AMC 8 2015 Problem 9
AMC8 2015 年第 9 题
数列与规律★★☆☆☆
On her first day of work, Janabel sold one widget. On day two, she sold three widgets. On day three, she sold five widgets, and on each succeeding day, she sold two more widgets than she had sold on the previous day. How many widgets in total had Janabel sold after working 20 days?
Janabel 在她上班的第一天卖出了 1 个器具，第二天卖出了 3 个器具，第三天卖出了 5 个器具，且在往后的每一天，她都比前一天多卖出 2 个器具，那么当她工作了 20 天后，共卖出多少个器具？
- A.39
- B.40
- C.210
- D.400
- E.401
正确答案 · Correct Answer
D
解析 · Solution
Solution 1
First, we recognize that the number of widgets Janabel sells each day forms an arithmetic sequence. On day 1, she sells 1 widget, on day 2, she sells 3 widgets, on day 3, she sells 5 widgets, and so on. This sequence has first term a = 1 and common difference d = 2.
Step 1: Find the number of widgets sold on day 20. The formula for the -th term of an arithmetic sequence is For n = 20: So Janabel sells 39 widgets on day 20.
Step 2: Calculate the total number of widgets sold in 20 days. The sum of the first terms of an arithmetic sequence is For n = 20: Therefore, the total number of widgets sold after 20 days is .
First, we have to find how many widgets she makes on Day 20. We can write the linear equation y=-1+2x to represent this situation. Then, we can plug in 20 for : y=-1+2(20) -- y=-1+40 -- y=39. The sum of 1,3,5,... 39 is . The sum is just the first 20 odd counting/natural numbers, which is .
We can easily find out she makes widgets on Day 20. Then, we make the sum of 1,3, 5,...,35,37,39 by adding in this way: (1+39)+(3+37)+(5+35)+...+(19+21), which include 10 pairs of 40. So, the sum of 1,3,5,... is .
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