Question:

<div class="solution"><p><strong>Step 1:</strong> <span class="math-inline">$\sum_{p=1}^n 2p=n^2+n$</span>. Inner term = <span class="math-inline">$\cot^{-1}(n^2+n+1)$</span>.</p><p><strong>Step 2:</strong> <span class="math-inline">$\cot^{-1}(n^2+n+1)=\tan^{-1}\!\frac{(n+1)-n}{1+n(n+1)}=\tan^{-1}(n+1)-\tan^{-1}n$</span>.</p><p><strong>Step 3:</strong> Telescoping: sum = <span class="math-inline">$\tan^{-1}20-\tan^{-1}1=\tan^{-1}\!\frac{19}{21}$</span>.</p><p><strong>Step 4:</strong> <span class="math-inline">$\cot(\tan^{-1}(19/21))=21/19$</span>.</p><p><strong>Answer: (1) 21/19</strong></p><div class="trap-box"><strong>Trap:</strong> Misreading the inner sum — once it's n²+n, the telescoping form is standard.</div><div class="key-concept"><strong>Key Concept:</strong> cot⁻¹ telescoping + cot of tan⁻¹ conversion</div></div>