The coefficient of x^5 in the expansion of fr | vedclass

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(D) Let $f(x) = \frac{x^2 + 1}{(x^2 + 4)(x - 2)}$.Using partial fractions,we write $\frac{x^2 + 1}{(x^2 + 4)(x - 2)} = \frac{Ax + B}{x^2 + 4} + \frac{

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$-\frac{1}{256}$

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(D) Let $f(x) = \frac{x^2 + 1}{(x^2 + 4)(x - 2)}$.Using partial fractions,we write $\frac{x^2 + 1}{(x^2 + 4)(x - 2)} = \frac{Ax + B}{x^2 + 4} + \frac{C}{x - 2}$.$x^2 + 1 = (Ax + B)(x - 2) + C(x^2 + 4)$.For $x = 2$: $2^2 + 1 = C(2^2 + 4) \implies 5 = 8C \implies C = \frac{5}{8}$.Comparing coefficients of $x^2$: $1 = A + C \implies A = 1 - \frac{5}{8} = \frac{3}{8}$.Comparing constants: $1 = -2B + 4C \implies 1 = -2B + 4(\frac{5}{8}) \implies 1 = -2B + \frac{5}{2} \implies 2B = \frac{3}{2} \implies B = \frac{3}{4}$.So,$f(x) = \frac{3x/8 + 3/4}{x^2 + 4} + \frac{5/8}{x - 2} = \frac{3}{8} \cdot \frac{x + 2}{x^2 + 4} - \frac{5}{16} \cdot \frac{1}{1 - x/2}$.Using the expansion $\frac{1}{1 - r} = \sum_{n=0}^{\infty} r^n$,we have $-\frac{5}{16} \sum_{n=0}^{\infty} (\frac{x}{2})^n = -\frac{5}{16} \sum_{n=0}^{\infty} \frac{x^n}{2^n}$.The coefficient of $x^5$ from this part is $-\frac{5}{16} \cdot \frac{1}{2^5} = -\frac{5}{512}$.For $\frac{3}{8} \cdot \frac{x + 2}{4(1 + x^2/4)} = \frac{3}{32} (x + 2) \sum_{n=0}^{\infty} (-1)^n (\frac{x^2}{4})^n = \frac{3}{32} (x + 2) \sum_{n=0}^{\infty} (-1)^n \frac{x^{2n}}{4^n}$.The $x^5$ term comes from $2 \cdot \frac{3}{32} \cdot (-1)^2 \frac{x^4}{4^2}$ (no $x^5$ here) and $x \cdot \frac{3}{32} \cdot (-1)^2 \frac{x^4}{4^2} = \frac{3}{32} \cdot \frac{x^5}{16} = \frac{3}{512} x^5$.Total coefficient = $\frac{3}{512} - \frac{5}{512} = -\frac{2}{512} = -\frac{1}{256}$.

The coefficient of $x^5$ in the expansion of $\frac{x^2 + 1}{(x^2 + 4)(x - 2)}$ is:

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