2√Sn=an+1n=1时,a1=14Sn=an^2+2an+1故4S(n-1)=a(n-1)^2+2a(n-1)+14an=4Sn-4S(n-1)=an^2+2an+1-[a(n-1)^2+2a(n-1)+1]即：[an-a(n-1)-2][an+a(n-1)]=0所以：（1）an-a(n-1)=2,{an}为等差数列,d=2an=1+2(n-1)=2n-1（2...