A) \[-\frac{2}{n}\]
B) \[-\frac{1}{n}\]
C) \[\frac{1}{n}\]
D) \[\frac{2}{n}\]
Correct Answer: A
Solution :
\[\log (1+x+{{x}^{2}})=\log \left\{ \frac{1-{{x}^{3}}}{1-x} \right\}\] \[=\log (1-{{x}^{3}})-\log (1-x)\] \[=\left[ -{{x}^{2}}-\frac{1}{2}{{({{x}^{3}})}^{2}}-\frac{1}{3}{{({{x}^{3}})}^{3}}-... \right]\] \[+\left[ x+\frac{1}{2}{{x}^{2}}+\frac{1}{3}{{x}^{3}}+..... \right]\] If\[n\]is a multiple of 3, then\[n=3m,\]where \[m\in Z\]. \[\therefore \]Coefficient of\[{{x}^{n}}=\]Coefficient of\[{{x}^{3m}}\] \[=-\frac{1}{m}+\frac{1}{3m}\] \[=-\frac{3}{n}+\frac{1}{n}=-\frac{2}{n}\]
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