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$\log (1 + x + x^{2} + . . . . . \infty) =$

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$x + \frac{x^{2}}{2} + \frac{x^{3}}{3} + \frac{x^{4}}{4} + \dots \dots \infty$

$1 + x + \frac{x^{2}}{2} + \frac{x^{3}}{3} + \frac{x^{4}}{4} + \dots \dots \infty$

$1 - x + \frac{x^{2}}{2} - \frac{x^{3}}{3} + \frac{x^{4}}{4} + \dots \dots \infty$

$x - \frac{x^{2}}{2} + \frac{x^{3}}{3} - \frac{x^{4}}{4} + \dots \dots \infty$

answer is A.

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Detailed Solution

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

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