1/[n*(n+1)]=1/n-1/(n+1)
所以每一项都可以写成1-1/[n*(n+1)]
比如9899/9900=1-1/9900
29/30=1-1/30
第一项分母2=1*2
第二项分母6=2*3
最后一项分母9900=99*100
所以总共由99项
所以原题=99-(1/2+1/6+1/12+1/20+1/30+……+1/9900 )
而
1/2+1/6+1/12+1/20+1/30+……+1/9900
=1-1/2+1/2-1/3+1/4-1/5+1/5-1/6+……+1/99-1/100
=1-1/100
所以原题=99-(1-1/100)=98+1/100
1/2+5/6+11/12+19/20+29/30+...+9701/9702+9899/9900
=(1-1/2)+(1-1/6)+(1-1/12)+(1-1/20)+(1-1/30)+……+(1-1/9702)+(1-1/9900)
=[1-(1\1-1/2) ]+[1-(1/2-1/3)]+[1-(1/3-1/4)]+[1-(1/4-1/5)]+[1-(1/5-1/6)]+……+[1-(1/98-1/99)]+[1-(1/99-1/100)]
=1*99-1+1/2-1/2+1/3-1/3+1/4-1/4+1/5-1/5+1/6-1/6+……1/98-1/98+1/99-1/99+1/100
=99-1+1/100
=98又1/100
10000/101
100又1/100
等于0