Question:

<div class="solution"><p><strong>Step 1:</strong> $\sum_{p=1}^n 2p=n^2+n$. Inner term = $\cot^{-1}(n^2+n+1)$.</p><p><strong>Step 2:</strong> $\cot^{-1}(n^2+n+1)=\tan^{-1}\!\frac{(n+1)-n}{1+n(n+1)}=\tan^{-1}(n+1)-\tan^{-1}n$.</p><p><strong>Step 3:</strong> Telescoping: sum = $\tan^{-1}20-\tan^{-1}1=\tan^{-1}\!\frac{19}{21}$.</p><p><strong>Step 4:</strong> $\cot(\tan^{-1}(19/21))=21/19$.</p><p><strong>Answer: (1) 21/19</strong></p><div class="trap-box"><strong>Trap:</strong> Misreading the inner sum -- once it's n^2+n, the telescoping form is standard.<div class="key-concept"><strong>Key Concept:</strong> cot⁻^1 telescoping + cot of tan⁻^1 conversion