$|x| \lt 1$, the coefficients of $x^2$ in the power series expansion of $\frac{x^4}{(x+1)(x-2)}$ is

Question

Aicharya · Accepted Answer

$\begin{aligned} & \frac{x^4}{(x+1)(x-2)}=\frac{x^4}{3}\left[\frac{1}{(x-2)}-\frac{1}{(x+1)}\right] \\ & =\frac{x^4}{3}\left[\frac{-1}{2}(1-x)^{-1}-(1+x)^{-1}\right]\end{aligned}$
$\begin{aligned} & =\frac{x^4}{3}\left[\frac{-1}{2}-\frac{x}{2}-\frac{x^2}{2}-\frac{x^3}{2}+\ldots-1+x-x^2+x^3-\ldots\right] \\ & =-\frac{x^4}{6}-\frac{x^5}{6}-\frac{x^6}{6}-\frac{x^7}{6}+\ldots-\frac{x^4}{3}+\frac{x^5}{3}+\ldots \end{aligned}$
So, coefficient of $x^2=0$.

$|x| \lt 1$, the coefficients of $x^2$ in the power series expansion of $\frac{x^4}{(x+1)(x-2)}$ is

Solution