log2(S1+1)=log2(a1+1)=1a1=1log2(Sn+1)=log2(2^n),Sn+1=2^n,Sn=2^n-1a(n)=Sn-S(n-1)=2^n-2^(n-1)-1+1=2^(n-1) (n>=2) a1=2^(1-1)=1=S1故a(n)=2^(n-1)a(n)=1*2^(n-1)所以该数列为首项1,公比2的等比数列...