Sequences & Series
Telescoping Series
Grade 11
Question:
<p>Let \( \alpha_n, \beta_n \) be the distinct roots of the equation \( x^2 + (n+1)x + n^2 = 0 \). If \[ \sum_{n=2}^{2021} \frac{1}{(\alpha_n + 1)(\beta_n + 1)} \] can be expressed in the form \( \dfrac{a}{b} \), where \( a \) and \( b \) are positive integers, the value of \( (b - a) \) is:</p>
<p>(a) 1</p>
<p>(b) 3</p>
<p>(c) 6</p>
<p>(d) 9</p>
Step-by-Step Solution
Key Concept: Use Vieta's formulas to express the product (αₙ + 1)(βₙ + 1) in terms of n, then decompose the resulting fraction using partial fractions to create a telescoping series.
<p><strong>Step 1: Apply Vieta's Formulas</strong></p><p>For x² + (n+1)x + n² = 0 with roots αₙ, βₙ:</p><p>• αₙ + βₙ = -(n+1)</p><p>• αₙβₙ = n²</p><p><strong>Step 2: Find (αₙ + 1)(βₙ + 1)</strong></p><p>(αₙ + 1)(βₙ + 1) = αₙβₙ + αₙ + βₙ + 1</p><p>= n² - (n+1) + 1 = n² - n = n(n-1)</p><p><strong>Step 3: Decompose using Partial Fractions</strong></p><p>1/[n(n-1)] = 1/(n-1) - 1/n</p><p><strong>Step 4: Sum the Telescoping Series</strong></p><p>∑₍ₙ₌₂₎²⁰²¹ [1/(n-1) - 1/n]</p><p>= [1/1 - 1/2] + [1/2 - 1/3] + ... + [1/2020 - 1/2021]</p><p>= 1 - 1/2021 = 2020/2021</p><p><strong>Step 5: Find b - a</strong></p><p>Here a = 2020, b = 2021 (in lowest terms, gcd(2020,2021) = 1)</p><p>∴ b - a = 2021 - 2020 = <strong>1</strong></p>
Correct Answer: A