已知 \(x_1, x_2, \dots, x_9\) 满足方程组
\[
\begin{cases}
\dfrac{x_1}{11} + \dfrac{x_2}{12} + \dots + \dfrac{x_9}{19} = 1 \\
\dfrac{x_1}{21} + \dfrac{x_2}{22} + \dots + \dfrac{x_9}{29} = 1 \\
\dfrac{x_1}{31} + \dfrac{x_2}{32} + \dots + \dfrac{x_9}{39} = 1 \\
\dfrac{x_1}{41} + \dfrac{x_2}{42} + \dots + \dfrac{x_9}{49} = 1 \\
\dfrac{x_1}{51} + \dfrac{x_2}{52} + \dots + \dfrac{x_9}{59} = 1 \\
\dfrac{x_1}{61} + \dfrac{x_2}{62} + \dots + \dfrac{x_9}{69} = 1 \\
\dfrac{x_1}{71} + \dfrac{x_2}{72} + \dots + \dfrac{x_9}{79} = 1 \\
\dfrac{x_1}{81} + \dfrac{x_2}{82} + \dots + \dfrac{x_9}{89} = 1 \\
\dfrac{x_1}{91} + \dfrac{x_2}{92} + \dots + \dfrac{x_9}{99} = 1 \\
\end{cases}
\]
试求 \(x_1 + x_2 + \dots + x_9\).

解答

构造 (关于 \(n\) 的) 多项式
\[
f(n) = \left[1 - \left(\dfrac{x_1}{n+1} + \dfrac{x_2}{n+2} + \dots + \dfrac{x_9}{n+9}\right)\right](n+1)(n+2)\cdots(n+9)
\]
则
\[
f(10) = f(20) = \dots = f(90) = 0
\]
这样 \(9\) 次多项式 \(f(n)\) 根全部确定, 因此可以假设
\[
f(n) = a(n-10) (n-20) \dots (n-90)
\]
对比最高次项 (\(9\) 次项) 的系数可得 \(a = 1\), 即
\[
\left[1 - \left(\dfrac{x_1}{n+1} + \dfrac{x_2}{n+2} + \dots + \dfrac{x_9}{n+9}\right)\right](n+1)(n+2)\cdots(n+9) = (n-10) (n-20) \dots (n-90)
\]
又对比 \(8\) 次项的系数可得
\[
(1 + 2 + \dots + 9) - (x_1 + x_2+ \dots + x_9) = (-10) + (-20) + \dots + (-90)
\]
所以 \(x_1 + x_2 + \dots + x_9 = 495\).