A) \[x-\frac{{{x}^{2}}}{2}+\frac{{{x}^{3}}}{3}-\,.....\]
B) \[x+\frac{{{x}^{2}}}{2}+\frac{{{x}^{3}}}{3}+\,.....\]
C) \[1-x+\frac{{{x}^{2}}}{2}-\frac{{{x}^{3}}}{3}+\,.....\]
D) \[1+x-\frac{{{x}^{2}}}{2}+\frac{{{x}^{3}}}{3}-\,.....\]
Correct Answer: B
Solution :
\[{{\log }_{e}}\frac{1}{(1-x)}={{\log }_{e}}{{(1-x)}^{-1}}=-{{\log }_{e}}^{(1-x)}\] \[=-\left[ -x-\frac{{{x}^{2}}}{2}-\frac{{{x}^{3}}}{3}-\frac{{{x}^{4}}}{4}-...\infty \right]\] \[=x+\frac{{{x}^{2}}}{2}+\frac{{{x}^{3}}}{3}+...\infty \]
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