Question

求方程的所有根（实根和非实根）的和。已知该方程没有重根。 $x^{2001} + \left(\tfrac{1}{2} - x\right)^{2001} = 0,$

Find the sum of the roots, real and non-real, of the equation $x^{2001} + \left(\tfrac{1}{2} - x\right)^{2001} = 0,$ given that there are no multiple roots.

答案

答案：500

知识点：多项式韦达定理二项式定理

难度评级：2300

解答：

用二项式定理展开 $\left(\frac{1}{2} - x\right)^{2001}$ 。它的最高次项 $(-x)^{2001} = -x^{2001}$ 会与方程中的 $x^{2001}$ 抵消，所以剩下的是一个 $2000$ 次多项式： $\begin{aligned} &2001 \cdot \frac{1}{2}\,x^{2000} \\ &\quad {}- \binom{2001}{2}\frac{1}{4}\,x^{1999} + \cdots = 0. \end{aligned}$

由韦达定理， $2000$ 个根的和为 $\begin{aligned} \frac{\binom{2001}{2}/4}{2001/2} &= \frac{2001 \cdot 2000/8}{2001/2} \\ &= \frac{2000}{4} = 500. \end{aligned}$

Expand $\left(\frac{1}{2} - x\right)^{2001}$ by the binomial theorem. Its leading term $(-x)^{2001} = -x^{2001}$ cancels the $x^{2001}$ in the equation, so what remains is a polynomial of degree $2000:$ $\begin{aligned} &2001 \cdot \frac{1}{2}\,x^{2000} \\ &\quad {}- \binom{2001}{2}\frac{1}{4}\,x^{1999} + \cdots = 0. \end{aligned}$

By Vieta's formulas, the sum of the $2000$ roots is $\begin{aligned} \frac{\binom{2001}{2}/4}{2001/2} &= \frac{2001 \cdot 2000/8}{2001/2} \\ &= \frac{2000}{4} = 500. \end{aligned}$

2001 AIME I 第 3 题

答案：500

难度评级：2300

解答：

其他年份的第 3 题