AMC 8 1990 Problem 16
AMC8 1990 年第 16 题
数列与规律★★★☆☆
题干整理中
- A.-990
- B.-10
- C.990
- D.1000
- E.1990
正确答案 · Correct Answer
D
解析 · Solution
Solution 1
This sum has terms. Grouping the sum as (1990-1980)+(1970-1960)+...+(30-20)+10 creates groups which each sum to 10, so the total sum is .
As in, this sum has 199 terms. Grouping the sum as 1990 + (-1980+1970) + (-1960+1950) +... + (-20+10) creates groups which each sum to -10, so the total sum is 1990 + (99)(-10) = 1000.
Grouping the terms as (1990 + 10) - (1980 + 20) +... + (1030 + 970) - (1020 + 980) + (1010 + 990) - 1000 shows that the sum is (2000 - 2000) +... + (2000 - 2000) + 2000 - 10000 = 0 +... + 0 + 1000 = 1000.
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