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  4. The coefficient of x6 in the power series expansion of (x4-12 x2+7/(x2+1)3) is

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

TS EAMCET 2019

Solution:

We have,
$\frac{x^{4}-12 x^{2}+7}{\left(x^{2}+1\right)^{3}}=\left(x^{4}-12 x^{2}+7\right)\left(x^{2}+1\right)^{-3}$
$=\left(x^{4}-12 x^{2}+7\right)\left(1-3 x^{2}+\frac{12 x^{4}}{2 !}-\frac{60}{3 !} x^{6} \ldots\right)$
$=\left(x^{4}-12 x^{2}+7\right)\left(1-3 x^{2}+6 x^{4}-10 x^{6} \ldots .\right)$
Coefficient of $x^{6}$ is $(-3-72-70)=-145$