Question

For $|x|<1$, the constant term in the expansion of $\frac{1}{(x-1{)}^2(x-2)}$ is

• A. 2
• B. 1
• C. 0
• D. $-\frac{1}{2}$

Answer 1

Answer: (D)

$\frac{1}{(x-1{)}^2(x-2)}=\frac{1}{-2(1-x{)}^2\left(1-\frac{x}{2}\right)}$
$=-\frac{1}{2}\left[(1-x{)}^{-2}{\left(1-\frac{x}{2}\right)}^{-1}\right]$
$=-\frac{1}{2}\left[(1+2x+\dots )\left(1+\frac{x}{2}+\dots \right)\right]$
$\therefore$ Coefficient of constant term is $-\frac{1}{2}$