1+2+3.......+999 997+999 998+999 999+1 000 000=?

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

[(1+1 000 000) x 1 000 000]/2
=500 000 500 000

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

=2999997
我本人係學校有學奧數

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

(1+1000000)X1000000/2
=500000500000