Sequence And Series Question 75
Question: The coefficient of $ {n^{-r}} $ in the expansion of $ {\log_{10}}( \frac{n}{n-1} ) $ is
Options:
A) $ \frac{1}{r,{\log_{e}}10} $
B) $ -\frac{1}{r{\log_{e}}10} $
C) $ -\frac{1}{r!{\log_{e}}10} $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
We have $ {\log_{10}}( \frac{n}{n-1} )={\log_{e}}( \frac{n}{n-1} ).{\log_{10}}e $ $ =-{\log_{e}}( \frac{n-1}{n} ).{\log_{10}}e $ $ =-{\log_{10}}e.{\log_{e}}( 1-\frac{1}{n} ) $ Therefore coefficient of $ {n^{-r}} $ is $ \frac{1}{r}{\log_{10}}e=\frac{1}{r{\log_{e}}10} $ .
BETA
