Sequence And Series Question 59
Question: $ \frac{1}{n^{2}}+\frac{1}{2n^{4}}+\frac{1}{3n^{6}}+……\infty = $
Options:
A) $ {\log_{e}}( \frac{n^{2}}{n^{2}+1} ) $
B) $ {\log_{e}}( \frac{n^{2}+1}{n^{2}} ) $
C) $ {\log_{e}}( \frac{n^{2}}{n^{2}-1} ) $
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
$ S=\frac{\frac{1}{n^{2}}}{1}+\frac{{{( \frac{1}{n^{2}} )}^{2}}}{2}+\frac{{{( \frac{1}{n^{2}} )}^{3}}}{3}+…. $ $ =-{\log_{e}}( 1-\frac{1}{n^{2}} )={\log_{e}}\frac{n^{2}}{n^{2}-1} $ . Trick: Putting $ n=2, $ the sum of the series upto 4 terms is $ \frac{1}{4}+\frac{1}{32}+\frac{1}{192}+\frac{1}{1024}=0.2874 $ … and option = - 0.223…., =0.223….., =0.2876….. Hence answer is (c).
BETA
