Sequences & Series
Mathematical Induction
Grade 11
Question:
<p>Let \(P(n) = \frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} + \cdots + \frac{1}{\sqrt{n}}\). Consider the following statements:<br><strong>Statement-1:</strong> \(P(n) = \sqrt{n(n+1)} < n+1\)<br><strong>Statement-2:</strong> \(P(k+1) > \sqrt{k+1}\) whenever \(P(k) > \sqrt{k}\)<br>Which of the following is correct?</p>
<p>Statement-1 is true, Statement-2 is true and Statement-2 is a correct explanation of Statement-1</p>
<p>Statement-1 is true, Statement-2 is true but Statement-2 is not a correct explanation of Statement-1</p>
<p>Statement-1 is true, Statement-2 is false</p>
<p>Statement-1 is false, Statement-2 is true</p>
Step-by-Step Solution
Key Concept: Recognize that P(n) is a sum of reciprocals of square roots (harmonic-like series), not a closed form. Use integral bounds to estimate P(n): ∫₁ⁿ dx/√x < P(n) < 1 + ∫₁ⁿ dx/√x, which gives 2√n - 1 < P(n) < 2√n, showing P(n) ≈ 2√n, not √(n(n+1)).
<p><strong>Step 1: Check Statement-1 directly</strong></p><p>For n=1: P(1) = 1/√1 = 1, but √(1·2) = √2 ≈ 1.414</p><p>Since 1 ≠ √2, Statement-1 is FALSE immediately.</p><p><strong>Step 2: Establish correct bound for P(n)</strong></p><p>Using integral test: ∫₁ⁿ dx/√x = 2√n - 2</p><p>This gives: 2√n - 2 < P(n) < 2√n (approximately P(n) ≈ 2√n - 1)</p><p><strong>Step 3: Analyze Statement-2</strong></p><p>Assume P(k) > √k. We need to verify if P(k+1) > √(k+1).</p><p>P(k+1) = P(k) + 1/√(k+1) > √k + 1/√(k+1)</p><p>We need: √k + 1/√(k+1) > √(k+1)</p><p>Rearranging: 1/√(k+1) > √(k+1) - √k = 1/(√(k+1) + √k)</p><p>This simplifies to: √(k+1) + √k > √(k+1)</p><p>Which is TRUE (always, since √k > 0)</p><p>Therefore Statement-2 is TRUE.</p><p><strong>Conclusion:</strong> Statement-1 is FALSE, Statement-2 is TRUE</p><p>∴ Answer: B (Only Statement-2 is correct)</p>
Correct Answer: B