Sequences & Series
Telescoping Series
Grade 11
Question:
<p><strong>54.</strong> Let \(\alpha_n\), \(\beta_n\) be the distinct roots of the equation \(x^2 + (n+1)x + n^2 = 0\). If \(\displaystyle\sum_{n=2}^{2021} \frac{1}{(\alpha_n + 1)(\beta_n + 1)}\) 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 find αₙ + βₙ and αₙβₙ, then express 1/[(αₙ+1)(βₙ+1)] in terms of n. Recognize the resulting telescoping series to find the sum.
<p><strong>Step 1: Apply Vieta's formulas</strong></p><p>For x² + (n+1)x + n² = 0:</p><p>• αₙ + βₙ = -(n+1)</p><p>• αₙβₙ = n²</p><p><strong>Step 2: Expand the denominator</strong></p><p>(αₙ+1)(βₙ+1) = αₙβₙ + αₙ + βₙ + 1</p><p>= n² + (-(n+1)) + 1</p><p>= n² - n - 1 + 1 = n² - n = n(n-1)</p><p><strong>Step 3: Use partial fractions</strong></p><p>$$\frac{1}{n(n-1)} = \frac{1}{n-1} - \frac{1}{n}$$</p><p><strong>Step 4: Sum the telescoping series</strong></p><p>$$\sum_{n=2}^{2021} \frac{1}{n(n-1)} = \sum_{n=2}^{2021}\left(\frac{1}{n-1} - \frac{1}{n}\right)$$</p><p>$$= \left(\frac{1}{1} - \frac{1}{2}\right) + \left(\frac{1}{2} - \frac{1}{3}\right) + \cdots + \left(\frac{1}{2020} - \frac{1}{2021}\right)$$</p><p>$$= 1 - \frac{1}{2021} = \frac{2020}{2021}$$</p><p><strong>Step 5: Find (b - a)</strong></p><p>$$\frac{a}{b} = \frac{2020}{2021}$$ where a = 2020, b = 2021</p><p>∴ b - a = 2021 - 2020 = <strong>1</strong></p>
Correct Answer: A