2000 AMC 10 第 12 题

先试着解答 2000 AMC 10 第 12 题,然后核对你的答案与精心整理的解答,解答来自 LIVE by Po-Shen Loh。你也可以参加完整限时模拟考试、查看全部 2000 AMC 10 解答,或核对答案。

所有题目均经美国数学协会(MAA)官方合法授权使用。

12.

图形 00,11,22 和 33 分别由 11,55,1313 和 2525 个互不重叠的单位正方形组成。如果继续这个规律,图形 100100 中会有多少个互不重叠的单位正方形?
Figures 0,0, 1,1, 2,2, and 33 consist of 1,1, 5,5, 13,13, and 2525 nonoverlapping unit squares, respectively. If the pattern were continued, how many nonoverlapping unit squares would there be in figure 100?100?
1040110401
1980119801
2020120201
3980139801
4080140801
答案:C
知识点:前n个奇数之和完全平方数找规律
小提示:
数列 1,5,13,25,…1, 5, 13, 25, \ldots 的相邻差为 4,8,12,…4, 8, 12, \ldots。
The differences 1,5,13,25,…1, 5, 13, 25, \ldots grow by 4,8,12,…4, 8, 12, \ldots
大提示:
图形 nn 有 n2+(n+1)2n^2 + (n+1)^2 个单位正方形。
Figure nn has n2+(n+1)2n^2 + (n+1)^2 unit squares
解答:
图形 nn 可以看作前 nn 个奇数之和加上前 n+1n+1 个奇数之和,所以单位正方形个数为 n2+(n+1)2n^2 + (n+1)^2。
对图形 100100 有 1002+1012=10000+10201=20201。 \begin{aligned} 100^2 + 101^2 &= 10000 + 10201 \\ &= 20201 \end{aligned}\text{。}
所以正确答案是 C。
Figure nn can be split into the sum of the first nn odd numbers and the first n+1n+1 odd numbers, giving n2+(n+1)2n^2 + (n+1)^2 unit squares.
For figure 100,100, this is 1002+1012=10000+10201=20201. \begin{aligned} 100^2 + 101^2 &= 10000 + 10201 \\ &= 20201. \end{aligned}
Thus, the correct answer is C.
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