1 * 3 / 1 + 3 * 5 / 1 + 5 * 7 / 1 + 7 * 9 / 1 +... + 2003 * 2005 / 1?

1 * 3 / 1 + 3 * 5 / 1 + 5 * 7 / 1 + 7 * 9 / 1 +... + 2003 * 2005 / 1?

1/[(2n-1)*(2n+1)]=[1/(2n-1)-1/(2n+1)]/2
Application of split term elimination method in summation of sequence
1*3/1+3*5/1+5*7/1+7*9/1+...+2003*2005/1=[1-1/3+1/3-1/5+...+1/2003-1/2005]/2=1002/2005

Calculation - 1 + 3 + (- 5) + 7 + (- 9) +... + 2003 + (- 2005) is an important process

-1+3+（-5）+7+（-9）+...+2003+（-2005）
Add head and tail
-（1+2005）+（3+2003）-（5+2001）+（7+1999）+…… -（1001+1005）+1003
＝-（1001+1005）+1003
＝-1003

How to calculate 1-3 5-7 9.2005-2007

Number of items 2N-1 = 2007, n = 1004
Divided into two arithmetic series a = 1 + 5 + 9 +... + 2005 (502 items in total)
=502(1+2005)/2=503506
B = - 3-7-11 -... - 2007 (502 items in total)
=-[502(3+2007)/2]=-504510
a+b=503505-504510=-1004
Or this calculation: (1-3) + (5-7) +... + (2005-2007) a total of 502 (- 2)
=-2*(1004/2)=-1004