The coefficient of x^4 in the power series ex | vedclass

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(C) We are given the expression $f(x) = \frac{x^2-1}{(x^2+1)(x^2+2)}$.Using partial fractions or binomial expansion,we can write:$f(x) = \frac{1}{2} (

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$-\frac{13}{8}$

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(C) We are given the expression $f(x) = \frac{x^2-1}{(x^2+1)(x^2+2)}$.Using partial fractions or binomial expansion,we can write:$f(x) = \frac{1}{2} (x^2-1) (1+x^2)^{-1} (1+\frac{x^2}{2})^{-1}$Expanding the terms using $(1+u)^{-1} = 1-u+u^2 - \dots$:$(1+x^2)^{-1} = 1-x^2+x^4 - \dots$$(1+\frac{x^2}{2})^{-1} = 1-\frac{x^2}{2}+\frac{x^4}{4} - \dots$Multiplying these two series:$(1-x^2+x^4)(1-\frac{x^2}{2}+\frac{x^4}{4}) = 1 - \frac{x^2}{2} + \frac{x^4}{4} - x^2 + \frac{x^4}{2} + x^4 = 1 - \frac{3}{2}x^2 + \frac{7}{4}x^4$Now multiply by $\frac{1}{2}(x^2-1)$:$\frac{1}{2}(x^2-1)(1 - \frac{3}{2}x^2 + \frac{7}{4}x^4) = \frac{1}{2} [x^2 - \frac{3}{2}x^4 - 1 + \frac{3}{2}x^2 - \frac{7}{4}x^4]$$= \frac{1}{2} [-1 + \frac{5}{2}x^2 - (\frac{3}{2} + \frac{7}{4})x^4]$$= \frac{1}{2} [-1 + \frac{5}{2}x^2 - \frac{13}{4}x^4]$The coefficient of $x^4$ is $\frac{1}{2} 	imes (-\frac{13}{4}) = -\frac{13}{8}$.

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

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