The values of x for which frac{x}{(x-1)^2(x-2 | vedclass

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(B) Given expression: $f(x) = \frac{x}{(x-1)^2(x-2)}$.Using partial fractions: $\frac{x}{(x-1)^2(x-2)} = \frac{A}{x-1} + \frac{B}{(x-1)^2} + \frac{C}{

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$|x| < 1, 1 - n - \frac{1}{2^n}$

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(B) Given expression: $f(x) = \frac{x}{(x-1)^2(x-2)}$.Using partial fractions: $\frac{x}{(x-1)^2(x-2)} = \frac{A}{x-1} + \frac{B}{(x-1)^2} + \frac{C}{x-2}$.Solving for constants: $x = A(x-1)(x-2) + B(x-2) + C(x-1)^2$.For $x=1$: $1 = B(1-2) \implies B = -1$.For $x=2$: $2 = C(2-1)^2 \implies C = 2$.Comparing coefficients of $x^2$: $0 = A + C \implies A = -2$.So,$f(x) = \frac{-2}{x-1} - \frac{1}{(x-1)^2} + \frac{2}{x-2} = \frac{2}{1-x} - \frac{1}{(1-x)^2} - \frac{1}{1-x/2}$.Expansion: $2(1+x+x^2+\dots+x^n+\dots) - (1+2x+3x^2+\dots+(n+1)x^n+\dots) - (1+\frac{x}{2}+\frac{x^2}{4}+\dots+\frac{x^n}{2^n}+\dots)$.Coefficient of $x^n$: $2 - (n+1) - \frac{1}{2^n} = 2 - n - 1 - \frac{1}{2^n} = 1 - n - \frac{1}{2^n}$.The expansion is valid for $|x| < 1$.

The values of $x$ for which $\frac{x}{(x-1)^2(x-2)}$ has a power series expansion and the coefficient of $x^n$ in such expansion are respectively:

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