1/1X3+1/3X5+1/5X7+..+1/2007X2009
2个回答

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