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

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(C) We have $(1+x)^2(8-x)^{-\frac{1}{3}} = (1+2x+x^2) \cdot 8^{-\frac{1}{3}} (1-\frac{x}{8})^{-\frac{1}{3}}$.Since $8^{-\frac{1}{3}} = \frac{1}{2}$,th

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$\frac{313}{576}$

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(C) We have $(1+x)^2(8-x)^{-\frac{1}{3}} = (1+2x+x^2) \cdot 8^{-\frac{1}{3}} (1-\frac{x}{8})^{-\frac{1}{3}}$.Since $8^{-\frac{1}{3}} = \frac{1}{2}$,the expression becomes $\frac{1}{2}(1+2x+x^2)(1-\frac{x}{8})^{-\frac{1}{3}}$.Using the binomial expansion $(1-z)^{-n} = 1 + nz + \frac{n(n+1)}{2!}z^2 + \dots$,we have $(1-\frac{x}{8})^{-\frac{1}{3}} = 1 + (\frac{1}{3})(\frac{x}{8}) + \frac{(-\frac{1}{3})(-\frac{1}{3}-1)}{2!}(-\frac{x}{8})^2 + \dots = 1 + \frac{x}{24} + \frac{2}{9} \cdot \frac{1}{2} \cdot \frac{x^2}{64} = 1 + \frac{x}{24} + \frac{x^2}{576}$.Now,multiply $\frac{1}{2}(1+2x+x^2)(1 + \frac{x}{24} + \frac{x^2}{576})$.The coefficient of $x^2$ is $\frac{1}{2} [1 \cdot \frac{1}{576} + 2 \cdot \frac{1}{24} + 1 \cdot 1] = \frac{1}{2} [\frac{1}{576} + \frac{1}{12} + 1] = \frac{1}{2} [\frac{1 + 48 + 576}{576}] = \frac{625}{1152}$.Wait,re-evaluating the expansion: $(1-\frac{x}{8})^{-\frac{1}{3}} = 1 + (\frac{1}{3})(\frac{x}{8}) + \frac{(1/3)(4/3)}{2}(\frac{x}{8})^2 = 1 + \frac{x}{24} + \frac{2}{9} \cdot \frac{x^2}{64} = 1 + \frac{x}{24} + \frac{x^2}{288}$.Coefficient of $x^2$ is $\frac{1}{2} [\frac{1}{288} + \frac{2}{24} + 1] = \frac{1}{2} [\frac{1}{288} + \frac{24}{288} + \frac{288}{288}] = \frac{1}{2} [\frac{313}{288}] = \frac{313}{576}$.

The coefficient of $x^2$ in the expansion of $(1+x)^2(8-x)^{-\frac{1}{3}}$ is

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