For |x|

Question

(D) Given the expression $\frac{1}{(x-1)^2(x-2)}$.We can rewrite this as $\frac{1}{(-1)^2(1-x)^2(-2)(1-\frac{x}{2})} = \frac{1}{1 \cdot (1-x)^2 \cdot

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$-\frac{1}{2}$

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(D) Given the expression $\frac{1}{(x-1)^2(x-2)}$.We can rewrite this as $\frac{1}{(-1)^2(1-x)^2(-2)(1-\frac{x}{2})} = \frac{1}{1 \cdot (1-x)^2 \cdot (-2) \cdot (1-\frac{x}{2})} = -\frac{1}{2}(1-x)^{-2}(1-\frac{x}{2})^{-1}$.Using the binomial expansion $(1-z)^{-n} = 1 + nz + \frac{n(n+1)}{2!}z^2 + \dots$,we have:$(1-x)^{-2} = 1 + 2x + 3x^2 + \dots$$(1-\frac{x}{2})^{-1} = 1 + \frac{x}{2} + \frac{x^2}{4} + \dots$Multiplying these series:$-\frac{1}{2}(1 + 2x + \dots)(1 + \frac{x}{2} + \dots) = -\frac{1}{2}(1 + \frac{x}{2} + 2x + \dots) = -\frac{1}{2}(1 + \frac{5x}{2} + \dots)$.The constant term is the term independent of $x$,which is $-\frac{1}{2} 	imes 1 = -\frac{1}{2}$.

For $|x| < 1$,the constant term in the expansion of $\frac{1}{(x-1)^2(x-2)}$ is

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