Sequence And Series Question 28
Question: $ \frac{1}{2}x^{2}+\frac{2}{3}x^{3}+\frac{3}{4}x^{4}+……\infty = $
Options:
A) $ \frac{x}{1+x}-{\log_{e}}(1-x) $
B) $ \frac{x}{1+x}+{\log_{e}}(1-x) $
C) $ \frac{x}{1-x}-{\log_{e}}(1-x) $
D) $ \frac{x}{1-x}+{\log_{e}}(1-x) $
Show Answer
Answer:
Correct Answer: D
Solution:
$ S=( 1-\frac{1}{2} )x^{2}+( 1-\frac{1}{3} )x^{3}+( 1-\frac{1}{4} )x^{4}+………… $ $ +( 1-\frac{1}{n+1} ){x^{n+1}}+……..\infty $ $ ={x^{2}+x^{3}+x^{4}+………}-{ \frac{x^{2}}{2}+\frac{x^{3}}{3}+………. } $ $ =\frac{x^{2}}{1-x}-{-{\log_{e}}(1-x)-x}=\frac{x}{1-x}+{\log_{e}}(1-x) $ .
BETA
