1+2+3+4+5+6...

1+2+3+4+5+6...+99999999999999999999999999999+10000000000000000000000000000
=(1+10000000000000000000000000000) x 10000000000000000000000000000 / 2

方程式是(第一項+最後的一項) x 項數 / 2

我想係

10000000000000000000000000000 X 10000000000000000000000000001
----------------------------------------------------------------------------------------------------------
2

Because 1+2+3+.........+n = n X n+1 over 2

2007-03-05 14:33:46 補充：
-----------2 = /2

呢條係奧數
將1+10000000000000000000000000000=10000000000000000000000000001
2+9999999999999999999999999999=10000000000000000000000000001
3+9999999999999999999999999998=10000000000000000000000000001
4+9999999999999999999999999997=10000000000000000000000000001
5+9999999999999999999999999996=10000000000000000000000000001
..................................
10000000000000000000000000000+1=100000000000000000000000000001
係將最頭ga項數同最尾ga項數加埋
再將第二個項數同尾二ga項數加埋
咁樣一路加上去
加哂後
每條式ga答案都係10000000000000000000000000001
而且仲有5000000000000000000000000000條式
最後5000000000000000000000000000x每條式ga答案
就知個答案係ge啦
答案係50000000000000000000000000005000000000000000000000000000
式:5000000000000000000000000000x10000000000000000000000000001
=50000000000000000000000000005000000000000000000000000000

其實好簡單,係有一個方程式嫁!
(a) Using the above formula, find
1. 1+2+3+4+5+......100
=(1+100)/2 = (頭項+尾項)/2
=5050
2. 1+2+3+4+5+......200
=(1+200)/2 =(頭項+尾項)/2
=20100

It is given that 1+2+3+4+5+......100
=(頭項+尾項)X項數/2
also=5050

Are you think it is so easy?

1+10000000000000000000000000000) x 10000000000000000000000000000 / 2

∵1+2+3+.....99+100 = 5500
∴1+2+3+4+5+6...+99999999999999999999999999999+10000000000000000000000000000
就應該 = 5500000000000000000000000000000000000000000000000000000000

(1+10000000000000000000000000000) x (10000000000000000000000000000) / 2

=5 x 10^55

1+2+3+4+5+6+.....n=n(n+1)/2