Math!

This is a A.S.

a= 1
d= 2

1 + 2(n-1) = 2003
2n-1 = 2003
2n = 2004
n = 1002

1002(2003+1)/2

=2004*501
=1004004

2003+2001+1999+1997+...+1007+1005-1003-1001-999-997-...-3-1=498996

2003+2001+1999+1997+...+1007+1005-1003-1001-999-997-...-3-1
=(2003-1003) + (2001-1001) + (1999 - 999) + ... + (1005 -5) -3 -1
=1000+1000+1000+...+1000-3-1
=499(1000)-3-1
=499000-3-1
=498996

[500(1005+2003)]/2-[503(1+1003)]/2
=499494

（2003+1005)/2x ((2003-1005)/2+1)-(1003+1)/2x((1003-1)/2+1)
=3008/2 x (998/2+1) - 1004/2 x (1002/2+1)
=1504 x (499+1) - 502 x (501+1)
=1504 x 500- 502 x 502
=752000 - 252004
=499996

=(2003-1003)x(2003-1003)/2
=1000x500
=500000

2008-01-17 17:59:57 補充：
更正=(2003-1003)x(2003+1)/2=1000x1002=1002000