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) \[\frac{{{(-1)}^{n}}}{n}{{\log }_{a}}e\]
Correct Answer: B
Solution :
We have \[{{\log }_{a}}(1+x)={{\log }_{e}}(1+x).{{\log }_{a}}e\]\[={{\log }_{a}}e\left( \sum\limits_{n=1}^{\infty }{{{(-1)}^{n-1}}\frac{{{x}^{n}}}{n}} \right)\] Therefore coefficient of \[{{x}^{n}}\]in \[{{\log }_{a}}(1+x)\] is\[\frac{{{(-1)}^{n-1}}}{n}{{\log }_{a}}e\].
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