有理数的减法 (28 21:28:50)1/1*3+1/3*5+1/5*7+.1/99*101

每项可拆成n/(2n-1)-(n+1)/(2n+1)
原式=1-2/3+2/3-3/5+……+50/99-51/101
=1-51/101=50/101

s=1/1*3+1/3*5+1/5*7+.......1/99*101
=(1/2)*[(1/1)-(1/3)]+(1/2)*[(1/3)-(1/5)]+...+(1/20)*[(1/99)-(1/101)]
=(1/2)*[(1/1)-(1/3)+(1/3)-(1/5)+...+(1/99)-(1/101)]
=(1/2)*[(1/1)-(1/101)]
=50/101

原式=0.5*（1-1/3+1/3-1/5+1/5-1/7.....+1/99-1/101)
=0.5*(1-1/101)
=50/101
这叫裂项相抵法

1/2*(1/1-1/101)=50/101

=(1/2)*(1-1/3)+(1/2)*(1/3-1/5)+(1/2)*(1/5-1/7)+……+(1/2)*(1/99-1/101)
=(1/2)*(1-1/3+1/3-1/5+1/5-1/7+……+1/99-1/101)
中间正负抵消
=(1/2)*(1-1/101)
=50/101

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

1/1*3+1/3*5+1/5*7+.......1/99*101
=1/2(1-1/3+1/3+…………+1/97-1/99+1/99-1/101)
=1/2(1-1/101)
=1/2*100/101
=50/101