<p>If <i>w</i> be an imaginary <i>n</i>th root of unity, then <i>∑</i><sub><i>r</i>=1</sub><sup><i>n</i></sup>(<i>a<sub>r</sub></i> + <i>b</i>)<i>w</i><sup><i>r</i>-1</sup> is equal to:</p>
<p>(a) <i>n</i>(<i>n</i> + 1)<i>a</i>/(2<i>w</i>)</p>
<p>(b) <i>nb</i>/(1 - <i>n</i>)</p>
<p>(c) <i>na</i>/(<i>w</i> - 1)</p>
<p>(d) None of these</p>
Step-by-Step Solution
Key Concept: When w is a primitive nth root of unity, the sum of all nth roots of unity equals zero: 1 + w + w² + ... + w^(n-1) = 0. Use this property to evaluate the given sum by separating it into two parts.
<p><strong>Step 1:</strong> Write the sum as ∑(r=1 to n)(ar + b)w^(r-1) = a∑(r=1 to n)r·w^(r-1) + b∑(r=1 to n)w^(r-1)</p><p><strong>Step 2:</strong> For the second sum: ∑(r=1 to n)w^(r-1) = 1 + w + w² + ... + w^(n-1). Since w is an nth root of unity, w^n = 1. If w ≠ 1, this geometric series equals (1 - w^n)/(1 - w) = 0/(1 - w) = 0.</p><p><strong>Step 3:</strong> For the first sum: ∑(r=1 to n)r·w^(r-1) = 1 + 2w + 3w² + ... + nw^(n-1). Multiply by (1-w) and use the geometric series formula to get ∑(r=1 to n)r·w^(r-1) = n/((1-w)²) · (1 - w^n) = 0 when w ≠ 1 and w^n = 1.</p><p><strong>Step 4:</strong> Therefore, the sum equals a(0) + b(0) = 0. However, this assumes w is a primitive nth root. The problem states w is 'an imaginary' nth root, which is ambiguous and may not guarantee w ≠ 1 or the standard formula applies uniformly across all choices of w.</p><p><strong>Step 5:</strong> Since the given options (A), (B), and (C) do not universally hold for all imaginary nth roots of unity without additional constraints, and the derivation shows the sum equals 0 for typical cases, the answer is none of the above.</p><p><strong>∴ Answer:</strong> D</p>
Correct Answer: D