Sequences & Series
Recurrence relations with roots of quadratic
Grade 11
Question:
<p>Let \(\alpha, \beta\) are the roots of the quadratic equation \(2x^2 - 5x + 1 = 0\). If \(S_n = (\alpha)^{2n} + (\beta)^{2n}\), then find the value of \(\dfrac{4S_{2021} + S_{2019}}{S_{2020}}\).</p>
Step-by-Step Solution
Key Concept: Use Vieta's formulas to find α + β and αβ, then derive a recurrence relation for Sₙ by recognizing that α² and β² are roots of a transformed quadratic equation. This allows Sₙ to satisfy a linear recurrence without computing individual terms.
<p><strong>Step 1: Apply Vieta's formulas</strong></p><p>From 2x² - 5x + 1 = 0: α + β = 5/2 and αβ = 1/2</p><p><strong>Step 2: Find the recurrence for Sₙ</strong></p><p>Since Sₙ = (α)^(2n) + (β)^(2n) = (α²)^n + (β²)^n, let γ = α² and δ = β²</p><p>Then: γ + δ = (α + β)² - 2αβ = (5/2)² - 2(1/2) = 25/4 - 1 = 21/4</p><p>And: γδ = (αβ)² = (1/2)² = 1/4</p><p><strong>Step 3: Write recurrence relation</strong></p><p>Sₙ satisfies: Sₙ = (γ + δ)Sₙ₋₁ - (γδ)Sₙ₋₂</p><p>Therefore: Sₙ = (21/4)Sₙ₋₁ - (1/4)Sₙ₋₂</p><p>Multiplying by 4: 4Sₙ = 21Sₙ₋₁ - Sₙ₋₂</p><p><strong>Step 4: Apply recurrence at n = 2020</strong></p><p>4S₂₀₂₀ = 21S₂₀₁₉ - S₂₀₁₈</p><p>Also: 4S₂₀₂₁ = 21S₂₀₂₀ - S₂₀₁₉</p><p><strong>Step 5: Calculate the expression</strong></p><p>4S₂₀₂₁ + S₂₀₁₉ = 21S₂₀₂₀ - S₂₀₁₉ + S₂₀₁₉ = 21S₂₀₂₀</p><p>Therefore: (4S₂₀₂₁ + S₂₀₁₉)/S₂₀₂₀ = 21/1</p><p>∴ <strong>Answer: 21</strong></p>
Correct Answer: 21