Q126 / 200 · AP EAMCET 2001 · Mathematics

$|x| < 1$, the coefficient of $x^3$ in the expansion of $\log \left(1+x+x^2\right)$ in ascending powers of $x$, is

  • A
    $\frac{2}{3}$
  • B
    $\frac{4}{3}$
  • Correct answer
    C
    $-\frac{2}{3}$
  • D
    $-\frac{4}{3}$

Solution

We have, $\log \left(1+x+x^2\right)=\log \frac{\left(1-x^3\right)}{(1-x)}$ $\begin{aligned} & =\log \left(1-x^3\right)-\log (1-x) \\ & =-\left[x^3+\frac{x^6}{2}+\frac{x^9}{3}+\ldots\right]+\left[x+\frac{x^2}{2}+\frac{x^3}{3}+\ldots\right] \\ & \text { Coefficient of } x^3=-1+\frac{1}{3}=-\frac{2}{3}\end{aligned}$

Hence, the correct answer is option: 'C'.

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