∑n=1∞ 12n-1 x2n =x21+(x2)23+(x2)35+......∞ =12 log 1+x21-x2

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$\sum_{n = 1}^{\infty} \frac{1}{2 n - 1} \cdot x^{2 n} =$

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$\frac{x}{2} \log {(\frac{1 + x}{1 - x})}^{x}$

$xlog \frac{1 + x}{1 - x}$

$\frac{x}{2} \log (\frac{1 + x}{1 - x})$

$\frac{1}{2} \log (\frac{1 + x^{2}}{1 - x^{2}})$

answer is B.

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Detailed Solution

$\begin{array}{rcl} \sum_{n = 1}^{\infty} \frac{1}{(2 n - 1)} x^{2 n} \\ = & \frac{x^{2}}{1} + \frac{{(x^{2})}^{2}}{3} + \frac{{(x^{2})}^{3}}{5} + . . . . . . \infty \\ = & \frac{1}{2} \log (\frac{1 + x^{2}}{1 - x^{2}}) \end{array}$

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