<p>The value of \(\displaystyle\sum_{k=1}^{10}\left(\sin\dfrac{2k\pi}{11} + i\cos\dfrac{2k\pi}{11}\right)\) is</p>
Step-by-Step Solution
Key Concept: Recognize that sin(2kπ/11) + i·cos(2kπ/11) = -i·e^(i·2kπ/11), and the sum telescopes as a geometric series of 11th roots of unity with one term excluded.
<p><strong>Step 1:</strong> Rewrite the general term using Euler's formula insight:</p><p>sin(2kπ/11) + i·cos(2kπ/11) = -i(cos(2kπ/11) + i·sin(2kπ/11)) = -i·e^(i·2kπ/11)</p><p><strong>Step 2:</strong> Factor out -i:</p><p>∑(k=1 to 10) = -i·∑(k=1 to 10) e^(i·2kπ/11)</p><p><strong>Step 3:</strong> Let ω = e^(i·2π/11). Then ∑(k=1 to 10) ω^k is the sum of all non-identity 11th roots of unity.</p><p><strong>Step 4:</strong> Since 1 + ω + ω² + ... + ω¹⁰ = 0 (sum of all 11th roots of unity), we have:</p><p>∑(k=1 to 10) ω^k = -1</p><p><strong>Step 5:</strong> Therefore: -i·(-1) = i</p><p>∴ Answer: <strong>i</strong> (Option D)</p>
Correct Answer: D