For |x|

Question

(B) We are given the expression $f(x) = \frac{x^4}{(x+1)(x-2)}$.Using partial fractions,$\frac{1}{(x+1)(x-2)} = \frac{A}{x+1} + \frac{B}{x-2}$.Solving

Vedclass · Accepted Answer

$0$

Vedclass · Answer

(B) We are given the expression $f(x) = \frac{x^4}{(x+1)(x-2)}$.Using partial fractions,$\frac{1}{(x+1)(x-2)} = \frac{A}{x+1} + \frac{B}{x-2}$.Solving for constants,$1 = A(x-2) + B(x+1)$.For $x = -1$,$1 = A(-3) \implies A = -\frac{1}{3}$.For $x = 2$,$1 = B(3) \implies B = \frac{1}{3}$.Thus,$f(x) = x^4 \left( \frac{1/3}{x-2} - \frac{1/3}{x+1} \right) = \frac{x^4}{3} \left( \frac{1}{x-2} - \frac{1}{x+1} \right)$.Expanding as power series for $|x| $f(x) = \frac{x^4}{3} \left[ -\frac{1}{2}(1 - \frac{x}{2})^{-1} - (1+x)^{-1} \right]$$f(x) = \frac{x^4}{3} \left[ -\frac{1}{2}(1 + \frac{x}{2} + \frac{x^2}{4} + \dots) - (1 - x + x^2 - \dots) \right]$.The lowest power of $x$ in this expansion is $x^4$. Therefore,the coefficients of $x^0$,$x^1$,and $x^2$ are all $0$.

For $|x| < 1$,the coefficient of $x^2$ in the power series expansion of $\frac{x^4}{(x+1)(x-2)}$ is

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