Question

If $|x|<1$, then the coefficient of $x^5$ 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

The correct option is B $-\frac{33}{32}$

We can write $\frac{3x}{(x-2)(x+1)}=\frac{A}{x-2}+\frac{B}{x+1}$
Thus $A=\frac{3\left(2\right)}{2+1}=2;B=\frac{3(-1)}{-1-2}=-\frac{-3}{-3}=1$
$\frac{3x}{(x-2)(x+1)}=\frac{2}{x-2}+\frac{1}{x+1}$
$=\frac{2}{-2(1-x/2)}+\frac{1}{x+1}=-(1-x/2{)}^{-1}+(1+x{)}^{-1}$
$=-(1+\frac{x}{2}+{\left(\frac{x}{2}\right)}^2+....+{\left(\frac{x}{2}\right)}^5+....)+(1-x+x^2-x^3+x^4-x^5+111)$
Coefficient of $x^5=\frac{-1}{32}-1=\frac{-33}{32}$