<p>The value of \(3\displaystyle\sum_{n=1}^{\infty} \left(\frac{1}{\pi}\sum_{k=1}^{\infty} \cot^{-1}\left(1+2\sqrt{\sum_{r=1}^{k} r^3}\right)\right)^n\) is less than:</p>
Step-by-Step Solution
Key Concept: Recognize that ∑r³ = [k(k+1)/2]², so 1 + 2√(∑r³) = (k+1)² + k², which allows telescoping via cot⁻¹(a²+b²) = cot⁻¹(a) - cot⁻¹(b). The inner sum telescopes to a constant x, then the outer geometric series evaluates to 3x/(1-x).
<p><strong>Step 1:</strong> Use the formula ∑(r=1 to k) r³ = [k(k+1)/2]²</p><p>So 1 + 2√(∑r³) = 1 + 2·k(k+1)/2 = 1 + k(k+1) = k² + k + 1</p><p><strong>Step 2:</strong> Apply the telescoping identity: cot⁻¹(k² + k + 1) = cot⁻¹(k) - cot⁻¹(k+1)</p><p>This follows from: cot⁻¹(a² + ab + b²) = cot⁻¹(a) - cot⁻¹(b) with a = k, b = k+1</p><p><strong>Step 3:</strong> The inner sum telescopes:</p><p>∑(k=1 to ∞) [cot⁻¹(k) - cot⁻¹(k+1)] = cot⁻¹(1) - lim(cot⁻¹(k+1)) = π/4 - 0 = π/4</p><p><strong>Step 4:</strong> Substitute into outer sum with x = (1/π)·(π/4) = 1/4:</p><p>3∑(n=1 to ∞) (1/4)ⁿ = 3 · (1/4)/(1 - 1/4) = 3 · (1/4)/(3/4) = 3 · 1/3 = 1</p><p>∴ Answer: B (The value equals 1, so it is less than options like 1.5, 2, etc.)</p>
Correct Answer: B