已知：数列{an}为1/(1×2),1/(2×3),1/(3×4),1/(4×5)，…1/[n(n+1)]，…求这个数列的前n项和．
sn=1/(1×2)+1/(2×3)+1/(3×4)+…+1/[(n-1)n]+1/[n(n+1)] =1/1-1/2+1/2-1/3+1/3-1/4+…+1/(n-1)-1/n+1/n-1/(n+1) =1-1/2+1/2-1/3+1/3-1/4+…+1/(n-1)-1/n+1/n-1/(n+1) =1-1/(n+1)=(n+1-1)/(n+1)=n/(n+1)
