A) \[{{(-1)}^{n}}\left[ \frac{{{2}^{n}}+1}{n} \right]\]
B) \[\frac{{{(-1)}^{n+1}}}{n}[{{2}^{n}}+1]\]
C) \[\frac{{{2}^{n}}+1}{n}\]
D) None of these
Correct Answer: B
Solution :
We have \[\log (1+3x+2{{x}^{2}})=\log (1+x)+\log (1+2x)\] = \[\sum\limits_{n=1}^{\infty }{{{(-1)}^{n-1}}}\frac{{{x}^{n}}}{n}+\sum\limits_{n=1}^{\infty }{{{(-1)}^{n-1}}}\frac{{{(2x)}^{n}}}{n}\] = \[\sum\limits_{n=1}^{\infty }{{{(-1)}^{n-1}}}\left( \frac{1}{n}+\frac{{{2}^{n}}}{n} \right)\,\,{{x}^{n}}\] = \[\sum\limits_{n=1}^{\infty }{{{(-1)}^{n-1}}}\left( \frac{1+{{2}^{n}}}{n} \right)\,\,{{x}^{n}}\] So, coefficient of \[{{x}^{n}}\] = \[{{(-1)}^{n-1}}\left( \frac{{{2}^{n}}+1}{n} \right)\] \[\Rightarrow \frac{{{(-1)}^{n+1}}({{2}^{n}}+1)}{n}\] , \[[\because \,\,{{(-1)}^{n}}={{(-1)}^{n+2}}=...]\].
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