1\(1乘3)+1/(3乘5)+1/(5乘7)+...+1/(99乘101)

1/（1*3）+1/（3*5）+1/（5*7）+……+1/（99*101）
=2[(1-1/3)+(1/3-1/5+(1/5-1/7)+……（1/99-1/101）]
=2[1-1/101]=200/101

2/1*3+2/3*5+2/5*7+......+2/99*101
=1-1/3+1/3-1/5+1/5-1/7+......+1/99-1/101
=1-1/101
=100/101

因为1/(1*3)=1/2*(1/1-1/3)1/(3*5)=1/2*(1/3-1/5)1/(5*7)=1/2*(1/5-1/7)同理推导可得1/(99*101)=1/2*(1/99-1/101)所以原式＝1/2*(1/1-1/3+1/3-1/5+1/5-1/7+.+1/97-1/99+1/99-1/101)=1/2*(1/1-1/101)=50/101...

1\(1乘3)+1/(3乘5)+1/(5乘7)+...+1/(99乘101)
=(1-1/3)/2+(1/3-1/5)/2+...+(1/99-1/101)/2
=(1-1/101)/2
=50/101