Question

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

• A. $\frac{1}{2}$ (2ⁿ⁺² – 3ⁿ⁺³ + 1)
• B. $\frac{1}{2}$ (3ⁿ⁺² – 2ⁿ⁺³ + 1)
• C. $\frac{1}{2}$ (2ⁿ⁺³ – 3ⁿ⁺² + 1)
• D. None of these

Answer 1

Answer: (B)

$\frac{1}{2}$ (3ⁿ⁺² – 2ⁿ⁺³ + 1)

Answer 2

We have, $\frac{1}{(1 - x)(1 - 2x)(1 - 3x)}$

= $\frac{1}{2(1 - x)} - \frac{4}{1 - 2x} + \frac{9}{2(1 - 3x)}$

[By resolving into partial fractions]

= $\frac{1}{2}$ (1 – x)^–1 –4 (1 – 2x)^–1 + $\frac{9}{2}$ (1 – 3x)^–1

= $\frac{1}{2}$ [1 + x + x² + …. + xⁿ +….] – 4 [1 + 2x + (2x)² + ….. + (2x)ⁿ + ….] + $\frac{9}{2}$ [1 + (3x) + (3x)² + ….. + (3x)ⁿ + …..]

\ Coefficient of xⁿ = $\frac{1}{2}$ [1 – 8.2ⁿ + 9.3ⁿ]

= $\frac{1}{2}$ [1 – 2ⁿ⁺³ + 3ⁿ⁺²].