1)Sn=a(an-1)
S(n-1)=a(a(n-1)-1)
两式相减,得
an=a(an-a(n-1))
即an=a/(a-1)*a(n-1)
即{an}时等比数列,公比为a/(a-1)
又a1=S1=a(a1-1),得a1=a/(a-1)
所以an=[a/(a-1)]^n
2)由题意可得,a/(a-1)=2+b
[a/(a-1)]^2>4+b=a/(a-1)+2
令a/(a-1)=t,即t^2-t-2>0
解得t2
即1+1/(a-1)2
得1/2