What is the coefficient of x^4 in the expansi | vedclass

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(B) First,we express the given fraction using partial fractions: $\frac{3x}{(x - 2)(x + 1)} = \frac{A}{x - 2} + \frac{B}{x + 1}$ $3x = A(x + 1) + B(x

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$\frac{15}{16}$

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(B) First,we express the given fraction using partial fractions: $\frac{3x}{(x - 2)(x + 1)} = \frac{A}{x - 2} + \frac{B}{x + 1}$ $3x = A(x + 1) + B(x - 2)$ For $x = 2$,$6 = 3A \implies A = 2$. For $x = -1$,$-3 = -3B \implies B = 1$. So,$\frac{3x}{(x - 2)(x + 1)} = \frac{2}{x - 2} + \frac{1}{x + 1} = \frac{2}{-2(1 - x/2)} + \frac{1}{1 + x} = -(1 - x/2)^{-1} + (1 + x)^{-1}$. Using the binomial expansion $(1 - z)^{-1} = 1 + z + z^2 + z^3 + z^4 + \dots$ and $(1 + z)^{-1} = 1 - z + z^2 - z^3 + z^4 - \dots$: The expression becomes $-(1 + \frac{x}{2} + \frac{x^2}{4} + \frac{x^3}{8} + \frac{x^4}{16} + \dots) + (1 - x + x^2 - x^3 + x^4 - \dots)$. The coefficient of $x^4$ is $-\frac{1}{16} + 1 = \frac{15}{16}$.

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

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