Thecoefficientof $x^3$ in the expansion of $(1-2 x)^{\frac{1}{2}}(1+3 x)^{\frac{1}{3}}$ is

Thecoefficientof $x^3$ in the expansion of $(1-2 x)^{\frac{1}{2}}(1+3 x)^{\frac{1}{3}}$ is
  1. $-\frac{20}{3}$
  2. $\frac{20}{3}$
  3. $\frac{17}{3}$
  4. $-\frac{17}{3}$

Solution

$(1-2 x)^{1 / 2}(1+3 x)^{-1 / 3}$ $=\left(1+\frac{1}{2}(-2 x)+\frac{1}{2} \frac{\left(\frac{1}{2}-1\right)}{2-1}(-2 x)^2+\frac{\frac{1}{2}\left(\frac{1}{2}-1\right)\left(\frac{1}{2}-2\right)}{3.2 .1}(-2 x)^3+\ldots\right)$ $\begin{aligned} & \times\left[1-\frac{1}{3}(3 x)+\frac{\left(-\frac{1}{3}\right)\left(-\frac{1}{3}-1\right)}{2.1}(3 x)^2+\right. \\ & \left.\frac{\left(-\frac{1}{3}\right)\left(-\frac{1}{3}-1\right)\left(-\frac{1}{3}-2\right)}{3.2 .1}(3 x)^3+\ldots\right] \end{aligned}$ Coefficient of $x^3$. $\begin{aligned} & =\frac{1}{6} \cdot\left(-\frac{1}{3}\right)\left(-\frac{4}{3}\right)\left(-\frac{7}{3}\right) \cdot 27-\frac{1}{2}\left(-\frac{1}{3}\right)\left(-\frac{4}{3}\right) \cdot 9 \\ & +\frac{1}{6} \times \frac{1}{2} \cdot\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)(-8)-\frac{1}{2} \times \frac{1}{2}\left(-\frac{1}{2}\right) \cdot 4=-\frac{20}{3} \end{aligned}$

Asked in: AP EAMCET 2023 (16 May Shift 2)

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