1/(1×2)+1/(2×3)+...+1/[n(n+1)]
=1-1/2+1/2-1/3+...+1/n -1(n+1)
=1- 1/(n+1)
=n/(n+1)那1/1×3 1/2×4 .... 1/n（n 2）呢？方法一样方法是一样的。1/(1×3)+1/(2×4)+...+1/[n(n+2)]=(1/2)[1-1/3+1/2-1/4+...+1/n -1/(n+2)]=(1/2)[1+1/2 -1/(n+1) -1/(n+2)]=3/4 -1/[2(n+1)] -1/[2(n+2)]