思路如下：
考虑通用性,研究一下1/[n(n+1)(n+2)]与1/n,1/(n+1),1/(n+2)的关系,可以知道下式成立：
1/[n(n+1)(n+2)]=1/2*[1/n+1/(n+2)]-1/(n+1),于是可以列出：
1/(1*2*3)=1/2(1+1/3)-1/2
1/(2*3*4)=1/2(1/2+1/4)-1/3
1/(3*4*5)=1/2(1/3+1/5)-1/4
1/(4*5*6)=1/2(1/4+1/6)-1/5
1/(5*6*7)=1/2(1/5+1/7)-1/6
.
1/(96*97*98)=1/2(1/96+1/98)-1/97
1/(97*98*99)=1/2(1/97+1/99)-1/98
1/(98*99*100)=1/2(1/98+1/100)-1/99
将上面98个式子加起来,研究等式右侧前后项抵消的关系,可以得到,
＝1/2*1/2+1/2*1/99+1/2*1/100-1/99
=9898/39600=4949/19800
=333300