Question

The coefficient of $x^4$ in the expansion of $\frac{(1-3x)^2}{(1-2x)}$ is

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

Answer: (D)

4

Answer 2

We write

$$\frac{(1-3x)^2}{(1-2x)} = \frac{1-6x+9x^2}{1-2x} = (1-6x+9x^2) \cdot \sum_{n=0}^{\infty} (2x)^n.$$

The $x^4$ term arises from:

• $1 \cdot (2^4 x^4):$ coefficient $2^4 = 16$,
• $-6x \cdot (2^3 x^3):$ coefficient $-6 \cdot 2^3 = -48$,
• $9x^2 \cdot (2^2 x^2):$ coefficient $9 \cdot 2^2 = 36$.

Adding these:

$$16 - 48 + 36 = 4.$$