Question

已知数列｛an｝的前n项和为Sn,满足log2(Sn+1)=n,(1)求数列的通项公式（2）求证此数列是等比数列.

Answer 1

log2(S1+1)=log2(a1+1)=1

a1=1

log2(Sn+1)=log2(2^n),

Sn+1=2^n,Sn=2^n-1

a(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