Sequence And Series Question 61
Question: $ \frac{m-n}{m+n}+\frac{1}{3}{{( \frac{m-n}{m+n} )}^{3}}+\frac{1}{5}{{( \frac{m-n}{m+n} )}^{5}}+……\infty = $
Options:
A) $ {\log_{e}}( \frac{m}{n} ) $
B) $ {\log_{e}}( \frac{n}{m} ) $
C) $ {\log_{e}}( \frac{m-n}{m+n} ) $
D) $ \frac{1}{2}{\log_{e}}( \frac{m}{n} ) $
Show Answer
Answer:
Correct Answer: D
Solution:
$ \frac{m-n}{m+n}+\frac{1}{3}{{( \frac{m-n}{m+n} )}^{3}}+…. $ = $ \frac{1}{2}{\log_{e}}( \frac{1+\frac{m-n}{m+n}}{1-\frac{m-n}{m+n}} )=\frac{1}{2}{\log_{e}}\frac{2m}{2n}=\frac{1}{2}{\log_{e}}( \frac{m}{n} ) $ .
BETA
