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

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Aicharya · Accepted Answer

Using binomial expansions, $ \begin{aligned} & \frac{\left(x^2-1ight)}{\left(x^2+1ight)\left(x^2+2ight)}=\frac{1}{2}\left(x^2-1ight)\left(1+x^2ight)^{-1}\left(1+\frac{x^2}{2}ight)^{-1} \ = & \frac{1}{2}\left[\left(x^2-1ight)\left(1-x^2+x^4 \ldotsight)\left(1-\frac{x^2}{2}+\frac{x^4}{4} \ldotsight)ight] \ = & \frac{1}{2}\left[\left(x^2-1ight)\left(1-\frac{x^2}{2}+\frac{x^4}{4}-x^2+\frac{x^4}{2}+x^4 \ldotsight)ight] \ = & \frac{1}{2}\left[\left(x^2-1ight)\left(1-\frac{3}{2} x^2+\frac{7}{4} x^4 \ldotsight)ight] \ = & \frac{1}{2}\left[-1+\frac{3}{2} x^2-\frac{7}{4} x^4+x^2-\frac{3}{2} x^4 \ldotsight] \end{aligned} $ taking coefficient of $x^4$, $ =\frac{1}{2}\left(-\frac{7}{4}-\frac{3}{2}ight)=\frac{1}{2}\left(\frac{-13}{4}ight)=\frac{-13}{8} 	ext {. } $

The coefficient of $x^4$ in the power series expansion of…

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