<p>If \(H_n = 1 + \dfrac{1}{2} + \cdots + \dfrac{1}{n}\), then the value of \(S_n = 1 + \dfrac{3}{2} + \dfrac{5}{3} + \cdots + \dfrac{99}{50}\) is</p>
Step-by-Step Solution
Key Concept: Decompose each term as (2k-1)/k = 2 - 1/k, then recognize that S_n becomes a sum involving harmonic numbers. The series telescopes into 2n - H_n where H_n is the nth harmonic number.
<p><strong>Step 1:</strong> Identify the general term. The numerators are 1, 3, 5, ..., 99 (odd numbers) and denominators are 1, 2, 3, ..., 50. So the kth term is (2k-1)/k where k goes from 1 to 50.</p><p><strong>Step 2:</strong> Decompose each term: (2k-1)/k = 2k/k - 1/k = 2 - 1/k</p><p><strong>Step 3:</strong> Write S_n as a sum:</p><p>S₅₀ = Σ(k=1 to 50) [(2k-1)/k] = Σ(k=1 to 50) [2 - 1/k]</p><p><strong>Step 4:</strong> Separate the sum:</p><p>S₅₀ = 2·Σ(k=1 to 50) 1 - Σ(k=1 to 50) (1/k)</p><p>S₅₀ = 2(50) - H₅₀</p><p>S₅₀ = 100 - H₅₀</p><p>where H₅₀ = 1 + 1/2 + 1/3 + ... + 1/50</p><p><strong>Step 5:</strong> If the answer choices are given in terms of H₅₀, then <strong>S₅₀ = 100 - H₅₀</strong>. If a numerical approximation is needed, H₅₀ ≈ 4.499, giving S₅₀ ≈ 95.5</p><p>∴ Answer: D</p>
Correct Answer: D