<p>Let <i>1/(a₁ + ω) + 1/(a₂ + ω) + 1/(a₃ + ω) + ... + 1/(aₙ + ω) = i</i> where <i>a₁, a₂, a₃, ..., aₙ ∈ ℝ</i> and <i>ω</i> is an imaginary cube root of unity. Then evaluate <i>∑ᵣ₌₁ⁿ (2aᵣ - 1)/(aᵣ² - aᵣ + 1)</i>.</p>
Step-by-Step Solution
Key Concept: Since ω is an imaginary cube root of unity, we have ω³ = 1 and 1 + ω + ω² = 0. The given condition with a sum equaling i constrains the real numbers aᵣ, allowing us to relate them through properties of cube roots of unity and derive the desired sum.
<p><strong>Step 1:</strong> Recall properties of ω (imaginary cube root of unity): ω = e^(2πi/3), ω³ = 1, and 1 + ω + ω² = 0. Also, ω² = ω̄ (conjugate) and ω² + ω + 1 = 0, so ω + ω² = -1.</p><p><strong>Step 2:</strong> Given: ∑ᵣ₌₁ⁿ 1/(aᵣ + ω) = i. Taking the conjugate of both sides: ∑ᵣ₌₁ⁿ 1/(aᵣ + ω̄) = -i, which means ∑ᵣ₌₁ⁿ 1/(aᵣ + ω²) = -i.</p><p><strong>Step 3:</strong> Multiply the original equation by (aᵣ + ω)(aᵣ + ω²): For each term, 1/(aᵣ + ω) + 1/(aᵣ + ω²) = (aᵣ + ω² + aᵣ + ω)/[(aᵣ + ω)(aᵣ + ω²)] = (2aᵣ + ω + ω²)/[(aᵣ + ω)(aᵣ + ω²)] = (2aᵣ - 1)/[aᵣ² + aᵣ(ω + ω²) + ωω²].</p><p><strong>Step 4:</strong> Simplify the denominator: Since ω + ω² = -1 and ω·ω² = ω³ = 1, we get aᵣ² + aᵣ(-1) + 1 = aᵣ² - aᵣ + 1.</p><p><strong>Step 5:</strong> Therefore: 1/(aᵣ + ω) + 1/(aᵣ + ω²) = (2aᵣ - 1)/(aᵣ² - aᵣ + 1).</p><p><strong>Step 6:</strong> Sum over all r: ∑ᵣ₌₁ⁿ [1/(aᵣ + ω) + 1/(aᵣ + ω²)] = ∑ᵣ₌₁ⁿ (2aᵣ - 1)/(aᵣ² - aᵣ + 1).</p><p><strong>Step 7:</strong> The left side equals: ∑ᵣ₌₁ⁿ 1/(aᵣ + ω) + ∑ᵣ₌₁ⁿ 1/(aᵣ + ω²) = i + (-i) = 0.</p><p><strong>∴ Answer:</strong> 0</p>
Correct Answer: 0