The coefficient of x^2 in the expansion of (1 | vedclass

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(B) Using the binomial expansion $(1+z)^n = 1 + nz + \frac{n(n-1)}{2!}z^2 + \dots$,we expand the two terms:$(1-3x)^{\frac{1}{3}} = 1 + \frac{1}{3}(-3x

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$\frac{3}{2}$

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(B) Using the binomial expansion $(1+z)^n = 1 + nz + \frac{n(n-1)}{2!}z^2 + \dots$,we expand the two terms:$(1-3x)^{\frac{1}{3}} = 1 + \frac{1}{3}(-3x) + \frac{\frac{1}{3}(\frac{1}{3}-1)}{2!}(-3x)^2 + \dots = 1 - x - x^2 + \dots$$(1+2x)^{-\frac{1}{2}} = 1 + (-\frac{1}{2})(2x) + \frac{(-\frac{1}{2})(-\frac{1}{2}-1)}{2!}(2x)^2 + \dots = 1 - x + \frac{3}{2}x^2 + \dots$Multiplying these expansions: $(1 - x - x^2 + \dots)(1 - x + \frac{3}{2}x^2 + \dots)$The coefficient of $x^2$ is obtained by: $(1 	imes \frac{3}{2}) + (-1 	imes -1) + (-1 	imes 1) = \frac{3}{2} + 1 - 1 = \frac{3}{2}$.

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

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