<p>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 \[ \sum_{n=1}^{10} \left( \frac{1}{1 - x_n} \right): \]</p>
Step-by-Step Solution
Key Concept: Recognize that x^10 + x^9 + ... + x + 1 = (x^11 - 1)/(x - 1), so roots are 11th roots of unity except 1. Use logarithmic differentiation or partial fractions on P'(x)/P(x) to find the sum of reciprocals of linear transformations.
<p><strong>Step 1:</strong> Identify the polynomial. Note that P(x) = x^10 + x^9 + ... + x + 1 = (x^11 - 1)/(x - 1), so the roots are the 11th roots of unity except 1: x_n = e^(2πik/11) for k = 1, 2, ..., 10.</p><p><strong>Step 2:</strong> Consider the sum S = ∑_{n=1}^{10} 1/(1 - x_n). Rewrite using the fact that if x_n are roots of P(x), then ∑ 1/(1 - x_n) can be evaluated via the derivative formula.</p><p><strong>Step 3:</strong> Compute P(1) = 1 + 1 + ... + 1 = 11 (sum of 11 terms).</p><p><strong>Step 4:</strong> Use the logarithmic derivative: For P(x) = ∏(x - x_n), we have P'(x)/P(x) = ∑ 1/(x - x_n).</p><p><strong>Step 5:</strong> Find P'(x) = 10x^9 + 9x^8 + ... + 2x + 1, so P'(1) = 10 + 9 + 8 + ... + 2 + 1 = 55.</p><p><strong>Step 6:</strong> Substitute x = 1 into ∑ 1/(1 - x_n): We have ∑_{n=1}^{10} 1/(1 - x_n) = -∑_{n=1}^{10} 1/(x_n - 1) = -P'(1)/P(1) = -55/11 = -5.</p><p>∴ Answer: <strong>-5</strong></p>
Correct Answer: C