數學奧林匹克問題(唔使計)

A^2 - B^2 = (A+B)(A-B) where A=2001, B=2000

Given A-B=1

Than

A^2-B^2 = (A+B) *1 = A+B

Hence,

2004^2-2003^2 = 2003+2004 = 2007



It also can proof that

If A-B = 2

=> A^2 - B^2 = 2 (A+B)

.......etc........

2007-11-21 09:43:23 補充：
It should be proof thatA^2 - B^2 = AB - B^2 + AAns.A^2 - B^2 = (A+B) (A-B) = A^2 - AB + AB - B^2 rearrange= AB - B^2 +A^2 - AB = AB - B^2 + A (A-B) Given 2001-2000 = 1Hence A-B =1 = AB - B^2 + A For every A-B =1A^2 - B^2 = AB - B^2 + A

2007-11-21 09:43:33 補充：
It should be proof thatA^2 - B^2 = AB - B^2 + AAns.A^2 - B^2 = (A+B) (A-B) = A^2 - AB + AB - B^2 rearrange= AB - B^2 +A^2 - AB = AB - B^2 + A (A-B) Given 2001-2000 = 1Hence A-B =1 = AB - B^2 + A For every A-B =1A^2 - B^2 = AB - B^2 + A

2007-11-21 09:43:38 補充：
It should be proof thatA^2 - B^2 = AB - B^2 + AAns.A^2 - B^2 = (A+B) (A-B) = A^2 - AB + AB - B^2 rearrange= AB - B^2 +A^2 - AB = AB - B^2 + A (A-B) Given 2001-2000 = 1Hence A-B =1 = AB - B^2 + A For every A-B =1A^2 - B^2 = AB - B^2 + A

其實現看題目
第 1 2 3 4 5 6 ----個數分別是 -
1 4 9 16 25 36 ---- -
既1^2 2^2 3^2 4^2 5^2 6^2-----------------------
所以第2000個數是 2000^2 ，，第2001個數是^2
問題問底2001個數和第2000個數的差 就是 2001^2-2000^2=4001

2001^2 - 2000^2
= (2001)(2000+1) - 2000^2
= 2001x2000 + 2001 - 2000^2
= 2001x2000 - 2000^2 + 2001
= 2000x(2001-2000) + 2001
= 2000x1 + 2001
= 4001

因為這串數係正方形數,條式應該係:

2001x2001-2000x2000
=4004001-4000000
=4001

2007-11-20 23:43:18 補充：
2001 2000=4001 所以3001x3001-3000x3000=3001 3000=6001