Sequences & Series
Summation of Series
Grade 11
Question:
<p>Let \(S_n = \tan^{-1}\left(\sin 1 \cdot \displaystyle\sum_{r=1}^{n} \sec(r-1)\sec r\right)\), then:</p>
<p>\(S_5 = 5 - \pi\)</p>
<p>\(S_5 = 5 - 2\pi\)</p>
<p>\(S_5 = 10 - 3\pi\)</p>
<p>\(S_{10} = 3\pi - 10\)</p>
Step-by-Step Solution
Key Concept: Recognize that sec(r-1)sec(r) can be decomposed using the telescoping identity: sec(r-1)sec(r) = tan(r) - tan(r-1), allowing the sum to collapse into a simple form.
<p><strong>Step 1:</strong> Decompose sec(r-1)sec(r) using the identity:</p><p>sec(r-1)sec(r) = [sin(r) - sin(r-1)]/[sin(1)·cos(r-1)cos(r)]</p><p>This can be rewritten as: sec(r-1)sec(r) = [tan(r) - tan(r-1)]/sin(1)</p><p><strong>Step 2:</strong> Apply summation from r=1 to n:</p><p>∑(r=1 to n) sec(r-1)sec(r) = (1/sin(1))∑(r=1 to n)[tan(r) - tan(r-1)]</p><p>This telescopes to: (1/sin(1))[tan(n) - tan(0)] = tan(n)/sin(1)</p><p><strong>Step 3:</strong> Substitute into S_n:</p><p>S_n = tan⁻¹(sin(1)·tan(n)/sin(1)) = tan⁻¹(tan(n)) = n</p><p>∴ Answer: <strong>S_n = n</strong></p>
Correct Answer: A