1+2+3.......+999 997+999 998+999 999+1 000 000=?
回答 (6)
2006-12-16 7:02 pm
✔ 最佳答案
呢條可以用arithmetic sequence 計S(n)=n(A1+An)/2
n=1000000, A=1, S(1000000)=1+2+3+....1000000
S(n)=[1000000(1+1000000)]/2
S(n)=500000500000
2006-12-16 11:06:27 補充：
唔明咩叫ap可以去呢個睇返http://hk.knowledge.yahoo.com/question/?qid=7006101305291唔好意思, 打錯左, 係arithmetic series先岩....中4定中5數學會教
2006-12-16 9:14 pm
[(1+1 000 000) x 1 000 000]/2
=500 000 500 000
=500 000 500 000
2006-12-16 7:07 pm
1+2+3.......+999 997+999 998+999 999+1 000 000=?
=(1+999)999997/2+1000000+(1+4)998/2+999999---------公式(1+項數)尾項/2
=500000500000
=(1+999)999997/2+1000000+(1+4)998/2+999999---------公式(1+項數)尾項/2
=500000500000
2006-12-16 6:52 pm
sum of AP (等差數列的總和)
(頭項+尾項) x (項數) / 2 = 總數
答案應該是5,000,050,000
(頭項+尾項) x (項數) / 2 = 總數
答案應該是5,000,050,000
2006-12-16 6:50 pm
(1+1000000)X1000000/2
=500000500000
=500000500000
收錄日期: 2021-04-18 20:29:37
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20061216000051KK00943