The coefficient of x^4 in the expansion of th | vedclass

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(B) Using partial fractions,$\frac{3x}{(x - 2)(x + 1)} = \frac{A}{x - 2} + \frac{B}{x + 1}$.Solving for $A$ and $B$,we get $A = 2$ and $B = 1$,so $\fr

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$15/16$

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(B) Using partial fractions,$\frac{3x}{(x - 2)(x + 1)} = \frac{A}{x - 2} + \frac{B}{x + 1}$.Solving for $A$ and $B$,we get $A = 2$ and $B = 1$,so $\frac{3x}{(x - 2)(x + 1)} = \frac{2}{x - 2} + \frac{1}{x + 1} = -\frac{1}{1 - x/2} + \frac{1}{1 + x}$.Expanding both terms using the binomial series $(1 - z)^{-1} = 1 + z + z^2 + z^3 + z^4 + \dots$ and $(1 + z)^{-1} = 1 - z + z^2 - z^3 + z^4 - \dots$:$-\left(1 + \frac{x}{2} + \frac{x^2}{4} + \frac{x^3}{8} + \frac{x^4}{16} + \dots\right) + \left(1 - x + x^2 - x^3 + x^4 - \dots\right)$.The coefficient of $x^4$ is $-\frac{1}{16} + 1 = \frac{15}{16}$.

The coefficient of $x^4$ in the expansion of the expression $\frac{3x}{(x - 2)(x + 1)}$ is

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