<p>If \(P(n) = \dfrac{1}{n+1} + \dfrac{1}{n+2} + \cdots + \dfrac{1}{2n} > \dfrac{13}{24}\), then \(P(n)\) is true for \(n \geq\) __________.</p>
Step-by-Step Solution
Key Concept: Recognize that P(n) is a telescoping-adjacent sum that increases with n. Use the comparison P(n) > 1/2 for all n ≥ 1, then find when P(n) crosses 13/24 by direct computation or by comparing consecutive differences.
<p><strong>Step 1:</strong> Establish the pattern by computing P(n) for small values.</p><p><strong>Step 1a:</strong> For n = 1: P(1) = 1/2 = 12/24 < 13/24 ✗</p><p><strong>Step 1b:</strong> For n = 2: P(2) = 1/3 + 1/4 = 4/12 + 3/12 = 7/12 = 14/24 > 13/24 ✓</p><p><strong>Step 2:</strong> Verify P(n) is increasing. Note that P(n+1) - P(n) adds terms 1/(2n+1) + 1/(2n+2) and removes 1/(n+1), giving P(n+1) - P(n) = 1/(2n+1) + 1/(2n+2) - 1/(n+1) > 0 (since 1/(2n+1) + 1/(2n+2) > 1/(n+1)).</p><p><strong>Step 3:</strong> Since P(n) is strictly increasing and P(2) > 13/24 while P(1) < 13/24, the inequality holds for all n ≥ 2.</p><p>∴ Answer: <strong>2</strong></p>
Correct Answer: 2