令n=1 S1=a1由an^2+2an=4Sn.(1)解得a1=24Sn-1=an-1^2+2an-1.(2)(1)-(2)得4an=an^2-an-1^2+2an-2an-1化简得（an+an-1)（an-an-1-2）=0.an+an-1≠0 则an-an-1-2=0an=an-1+2{an}是首项a1=2公差d=2 的等差数列an=2n...