Question:

Step 1: Identify the polynomial and its roots. The polynomial $P(x) = x^{10} + x^9 + \cdots + x + 1$ can be recognized as a geometric series. This polynomial equals $\dfrac{x^{11}-1}{x-1}$ for $x \neq 1$. The roots of this polynomial are the 11th roots of unity excluding 1, which 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: Apply the logarithmic derivative technique. For a polynomial with roots $x_1, x_2, \ldots, x_{10}$, we can write: $$P(x) = (x - x_1)(x - x_2) \cdots (x - x_{10})$$ Taking the logarithmic derivative: $$\frac{P'(x)}{P(x)} = \sum_{n=1}^{10} \frac{1}{x - x_n}$$ Step 4: Evaluate the logarithmic derivative at $x = 1$. Substituting $x = 1$ into the logarithmic derivative: $$\frac{P'(1)}{P(1)} = \sum_{n=1}^{10} \frac{1}{1 - x_n} = S$$ Step 5: Calculate $P(1)$. Since $P(x) = 1 + x + x^2 + \cdots + x^{10}$: $$P(1) = 1 + 1 + 1 + \cdots + 1 = 11$$ Step 6: Calculate $P'(x)$ and $P'(1)$. Taking the derivative of $P(x) = 1 + x + x^2 + \cdots + x^{10}$: $$P'(x) = 1 + 2x + 3x^2 + \cdots + 10x^9$$ Evaluating at $x = 1$: $$P'(1) = 1 + 2 + 3 + \cdots + 10 = \frac{10 \cdot 11}{2} = 55$$ Step 7: Calculate the final answer. Using the result from Step 4: $$S = \frac{P'(1)}{P(1)} = \frac{55}{11} = 5$$ Therefore, the value of $\displaystyle\sum_{n=1}^{10}\left(\frac{1}{1 - x_n}\right) = 5$. The correct answer is **Option 1: 5**.