數學問題～幫我解決．
回答 (2)
2006-11-28 6:17 am
✔ 最佳答案
若等差級數25+18+11+4...的總和為-830,求該數列的總項數。 T(1) = 25
等差=18 - 25= -7
設 n 是該數列的總項數
n[2(25) + (n-1)(-7)]/2 = -830
n(50 -7n + 7) = -1660
57n -7n2 = -1660
7n2 - 57n - 1660 = 0
(7n+83)(n-20) = 0
7n+83= 0 or n-20=0
n = -83/7 (捨去) or n = 20
該數列的總項數是20項
2006-11-28 6:17 pm
(n) (Item) (Sum of accumulated item)
12525
21843
31154
4458
5-355
6-1045
7-1728
8-244
9-31-27
10-38-65
11-45-110
12-52-162
13-59-221
14-66-287
15-73-360
16-80-440
17-87-527
18-94-621
19-101-722
20-108-830
21-115-945
22-122-1067
23-129-1196
24-136-1332
I think n=20, I used EXCEL to calculate.
12525
21843
31154
4458
5-355
6-1045
7-1728
8-244
9-31-27
10-38-65
11-45-110
12-52-162
13-59-221
14-66-287
15-73-360
16-80-440
17-87-527
18-94-621
19-101-722
20-108-830
21-115-945
22-122-1067
23-129-1196
24-136-1332
I think n=20, I used EXCEL to calculate.
收錄日期: 2021-04-25 16:52:22
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20061127000051KK05057