解.裂项法.
Sn=1/[n(n+1)(n+2)]=(1/2){1/[n)n+1)]-1/[(n+1)(n+2)]}
=(1/2)[1/n-1/(n+1)-1/(n+1)+1/(n+2)]
=(1/2)[1/n-2/(n+1)+1/(n+2)]
=1/1x2x3+1/2x3x4+1/3x4x5+...+1x/n(n+1)(n+2)
=(1/2)[1/1-2/2+1/3+1/2-2/3+1/4+1/3-2/4+1/5+/4-2/5+1/6
+.+1/n-2/(n+1)+1/(n+2)]
=(1/2)[1-1/2-1/(n+1)+1/(n+2)]
=(n^2+3n)/[4(n+1)(n+2)]
其他类似的题目比如：
1/1x2x3+1/2x3x4+1/3x4x5+------+1/98x99x100
=(1/2)*(1/1*2-1/2*3)+(1/2)*(1/2*3-1/3*4)+...+(1/2)(1/98*99-1/99*100)
=(1/2)*(1/1*2-1/2*2+1/2*3-1/3*4+...+1/98*99-1/99*100)
=(1/2)*(1/2-1/9900)
=(1/2)*(4949/9900)
=4949/19800.