If $|x| < 1$, then the coefficient of $x^5$ in the expansion of $\frac{3 x}{(x-2)(x+1)}$ is
- $\frac{33}{32}$
- $\frac{-33}{32}$
- $\frac{31}{32}$
- $\frac{-31}{32}$
Solution
$\begin{aligned} & 3 x=A(x+1)+B(x-2) \\ & \text { At } x=2, \\ & \therefore \quad A=\frac{3(2)}{2+1} \Rightarrow A=\frac{3(2)}{3}=A=2 \\ & \text { At } x=-1, B=\frac{3(-1)}{-1-2} \\ & B=\frac{-3}{-3} \Rightarrow B=1 \end{aligned}$ Substitute the values of $A$ and $B$ in Eq. (i), we get $\begin{aligned} & \frac{3 x}{(x-2)(x+1)}=\frac{2}{x-2}+\frac{1}{x+1} \\ & =\frac{2}{-2\left(1-\frac{x}{2}\right)}+\frac{1}{(x+1)}=-\left(1-\frac{x}{2}\right)^{-1}+(1+x)^{-1} \\ & =-\left[1+\frac{x}{2}+\left(\frac{x}{2}\right)^2+\ldots+\left(\frac{x}{2}\right)^5+\ldots .\right] \end{aligned}$ $ +\left(1-x+x^2-x^3+x^4-x^5+\ldots .\right) $ $\therefore \quad$ Coefficient of $x^5=-\left(\frac{1}{2}\right)^5-1=\frac{-1}{32}-1=-\frac{33}{32}$
Asked in: AP EAMCET 2015