Question

If $\left|\begin{array}{c}x\end{array}\right|<1$ then the coefficient of $x^3$ in the expansion of $\frac{3x}{(x-2)(x+1)}$ is

• A. $\frac{33}{32}$
• B. $-\frac{33}{32}$
• C. $\frac{31}{32}$
• D. $-\frac{31}{32}$

Answer 1

Answer: (B)

$-\frac{33}{32}$

$\frac{3x}{(x-2)(x+1)}=\frac{A}{x-2}+\frac{B}{x+1}$
$A=\frac{3\left(2\right)}{2+1}=2$ ; $B=\frac{3(-1)}{-1-2}=\frac{-3}{-3}=1$
$\frac{3x}{(x-2)(x+1)}=\frac{2}{x-2}+\frac{1}{x+1}$
= $\frac{2}{-2(1-\frac{x}{2})}+\frac{1}{x+1}=-(1-\frac{x}{2}{)}^{-1}+(1+x{)}^{-1}$
= -$\left(\begin{array}{c}1+\frac{x}{2}+(\frac{x}{2}{)}^3+....+(\frac{x}{2}{)}^3+....\end{array}\right)+(1-x+x^2-x^3+x^4-x^5+...)$

Coefficient of $x^5=\frac{-1}{32}-1=\frac{-33}{32}$