1/1*3+1/3*5+1/5*7+1/7*9.+1/2001*2003

问题描述:

1/1*3+1/3*5+1/5*7+1/7*9.+1/2001*2003

1/1*3=(1/2)*(1/1-1/3)同理1/3*5=(1/2)*(1/3-1/5)所以原式=1/1-1/3+1/3-1/5+1/5-·····-1/2001+/2001-1/2003=1-1/2003=2002/2003

原式=(2/1*3+2/3*5+2/5*7+2/7*9+……+2/2001*2003)/2
=(1-1/3+1/3-1/5+1/5-1/7+1/7-1/9+……1/2001-1/2003)/2
=(1-1/2003)/2
=1001/2003