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

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Question

(D) Let $\frac{x^2 + 1}{(x^2 + 4)(x - 2)} = \frac{Ax + B}{x^2 + 4} + \frac{C}{x - 2}$.By partial fraction decomposition,$x^2 + 1 = (Ax + B)(x - 2) + C

Vedclass · Accepted Answer

$-1/256$

Vedclass · Answer

(D) Let $\frac{x^2 + 1}{(x^2 + 4)(x - 2)} = \frac{Ax + B}{x^2 + 4} + \frac{C}{x - 2}$.By partial fraction decomposition,$x^2 + 1 = (Ax + B)(x - 2) + C(x^2 + 4)$.Comparing coefficients,we get $A + C = 1$,$-2A + B = 0$,and $-2B + 4C = 1$.Solving these,we find $A = 3/8$,$B = 3/4$,and $C = 5/8$.Thus,the expression is $\frac{3x/8 + 3/4}{x^2 + 4} + \frac{5/8}{x - 2}$.Rewriting for expansion: $\frac{1}{4}(\frac{3x}{8} + \frac{3}{4})(1 + \frac{x^2}{4})^{-1} - \frac{5}{16}(1 - \frac{x}{2})^{-1}$.Using the binomial expansion $(1+y)^{-1} = 1 - y + y^2 - y^3 + \dots$ and $(1-y)^{-1} = 1 + y + y^2 + y^3 + \dots$:Term from first part: $\frac{1}{4}(\frac{3x}{8} + \frac{3}{4})(1 - \frac{x^2}{4} + \frac{x^4}{16} - \dots)$. The $x^5$ term comes from $\frac{3x}{8} 	imes (\frac{x^4}{16}) = \frac{3x^5}{128}$. Multiplying by $1/4$ gives $\frac{3x^5}{512}$.Term from second part: $-\frac{5}{16}(1 + \frac{x}{2} + \frac{x^2}{4} + \frac{x^3}{8} + \frac{x^4}{16} + \frac{x^5}{32} + \dots)$. The $x^5$ term is $-\frac{5}{16} 	imes \frac{x^5}{32} = -\frac{5x^5}{512}$.Summing the coefficients: $\frac{3}{512} - \frac{5}{512} = -\frac{2}{512} = -\frac{1}{256}$.

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

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