Q55 / 160 · AP EAMCET 2023 (16 May Shift 1) · Mathematics

The coefficient of $x^f$ in the expansion of $\frac{1}{\sqrt[3]{(1-2 x)^2}}$ is

  • A
    $\frac{2.5 .8 \ldots(3 r-1)}{r !}(-1)^r\left(\frac{2}{3}\right)^r$
  • B
    $\frac{2.5 .8 \ldots(3 r-1)}{r !}(-1)^r\left(\frac{3}{2}\right)^r$
  • Correct answer
    C
    $\frac{2.5 \cdot 8 \ldots(3 r-1)}{r !}\left(\frac{2}{3}\right)^r$
  • D
    $\frac{2.5 .8 \ldots(3 \mathrm{r}-1)}{\mathrm{r} !}\left(\frac{3}{2}\right)^{\mathrm{r}}$

Solution

$\frac{1}{\sqrt[3]{(1-2 x)^2}}=\frac{1}{(1-2 x)^{2 / 3}}=(1-2 x)^{-2 / 3}$ $=1+\left(\frac{2}{3}\right)(2 x)+\frac{\frac{2}{3}\left(\frac{2}{3}+1\right)}{2 !}(2 x)^2$ $+\frac{\frac{2}{3}\left(\frac{2}{3}+1\right)\left(\frac{2}{3}+2\right)}{3 !}(2 x)^3+\ldots \ldots$ $+\frac{\frac{2}{3}\left(\frac{2}{3}+1\right)\left(\frac{2}{3}+2\right) \ldots\left(\frac{2}{3}+(r-1)\right)}{r !}(2 x)^r+\ldots$ Here, the coefficient of $x^r$ is : $=\frac{\frac{2}{3}\left(\frac{2}{3}+1\right)\left(\frac{2}{3}+2\right) \ldots\left(\frac{2}{3}+(r-1)\right)}{r !} 2^r$ $=\frac{\frac{2}{3} \cdot \frac{5}{3} \cdot \frac{8}{3} \cdot \ldots \ldots \cdot\left(\frac{3 r-1}{3}\right)}{r !} \cdot 2^r=\frac{2 \cdot 5 \cdot 8 \cdot \ldots \ldots(3 r-1)}{r ! \cdot 3^r} \cdot 2^r$ $=\frac{2 \cdot 5 \cdot 8 \cdot \ldots \ldots(3 r-1)}{r !} \cdot\left(\frac{2}{3}\right)^r$

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

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