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1 votes
1 votes

If $0 <x<1$, then the sum of the infinite series $\frac{1}{2}x^2+\frac{2}{3}x^3+\frac{3}{4}x^4+ \cdots$ is

  1. $\log \frac{1+x}{1-x}$
  2. $\frac{x}{1-x} + \log(1+x)$
  3. $\frac{1}{1-x} + \log(1-x)$
  4. $\frac{x}{1-x} + \log(1-x)$
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1 Answer

3 votes
3 votes
$\log(1+x)=x-\frac{x^{2}}{2} +\frac{x^{3}}{3} - \frac{x^{4}}{4}+...$

$\log(1-x)= -x-\frac{x^{2}}{2} - \frac{x^{3}}{3} - \frac{x^{4}}{4}-...$

$\frac{1}{(1-x)}=1+x+x^{2}+x^{3}+x^{4}+...$

$ \frac{x}{(1-x)}=x+x^{2}+x^{3}+x^{4}+x^{5}+...$

Let us check option D:

$\log(1-x) +  \frac{x}{(1-x)} = -x-\frac{x^{2}}{2} -\frac{x^{3}}{3} - \frac{x^{4}}{4}-...$  $+$ $x+x^{2}+x^{3}+x^{4}+x^{5}...$

$=\frac{x^{2}}{2} +\frac{2x^{3}}{3} +\frac{3x^{4}}{4}+...$

which matches with given question..

Hence option d is the answer.
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