2001 AIME I 第 3 题

Question

求方程

x^{2001} + \left(\tfrac{1}{2} - x\right)^{2001} = 0

的所有根（实根和非实根）之和。已知该方程没有重根。

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

展开

\left(\frac{1}{2} - x\right)^{2001}

时，

x^{2001}

项会被抵消，留下一个

2000

次多项式。

Expanding

\left(\frac{1}{2} - x\right)^{2001}

cancels the

x^{2001}

term, leaving a polynomial of degree

2000

根之和等于

x^{1999}

项系数的相反数除以

x^{2000}

项系数。

The sum of the roots is minus the coefficient of

x^{1999}

divided by the coefficient of

x^{2000}

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}\text{。}

由韦达定理，

2000

个根的和为

\begin{aligned} \frac{\frac{\binom{2001}{2}}{4}}{\frac{2001}{2}} &= \frac{\frac{2001 \cdot 2000}{8}}{\frac{2001}{2}} \\ &= \frac{2000}{4} = 500 \end{aligned}\text{。}

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{\frac{\binom{2001}{2}}{4}}{\frac{2001}{2}} &= \frac{\frac{2001 \cdot 2000}{8}}{\frac{2001}{2}} \\ &= \frac{2000}{4} = 500. \end{aligned}

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