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

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(B) Using the binomial expansion $(1+z)^n = 1 + nz + \frac{n(n-1)}{2!}z^2 + \frac{n(n-1)(n-2)}{3!}z^3 + \dots$ Here,$n = \frac{1}{2}$ and $z = -\frac{

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$\frac{-27}{1024}$

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(B) Using the binomial expansion $(1+z)^n = 1 + nz + \frac{n(n-1)}{2!}z^2 + \frac{n(n-1)(n-2)}{3!}z^3 + \dots$ Here,$n = \frac{1}{2}$ and $z = -\frac{3}{4}x$. The term containing $x^3$ is given by $\frac{n(n-1)(n-2)}{3!}z^3$. Substituting the values: $\frac{\frac{1}{2}(\frac{1}{2}-1)(\frac{1}{2}-2)}{3!} (-\frac{3}{4}x)^3$ $= \frac{\frac{1}{2} 	imes (-\frac{1}{2}) 	imes (-\frac{3}{2})}{6} 	imes (-\frac{27}{64}x^3)$ $= \frac{3/8}{6} 	imes (-\frac{27}{64}x^3)$ $= \frac{1}{16} 	imes (-\frac{27}{64}x^3) = -\frac{27}{1024}x^3$. Thus,the coefficient of $x^3$ is $-\frac{27}{1024}$.

The coefficient of $x^3$ in the expansion of $(1-\frac{3}{4} x)^{\frac{1}{2}}$ is

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