1/1*3+1/3*5+1/5*7+…+1/2001*2003=?

1/1*3=(1/2)*(1-1/3)后面同样分解,最后结果1001/2003

1/1*3=(1/2)*(1-1/3)
1/3*5=(1/2)*(1/3-1/5)
..
1/2001*2003=(1/2)*(1/2001-1/2003)

所以：
1/1*3+1/3*5+1/5*7+…+1/2001*2003
＝(1/2)*(1-1/3+1/3-1/5+..+1/2001-1/2003)
=(1/2)(1-1/2003)
=1001/2003

1001/2003

通项为an=1/((2n-1)(2n+1))=1/2(1/(2n-1)-1/(2n+1));
所以原式＝1/2(1-1/3+1/3-1/5+....+1/2001-1/2003)=1/2 * (1-1/2003)=1001/1003