已知数列An满足2根号下Sn=An+1,且An>0,求a1.a2,a3

由题意得,Sn=[(an+1)/2]^2①则S(n+1)=[(a(n+1)+1)/2]^2②②-①得(结合a(n+1)=S(n+1)-Sn)a(n+1)=[(a(n+1)+1)/2]^2-[(an+1)/2]^2化简得[a(n+1)+an][a(n+1)-an-2]=0因为{an}是正项数列所以a(n+1)+an＞0所以a(n+1)-an-2=...