Sequences & Series
Sum of Series / Binomial Expansion
Grade 11
Question:
<p>Given that \(f(x) = \dfrac{1}{(1-ax)(1-bx)} = (1-ax)^{-1}(1-bx)^{-1}\). The coefficient of \(x^n\) in the expansion of \(f(x)\) is:</p>
<p>\(\dfrac{a^n + b^n}{a - b}\)</p>
<p>\(\dfrac{a^n - b^n}{a + b}\)</p>
<p>\(\dfrac{a^n b^n}{a - b}\)</p>
<p>\(\dfrac{a^{n+1} - b^{n+1}}{a - b}\)</p>
Step-by-Step Solution
Key Concept: Use the binomial series expansion (1-u)^(-1) = 1 + u + u² + ... for each factor, then multiply the two series and collect the coefficient of x^n using the convolution principle.
<p><strong>Step 1:</strong> Expand each factor using the binomial series (1-u)^(-1) = Σ u^k for |u| < 1:</p><p>(1-ax)^(-1) = 1 + ax + a²x² + a³x³ + ... + a^n x^n + ...</p><p>(1-bx)^(-1) = 1 + bx + b²x² + b³x³ + ... + b^n x^n + ...</p><p><strong>Step 2:</strong> Multiply the two series and find the coefficient of x^n by collecting all terms that produce x^n:</p><p>Coefficient of x^n = (coefficient of x^0 in first) × (coefficient of x^n in second) + (coefficient of x^1 in first) × (coefficient of x^(n-1) in second) + ... + (coefficient of x^n in first) × (coefficient of x^0 in second)</p><p>= 1·b^n + a·b^(n-1) + a²·b^(n-2) + ... + a^(n-1)·b + a^n·1</p><p><strong>Step 3:</strong> This is a geometric series sum:</p><p>Coefficient of x^n = b^n + ab^(n-1) + a²b^(n-2) + ... + a^(n-1)b + a^n</p><p>= Σ(k=0 to n) a^k b^(n-k)</p><p><strong>Step 4:</strong> If a ≠ b, use the geometric series formula:</p><p>= (a^(n+1) - b^(n+1))/(a - b)</p><p>∴ Answer: D</p>
Correct Answer: D