Sequence And Series Question 37
Question: $ \frac{1}{x+1}+\frac{1}{2,{{(x+1)}^{2}}}+\frac{1}{3,{{(x+1)}^{3}}}+….\infty = $
Options:
A) $ {\log_{e}}( 1+\frac{1}{x} ) $
B) $ {\log_{e}}( 1-\frac{1}{x} ) $
C) $ {\log_{e}}( \frac{x}{x+1} ) $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
Given series is $ \frac{1}{x+1}+\frac{1}{2{{(x+1)}^{2}}}+\frac{1}{3{{(x+1)}^{3}}}+…….\infty $ $ =-{\log_{e}}( 1-\frac{1}{x+1} )=-{\log_{e}}( \frac{x}{x+1} ) $ $ ={\log_{e}}( \frac{x+1}{x} )={\log_{e}}( 1+\frac{1}{x} ) $ .
BETA
