数学题1/1*3+1/3*5+1/5*7+．．．．1/97*99+1/99*101＝?

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

1/[n*(n+2)]=1/2(1/n-1/n+2)所以原式可一直化简为1/2(1-1/101)=50/101

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

这是裂项法
1/3=（1-1/3）/2
1/3*5=（1/3-1/5）/2
所以都拆了 中间约去只剩首尾
即为（1-1/101）/2=50/101

由于1/(2n-1)(2n+1)=[1/(2n-1)-1/(2n+1)]/2
所以1/3+1/3*5+1/5*7.1/99*101
=[(1-1/3)+(1/3-1/5)+...+(1/99-1/101)]/2
=(1-1/101)/2
=50/101