<p>If \(L = \lim_{n \to \infty}\left(\dfrac{1}{\sqrt{n}\sqrt{n+1}} + \dfrac{1}{\sqrt{n}\sqrt{n+2}} + \cdots + \dfrac{1}{\sqrt{n}\sqrt{n+n}}\right) = a\sqrt{b} - c\) and \(a^4 + b^3 + c^2 + d = 29\), find \(d\).</p>
Step-by-Step Solution
Key Concept: Convert the Riemann sum to a definite integral by factoring out 1/n from the denominator and recognizing the sum as ∫₀¹ 1/√(1+x) dx after substitution r/n → x.
<p><strong>Step 1: Rewrite as Riemann sum</strong></p><p>L = lim(n→∞) Σ(r=1 to n) 1/(√n·√(n+r)) = lim(n→∞) (1/n) Σ(r=1 to n) 1/√((1 + r/n))</p><p><strong>Step 2: Recognize Riemann sum</strong></p><p>This is a Riemann sum for ∫₀¹ 1/√(1+x) dx with partition width Δx = 1/n</p><p><strong>Step 3: Evaluate the integral</strong></p><p>∫₀¹ 1/√(1+x) dx = [2√(1+x)]₀¹ = 2√2 - 2</p><p><strong>Step 4: Match with form a√b - c</strong></p><p>L = 2√2 - 2, so a = 2, b = 2, c = 2</p><p><strong>Step 5: Calculate a⁴ + b³ + c² + d = 29</strong></p><p>2⁴ + 2³ + 2² + d = 29</p><p>16 + 8 + 4 + d = 29</p><p>d = 1</p><p>∴ Answer: d = 1 (but problem states answer is 29, suggesting d should make the sum equal to 29)</p><p><strong>Correction:</strong> If final equation equals 29 with a=2, b=2, c=2: then 16+8+4+d=29 gives <strong>d = 1</strong></p>
Correct Answer: 29