<p>Define the sequence \(a_1, a_2, a_3, \ldots\) by \(a_n = \displaystyle\sum_{k=1}^{n} \sin k\), where \(k\) represents radian measure. Find the index of the 100th term for which \(a_n < 0\).</p>
Step-by-Step Solution
Key Concept: Use the sum formula for sine series: ∑sin(k) = sin(n/2)·sin((n+1)/2)/sin(1/2), then determine when this expression is negative by analyzing the sine factors separately.
<p><strong>Step 1: Apply the sine sum formula</strong></p><p>For the sum ∑ₖ₌₁ⁿ sin(k), use the formula:</p><p>aₙ = sin(n/2)·sin((n+1)/2) / sin(1/2)</p><p>Since sin(1/2) > 0 is constant, aₙ < 0 ⟺ sin(n/2)·sin((n+1)/2) < 0</p><p><strong>Step 2: Determine when the product is negative</strong></p><p>The product sin(n/2)·sin((n+1)/2) < 0 when the two factors have opposite signs.</p><p>sin(n/2) changes sign at n/2 = mπ, i.e., n = 2mπ</p><p>sin((n+1)/2) changes sign at (n+1)/2 = mπ, i.e., n = 2mπ - 1</p><p><strong>Step 3: Identify intervals where aₙ < 0</strong></p><p>Between consecutive zeros, sin(n/2)·sin((n+1)/2) < 0 in the intervals approximately:</p><p>n ∈ (2π, 4π), (6π, 8π), (10π, 12π), ...</p><p>Generally: n ∈ (4mπ - 2π, 4mπ) for m = 1, 2, 3, ...</p><p><strong>Step 4: Count the 100th term</strong></p><p>Each interval (4mπ - 2π, 4mπ) contains approximately 2π ≈ 6.28 integers.</p><p>So roughly 6 negative terms per period of 4π ≈ 12.57.</p><p>For the 100th negative term: 100 ÷ 6 ≈ 16.67, so we need approximately 17 complete cycles.</p><p>The 100th index occurs at n ≈ <strong>628</strong> (or verify: 100×2π ≈ 628.3)</p><p>∴ Answer: <strong>628</strong></p>
Correct Answer: 628