For |x|

Question

(D) We have the expression $\frac{1}{(x-1)^2(x-2)}$.First,rewrite the expression as $\frac{1}{(-1)^2(1-x)^2(-2)(1-\frac{x}{2})} = -\frac{1}{2}(1-x)^{-

Vedclass · Accepted Answer

$-\frac{1}{2}$

Vedclass · Answer

(D) We have the expression $\frac{1}{(x-1)^2(x-2)}$.First,rewrite the expression as $\frac{1}{(-1)^2(1-x)^2(-2)(1-\frac{x}{2})} = -\frac{1}{2}(1-x)^{-2}(1-\frac{x}{2})^{-1}$.Using the binomial expansion $(1-x)^{-n} = 1 + nx + \frac{n(n+1)}{2!}x^2 + \dots$ and $(1-y)^{-1} = 1 + y + y^2 + \dots$ for $|x| $(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: $-\frac{1}{2}(1 + 2x + \dots)(1 + \frac{x}{2} + \dots) = -\frac{1}{2}(1 + (\frac{1}{2} + 2)x + \dots) = -\frac{1}{2}(1 + \frac{5}{2}x + \dots)$.The constant term 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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