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反复拆分硬币堆的乘积和不变

Coin Split Problem

专题
Probability / 概率
难度
L4

题目详情

你有 1000 枚硬币。

先把它们分成两堆，分别为 xx 与 yy 枚，记录乘积 xyxy。

然后继续把每一堆再分成两堆，并把新的乘积都加到总和中。

不断重复，直到所有堆都变成单枚硬币（但单枚不再产生乘积项）。

证明：无论如何分堆，所有中间乘积之和恒定不变，并求该和值。

You have 1000 coins. First, split them into two piles with xx and yy coins, and record the product xyxy. Then split each of those piles again, and add the new products. You keep splitting piles until you end up with single coins (but you do not add the products of singletons). Show that no matter how you split, the sum of all these intermediate products is the same. What is that sum?

解析

总和恒为

1000⋅9992=499,500.\frac{1000\cdot 999}{2}=499{,}500.

更一般地，nn 枚硬币时恒为 n(n−1)2\frac{n(n-1)}{2}。

可用归纳证明：把一堆 nn 分成 aa 与 bb 后，剩余过程中分别贡献 a(a−1)2\frac{a(a-1)}{2} 与 b(b−1)2\frac{b(b-1)}{2}，再加上第一次的 abab，恰好等于 n(n−1)2\frac{n(n-1)}{2}。

Original Explanation

The final sum is always 1000×9992=499,500.\frac{1000 \times 999}{2} = 499{,}500. In general, splitting nn coins repeatedly yields a total sum n(n−1)2\tfrac{n(n-1)}{2}. One intuitive way: each “cut” can be seen as increasing the count of piles by 1 (from 1 pile of nn coins to nn piles of 1 coin), and each cut’s product-sum leads to the same final result via induction.