The coefficient of x^6 in the power series ex | vedclass

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(C) We have,$\frac{x^4-12x^2+7}{(x^2+1)^3} = (x^4-12x^2+7)(1+x^2)^{-3}$.Using the binomial expansion $(1+u)^{-n} = 1 - nu + \frac{n(n+1)}{2!}u^2 - \fr

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-$145$

Vedclass · Answer

(C) We have,$\frac{x^4-12x^2+7}{(x^2+1)^3} = (x^4-12x^2+7)(1+x^2)^{-3}$.Using the binomial expansion $(1+u)^{-n} = 1 - nu + \frac{n(n+1)}{2!}u^2 - \frac{n(n+1)(n+2)}{3!}u^3 + \dots$,where $u = x^2$ and $n = 3$:$(1+x^2)^{-3} = 1 - 3x^2 + 6x^4 - 10x^6 + \dots = 1 - 3x^2 + 6x^4 - 10x^6 + \dots$Now,multiply by $(x^4 - 12x^2 + 7)$:$(x^4 - 12x^2 + 7)(1 - 3x^2 + 6x^4 - 10x^6 + \dots)$The terms containing $x^6$ are:$x^4 	imes (6x^4)$ is not $x^6$,but $x^4 	imes (-3x^2) = -3x^6$$-12x^2 	imes (6x^4) = -72x^6$$7 	imes (-10x^6) = -70x^6$Summing these coefficients: $-3 - 72 - 70 = -145$.

The coefficient of $x^6$ in the power series expansion of $\frac{x^4-12x^2+7}{(x^2+1)^3}$ is

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