每项的差组成等差数列,另Bn=A(n+1)-An,则Bn为首项为3,公差为2的等差数列,那么Bn=2n+1,即A(n+1)-An=2n+1,A2-A1=2*1+1A3-A2=2*2+1A4-A3=2*3+1…………An-A(n-1)=2*(n-1)+1 (共n-1项)叠加可得：An-A1=2*1+2*2+2*3+……...