} )}^{2}}+\frac{1}{3}{{( \frac{1}{4} )}^{3}}+…. $ = $ {\log_{e}}\frac{4}{5}+{\log_{e}}( 1+\frac{1}{4} )={\log_{e}}\frac{4}{5}+{\log_{e}}\frac{5}{4}=0 $ .
BETA
