Sequence And Series Question 76
Question: The coefficient of $ x^{n} $ in the expansion of $ {\log_{e}}(1+3x+2x^{2}) $ is
[UPSEAT 2001]
Options:
A) $ {{(-1)}^{n}}[ \frac{2^{n}+1}{n} ] $
B) $ \frac{{{(-1)}^{n+1}}}{n}[2^{n}+1] $
C) $ \frac{2^{n}+1}{n} $
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
We have $ \log (1+3x+2x^{2})=\log (1+x)+\log (1+2x) $ = $ \sum\limits_{n=1}^{\infty }{{{(-1)}^{n-1}}}\frac{x^{n}}{n}+\sum\limits_{n=1}^{\infty }{{{(-1)}^{n-1}}}\frac{{{(2x)}^{n}}}{n} $ = $ \sum\limits_{n=1}^{\infty }{{{(-1)}^{n-1}}}( \frac{1}{n}+\frac{2^{n}}{n} )x^{n} $ = $ \sum\limits_{n=1}^{\infty }{{{(-1)}^{n-1}}}( \frac{1+2^{n}}{n} )x^{n} $ So, coefficient of $ x^{n} $ = $ {{(-1)}^{n-1}}( \frac{2^{n}+1}{n} ) $
$ \Rightarrow \frac{{{(-1)}^{n+1}}(2^{n}+1)}{n} $ , $ [\because {{(-1)}^{n}}={{(-1)}^{n+2}}=…] $ .
BETA
