Binomial Theorem
Expansion of rational functions
Grade 11
Question:
<p>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</p>
<p>(1) \(\dfrac{b^n - a^n}{b - a}\)</p>
<p>(2) \(\dfrac{a^n - b^n}{b - a}\)</p>
<p>(3) \(\dfrac{a^{n+1} - b^{n+1}}{b - a}\)</p>
<p>(4) \(\dfrac{b^{n+1} - a^{n+1}}{b - a}\)</p>
Step-by-Step Solution
Key Concept: Use partial fraction decomposition to split the denominator into two geometric series, then combine the coefficients of x^n from each series using the convolution principle.
<p><strong>Step 1:</strong> Apply partial fraction decomposition:</p><p>$$\frac{1}{(1-ax)(1-bx)} = \frac{A}{1-ax} + \frac{B}{1-bx}$$</p><p>Multiplying both sides by $(1-ax)(1-bx)$:</p><p>$$1 = A(1-bx) + B(1-ax)$$</p><p>Setting $x = \frac{1}{a}$: $1 = A(1-\frac{b}{a})$, so $A = \frac{a}{a-b}$</p><p>Setting $x = \frac{1}{b}$: $1 = B(1-\frac{a}{b})$, so $B = \frac{b}{b-a}$</p><p><strong>Step 2:</strong> Expand each term as a geometric series:</p><p>$$\frac{1}{(1-ax)(1-bx)} = \frac{a}{a-b}\sum_{k=0}^{\infty}(ax)^k + \frac{b}{b-a}\sum_{k=0}^{\infty}(bx)^k$$</p><p><strong>Step 3:</strong> Extract the coefficient of $x^n$:</p><p>$$a_n = \frac{a}{a-b} \cdot a^n + \frac{b}{b-a} \cdot b^n = \frac{a^{n+1}}{a-b} - \frac{b^{n+1}}{a-b}$$</p><p>$$a_n = \frac{a^{n+1} - b^{n+1}}{a-b}$$</p><p>∴ Answer: D</p>
Correct Answer: D