Trigonometry & Inverse Trigonometry
General
Grade 12
Question:
<p><span class="math-inline">\(\displaystyle\lim_{n\to\infty}\sum_{r=1}^{n}\tan^{-1}\!\frac{2r+1}{r^4+2r^3+r^2+1}=\)</span></p>
Step-by-Step Solution
Key Concept: General
<div class="solution"><p><strong>Key Idea:</strong> Factor denominator: <span class="math-inline">$r^4+2r^3+r^2+1=1+(r^2+r)^2$</span>. Numerator <span class="math-inline">$=(r+1)^2-r^2$</span>.</p><p><strong>Step 1:</strong> <span class="math-inline">$T_r=\tan^{-1}(r+1)^2-\tan^{-1}r^2$</span>.</p><p><strong>Step 2:</strong> Telescoping: sum <span class="math-inline">$=\tan^{-1}(n+1)^2-\tan^{-1}(1)\xrightarrow{n\to\infty}\pi/2-\pi/4=\pi/4$</span>.</p><p><strong>Answer: (A) <span class="math-inline">$\pi/4$</span></strong></p><div class="trap-box"><strong>Trap:</strong> Ignoring the subtracted initial term <span class="math-inline">$\tan^{-1}(1)=\pi/4$</span>.</div><div class="key-concept"><strong>Key Concept:</strong> Quartic denominator hides <span class="math-inline">$1+(r^2+r)^2$</span> — numerator is difference of squares</div></div>
Correct Answer: 1