Sequence And Series Question 74
Question: The coefficient of $ x^{n} $ in the expansion of $ {\log_{a}}(1+x) $ is
Options:
A) $ \frac{{{(-1)}^{n-1}}}{n} $
B) $ \frac{{{(-1)}^{n-1}}}{n}{\log_{a}}e $
C) $ \frac{{{(-1)}^{n-1}}}{n}{\log_{e}}a $
D) $ \frac{{{(-1)}^{n}}}{n}{\log_{a}}e $
Show Answer
Answer:
Correct Answer: B
Solution:
We have $ {\log_{a}}(1+x)={\log_{e}}(1+x).{\log_{a}}e $ $ ={\log_{a}}e( \sum\limits_{n=1}^{\infty }{{{(-1)}^{n-1}}\frac{x^{n}}{n}} ) $ Therefore coefficient of $ x^{n} $ in $ {\log_{a}}(1+x) $ is $ \frac{{{(-1)}^{n-1}}}{n}{\log_{a}}e $ .
BETA
