<p><strong>52.</strong> Let \(x_1, x_2, \ldots, x_{10}\) be the roots of the polynomial equation \(x^{10} + x^9 + \cdots + x + 1 = 0\). Then the value of \(\displaystyle\sum_{n=1}^{10} \left(\frac{1}{1 - x_n}\right)\):</p>
Step-by-Step Solution
Key Concept: Recognize that the polynomial is a geometric series: x^10 + x^9 + ... + x + 1 = (x^11 - 1)/(x - 1) for x ≠ 1. The roots are the 11th roots of unity excluding 1, so x_n = e^(2πik/11) for k = 1,2,...,10. Use Vieta's formulas or logarithmic differentiation on the polynomial to find the sum.
<p><strong>Step 1:</strong> Recognize the polynomial structure. We have P(x) = x^10 + x^9 + ... + x + 1 = (x^11 - 1)/(x - 1). The roots x_1, x_2, ..., x_10 are the 11th roots of unity except 1.</p><p><strong>Step 2:</strong> Consider S = Σ(1/(1-x_n)). Rewrite: S = Σ 1/(1-x_n). Note that P(x) = (x-x_1)(x-x_2)...(x-x_10).</p><p><strong>Step 3:</strong> Take logarithmic derivative: P'(x)/P(x) = Σ 1/(x-x_n). Evaluate at x = 1: lim(x→1) P'(x)/P(x).</p><p><strong>Step 4:</strong> From P(x) = (x^11-1)/(x-1), using L'Hôpital or direct expansion: P'(x) = 10x^9 + 9x^8 + ... + 1. At x = 1: P'(1) = 10 + 9 + 8 + ... + 1 = 55.</p><p><strong>Step 5:</strong> Also, lim(x→1) P(x)/(x-1) = 11 (by L'Hôpital on (x^11-1)/(x-1)). So P(1) = 11·0 is indeterminate; instead use lim(x→1) [(x^11-1)/(x-1)^2]·(x-1) evaluated carefully.</p><p><strong>Step 6:</strong> Direct approach: Σ 1/(1-x_n) = -Σ 1/(x_n-1). Note that d/dx[P(x)] evaluated strategically, or use: Σ 1/(1-x_n) = Σ [1 + x_n + x_n^2 + ...] (geometric series). By symmetry and Vieta relations on the transformed polynomial Q(y) = P(1-1/y), the sum equals 55/11 = 5 after careful calculation.</p><p>∴ Answer: C</p>
Correct Answer: C