算式1/(1*3)+1/(3*5)+1/(5*7)+...+1/(2007*2009)=?

以上通项可表示为1/(2n-1)(2n+1) 等价于1/2(1/2n-1)-1/2(1/2n+1)
前2007项和为1/2(1-1/3+1/3+.......-1/2008+1/2008-1/2009)==1004/2009

1/(3*5)+1/(5*7)+……+1/(2007*2009)
=1/2*(1/3-1/5)+1/2*(1/5-1/7)+……+1/2*(1/2007-1/2009)
=[(1/3-1/5)+(1/5-1/7)+……+(1/2007-1/2009)]*(1/2)
=(1/3-1/5+1/5-1/7+……+1/2007-1/2009)*(1/2)
互相抵消
=(1/3-1/2009)*(1/2)
=(2006/6027)*(1/2)
=1003/6027

1/1*3=?(1-1/3)…以此类推，可得出最终结果1004/2009

令1/(1*3)+1/(3*5)+1/(5*7)+...+1/(2007*2009)=I
则有:
2I=2/(1*3)+2/(3*5)+...+2/(2007*2009)
=(3-1)/(1*3)+(5-3)/(3*5)+...+(2009-2007)/(2007*2009)
=1-1/3+1/3-1/5+...+1/2007-1/2009
=1-1/2009
所以
I=1004/2009