<p>The least positive integer \(n\) such that \(1 - \dfrac{2}{3} - \dfrac{2}{3^2} - \cdots - \dfrac{2}{3^{n-1}} < \dfrac{1}{100}\), is</p>
Step-by-Step Solution
Key Concept: Recognize this as a geometric series sum problem; rewrite the inequality as finding when a geometric series with first term 2/3 and common ratio 1/3 exceeds 1/3, then apply the geometric series formula S = a(1-r^n)/(1-r).
<p><strong>Step 1:</strong> Rewrite the inequality. We need: <br/>1 - (2/3 + 2/3² + ... + 2/3^(n-1)) < 1/3</p><p><strong>Step 2:</strong> This simplifies to: 2/3 + 2/3² + ... + 2/3^(n-1) > 2/3</p><p><strong>Step 3:</strong> Recognize the left side as a geometric series with first term a = 2/3, common ratio r = 1/3, and n terms.</p><p><strong>Step 4:</strong> Apply formula: S_n = (2/3)(1 - (1/3)^n)/(1 - 1/3) = (2/3) · (1 - (1/3)^n) · (3/2) = 1 - (1/3)^n</p><p><strong>Step 5:</strong> Solve: 1 - (1/3)^n > 2/3, which gives (1/3)^n < 1/3</p><p><strong>Step 6:</strong> Therefore (1/3)^n < (1/3)^1, so n > 1. The least positive integer is n = <strong>2</strong>.</p><p><strong>Verification:</strong> For n=2: 1 - 2/3 = 1/3 ✗ (not strictly less). For n=3: 1 - 2/3 - 2/9 = 1/9 < 1/3 ✓</p><p>∴ Answer: n = 3 (Option D)</p>
Correct Answer: D