Sequences & Series Question - None

Question

If the expansion in powers of x of the function $\dfrac{1}{(1-ax)(1-bx)}$ is $a_0 + a_1 x + a_2 x^2 + a_3 x^3 + \cdots$, then $a^n$ is

Accepted Answer

Step 1: Decompose using partial fractions:

$$\frac{1}{(1-ax)(1-bx)} = \frac{A}{1-ax} + \frac{B}{1-bx}$$

Multiplying both sides by (1-ax)(1-bx):

$$1 = A(1-bx) + B(1-ax)$$

Setting x = 1/a: $1 = A(1-b/a)$, so $A = \frac{a}{a-b}$

Setting x = 1/b: $1 = B(1-a/b)$, so $B = \frac{b}{b-a} = \frac{-b}{a-b}$

Step 2: Expand each fraction as geometric series:

$$\frac{1}{(1-ax)(1-bx)} = \frac{a}{a-b}\sum_{n=0}^{\infty}(ax)^n + \frac{-b}{a-b}\sum_{n=0}^{\infty}(bx)^n$$

Step 3: Extract coefficient of x^n:

$$a_n = \frac{a}{a-b} \cdot a^n + \frac{-b}{a-b} \cdot b^n = \frac{a^{n+1} - b^{n+1}}{a-b}$$

∴ Answer: $a_n = \dfrac{a^{n+1} - b^{n+1}}{a-b}$ (Option D)

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