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. $-\frac{1}{2}$

Hint:

To find the constant term, set $x=0$ in the expression.

Answer 1

The Correct Option is D

Step 1: Rearranging for Binomial Form
$\frac{1}{(1-x)^{2}(-2)(1-\frac{x}{2})} = -\frac{1}{2}(1-x)^{-2}(1-\frac{x}{2})^{-1}$.
Step 2: Expanding
$(1-x)^{-2} = 1 + 2x + ...$
$(1-\frac{x}{2})^{-1} = 1 + \frac{x}{2} + ...$
Step 3: Finding Constant Term
Constant term = $-\frac{1}{2} \times 1 \times 1 = -\frac{1}{2}$.
Final Answer: (d)