解: a(n)=1/n(n+1)(n+2)=0.5[1/n(n+1)]-0.5[1/(n+1)(n+2)] 对于数列b(n)=0.5/n(n+1),可采取裂项求和法: S1(n)=0.5[1-(1/2)+(1/2)-(1/3)+.+(1/n)-1/(n+1)]=0.5[1-1/(n+1)]=0.5n/(n+1) 对于数列c(n)=0.5/(n+1)(n+2),采用相同的方法得到: S2(n)=S1(n+1)-b(1)=0.5(n+1)/(n+2)-0.25 故有原数列前n项和: S(n)=S1(n)-S2(n)=0.5[n/(n+1)-(n+1)/(n+2)+0.5] 原式=S(8)=0.5[(8/9)-(9/10)+0.5]=11/45