Question:

Step 1: Identify the roots of the polynomial. The polynomial $P(x) = x^{10} + x^9 + \cdots + x + 1$ can be recognized as a geometric series. We can write: $$P(x) = \frac{x^{11} - 1}{x - 1}$$ This means the roots of $P(x)$ are the 11th roots of unity except $x = 1$. Therefore, the roots are $x_n = e^{2\pi i n/11}$ for $n = 1, 2, \ldots, 10$. Step 2: Set up the sum we need to evaluate. We need to find: $$S = \sum_{n=1}^{10} \frac{1}{1-x_n}$$ Step 3: Use the logarithmic derivative technique. For a polynomial with roots $x_1, x_2, \ldots, x_{10}$, we have the identity: $$\frac{P'(x)}{P(x)} = \sum_{n=1}^{10} \frac{1}{x - x_n}$$ Evaluating this at $x = 1$: $$\frac{P'(1)}{P(1)} = \sum_{n=1}^{10} \frac{1}{1 - x_n}$$ Step 4: Calculate $P(1)$. Since $P(x) = x^{10} + x^9 + \cdots + x + 1$, we have: $$P(1) = 1 + 1 + \cdots + 1 \text{ (11 terms)} = 11$$ Step 5: Calculate $P'(1)$. Taking the derivative: $$P'(x) = 10x^9 + 9x^8 + 8x^7 + \cdots + 2x + 1$$ Evaluating at $x = 1$: $$P'(1) = 10 + 9 + 8 + \cdots + 2 + 1 = \sum_{k=1}^{10} k = \frac{10 \cdot 11}{2} = 55$$ Step 6: Compute the final answer. Using the result from Step 3: $$\sum_{n=1}^{10} \frac{1}{1-x_n} = \frac{P'(1)}{P(1)} = \frac{55}{11} = 5$$ Therefore, the value of $\displaystyle\sum_{n=1}^{10}\left(\frac{1}{1-x_n}\right) = 5$. The correct answer is **Option 3: 5**. <div class="key-concept"><strong>Key Concept:</strong> Using logarithmic differentiation of the polynomial whose roots are given to evaluate symmetric sums of the form $\Sigma \frac{1}{a - x_n}$.</div> <div class="trap-box"><strong>Trap:</strong> Students may confuse the sign or forget that $P'(x)/P(x) = \sum 1/(x-x_n)$, not $\sum 1/(x_n-x)$.</div>