Sequence And Series Question 563
Question: If $ y=2x^{2}-1 $ , then $ [ \frac{1}{y}+\frac{1}{3y^{3}}+\frac{1}{5y^{5}}+…. ] $ is equal to
Options:
A) $ \frac{1}{2}[ \frac{1}{x^{2}}-\frac{1}{2x^{4}}+\frac{1}{3x^{6}}-….. ] $
B) $ \frac{1}{2}[ \frac{1}{x^{2}}+\frac{1}{2x^{4}}+\frac{1}{3x^{6}}+….. ] $
C) $ \frac{1}{2}[ \frac{1}{x^{2}}+\frac{1}{3x^{6}}+\frac{1}{5x^{10}}+….. ] $
D) $ \frac{1}{2}[ \frac{1}{x^{2}}-\frac{1}{3x^{6}}+\frac{1}{5x^{10}}-….. ] $
Show Answer
Answer:
Correct Answer: B
Solution:
Given that y = 2x2?1, then $ [ \frac{1}{y}+\frac{1}{3y^{3}}+\frac{1}{5y^{5}}+…. ],=\frac{1}{2}[ {\log_{e}}( \frac{1+\frac{1}{y}}{1-\frac{1}{y}} ) ] $ $ =\frac{1}{2}{\log_{e}}( \frac{y+1}{y-1} )=-\frac{1}{2}{\log_{e}}( \frac{y-1}{y+1} ) $ $ =-\frac{1}{2}{\log_{e}}( 1-\frac{2}{1+y} )=-\frac{1}{2}{\log_{e}}( 1-\frac{1}{x^{2}} ) $ $ =\frac{1}{2}[ \frac{1}{x^{2}}+\frac{1}{2x^{4}}+\frac{1}{3x^{6}}+…. ] $ .
BETA
