frac{1}{(2 + x)^4} = | vedclass

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(B) We have $\frac{1}{(2 + x)^4} = (2 + x)^{-4}$.Factor out $2$ from the expression: $(2 + x)^{-4} = [2(1 + \frac{x}{2})]^{-4} = 2^{-4}(1 + \frac{x}{2

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$\frac{1}{16}\left( 1 - 2x + \frac{5}{2}x^2 - \dots \right)$

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(B) We have $\frac{1}{(2 + x)^4} = (2 + x)^{-4}$.Factor out $2$ from the expression: $(2 + x)^{-4} = [2(1 + \frac{x}{2})]^{-4} = 2^{-4}(1 + \frac{x}{2})^{-4} = \frac{1}{16}(1 + \frac{x}{2})^{-4}$.Using the binomial expansion $(1 + z)^n = 1 + nz + \frac{n(n-1)}{2!}z^2 + \dots$,where $n = -4$ and $z = \frac{x}{2}$:$(1 + \frac{x}{2})^{-4} = 1 + (-4)(\frac{x}{2}) + \frac{(-4)(-5)}{2}(\frac{x}{2})^2 + \dots$$= 1 - 2x + \frac{20}{2 	imes 4}x^2 + \dots = 1 - 2x + \frac{5}{2}x^2 + \dots$Therefore,$\frac{1}{(2 + x)^4} = \frac{1}{16}(1 - 2x + \frac{5}{2}x^2 - \dots)$.

$\frac{1}{(2 + x)^4} = $

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