The coefficient of $x^3$ in the expansion of $(1-x)^{3 / 2}$, $(|x| < 1)$ is

The coefficient of $x^3$ in the expansion of $(1-x)^{3 / 2}$, $(|x| < 1)$ is
  1. $-\frac{3}{16}$
  2. $\frac{1}{16}$
  3. $\frac{1}{8}$
  4. $\frac{3}{16}$

Solution

The expansion of $(1-x)^{3 / 2}$ is $ \begin{aligned} & \Rightarrow(1-x)^{3 / 2}=1+\frac{3}{2}(-x)+\frac{\left(\frac{3}{2}\right)\left(\frac{3}{2}-1\right)(x)^2}{2 !} \\ & +\frac{\frac{3}{2}\left(\frac{3}{2}-1\right)\left(\frac{3}{2}-2\right)}{3 !}(-x)^3+\ldots \\ & =1-\frac{3}{2} x+\frac{3}{8} x^2+\frac{1}{16} x^3+\ldots \\ & \therefore \text { Coefficient of } x^3=\frac{1}{16} \end{aligned} $

Asked in: AP EAMCET 2023 (18 May Shift 2)

Practice more Binomial Theorem questions on Aicharya