求解.数学题

1/(1*2)+1/(2*3)+1/(3*4)+1/(4*5)+....+1/(99*100)
=1-1/2+1/2-1/3+....+1/99-1/100
=1-1/100
=99/100

1/(1*2)+1/(2*3)+1/(3*4)+1/(4*5)+....+1/(99*100)
=(1-1/2)+(1/2-1/3)+(1/3-1/4)+(1/4-1/5)+......+(1/99-1/100)
=1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+......+1/99-1/100
=1-1/100
=99/100

=（1-1/2）+（1/2-1/3）+（1/3-1/4）+。。。。。。。
=1-1/2+1/2-1/3+1/3-1/4+。。。。
=1-1/100
=99/100

1/a(a+1)=1/a-1/(a+1)
原式=(1-1/2)+(1/2-1/3)+(1/3-1/4)+```+(1/99-1/100)
=1-1/2+1/2-1/3+1/3-1/4+1/4-```-1/99+1/99-1/100
=1-(1/2-1/2)-(1/3-1/3)-(1/4-1/4)-```-(1/99-1/99)-1/100
=1-0-0-0-```-0-1/100
=1-1/100
=99/100

呃
，，说起来好麻烦1/(1*2)可以变形为1-1/2
1/(2*3)可以变形为1/2-1/3
1/(3*4)可以变形为1/3-1/4
依此类推
而1/(1*2)+1/(2*3)=1-1/2+1/2-1/3
就消掉了1/2项
依此类推，中间的都可以消掉
最后剩1-1/100=99/100

1/(1*2)+1/(2*3)+1/(3*4)+1/(4*5)+....+1/(99*100)
=2-1/2X1+3-2/3X2+....100-99/100X99
=1-1/2+3/6-2/6+....+10