A) \[\frac{x}{1+x}+\log (1+x)\]
B) \[\frac{x}{1-x}+\log (1-x)\]
C) \[-\frac{x}{1+x}+\log (1+x)\]
D) None of these
Correct Answer: B
Solution :
We have,\[\frac{{{x}^{2}}}{2}+\frac{2}{3}{{x}^{3}}+\frac{3}{4}{{x}^{4}}+\frac{4}{5}{{x}^{5}}+...\] \[=\sum\limits_{n=1}^{\infty }{\frac{n}{n+1}}\cdot {{x}^{n+1}}=\sum\limits_{n=1}^{\infty }{\frac{n+1-1}{n+1}\cdot {{x}^{n+1}}}\] \[=\sum\limits_{n=1}^{\infty }{\left( 1-\frac{1}{n+1} \right)\cdot {{x}^{n+1}}}\] \[=\sum\limits_{n=1}^{\infty }{{{x}^{n+1}}-\sum\limits_{n=1}^{\infty }{\frac{{{x}^{n+1}}}{n+1}}}\] \[=\frac{{{x}^{2}}}{1-x}+x-\sum\limits_{n=0}^{\infty }{\cdot \frac{{{x}^{n+1}}}{n+1}}\] \[=\frac{{{x}^{2}}}{1-x}+x+\log (1-x)\] \[=\frac{x}{1-x}+\log (1-x)\]
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