The coefficient of x^3 in the expansion of (1 | vedclass

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(B) The binomial expansion of $(1-x)^n$ for any index $n$ is given by $1 + nx + \frac{n(n-1)}{2!}x^2 + \frac{n(n-1)(n-2)}{3!}x^3 + \dots$ For $n = \fr

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$\frac{1}{16}$

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(B) The binomial expansion of $(1-x)^n$ for any index $n$ is given by $1 + nx + \frac{n(n-1)}{2!}x^2 + \frac{n(n-1)(n-2)}{3!}x^3 + \dots$ For $n = \frac{3}{2}$,the term containing $x^3$ is $\frac{\frac{3}{2}(\frac{3}{2}-1)(\frac{3}{2}-2)}{3!}(-x)^3$. Calculating this: $\frac{\frac{3}{2} 	imes \frac{1}{2} 	imes (-\frac{1}{2})}{6} 	imes (-x^3) = \frac{-\frac{3}{8}}{6} 	imes (-x^3) = \frac{3}{48}x^3 = \frac{1}{16}x^3$. Thus,the coefficient of $x^3$ is $\frac{1}{16}$.

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

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