1/1X3+1/3X5+1/5X7+..+1/2007X2009

1/1X3+1/3X5+1/5X7+..+1/2007X2009
1/1X3+1/3X5+1/5X7+..+1/2007X2009

1/1X3+1/3X5+1/5X7+..+1/2007X2009
裂项相消,把每一项拆分成2项,使得前后项相等消掉.
1/ax(a+2)=[1/a-1/(a+2)]/2
所以1/1X3=[1/1-1/3]/2,1/3X5=[1/3-1/5]/2 .1/2007X2009=[1/2007-1/2009]/2
原式=[1/1-1/3+1/3-1/5+1/5-1/7.1/2005-1/2007+1/2007-1/2009]/2
=[1/1-1/2009]/2
=1004/2009

1/1X3+1/3X5+1/5X7+..+1/2007X2009
=(1-1/3)/2+(1/3-1/5)/2+(1/5-1/7)/2+..+(1/2007-1/2009)/2
=(1-1/3+1/3-1/5+1/5-1/7+..+1/2007-1/2009)/2
=(1-1/2009)/2
=2008/2009÷2
=1004/2009