∵log2（Sn-1）=n+1,
∴2n+1=Sn-1,得Sn=2n+1+1,
于是当n≥2时,an=Sn-Sn-1=2n-2n+1,
当n=1时,a1=S1=3,
综上,an=3,n=1时,an=2n-2n+1,n大于等于2时