设公差为d
log2(a3+1)-log2(a1+1)=2d
d=[log2(a3+1)-log2(a1+1)]/2
=[log2(3+1)-log2(0+1)]/2
=(2-0)/2
=1
log2(an +1)=log2(a1+1)+1×(n-1)=n-1
an +1=2^(n-1)
an=2^(n-1) -1
n=1时,a1=2^0 -1=1-1=0；n=3时,a3=2²-1=3,均满足通项公式
数列{an}的通项公式为an=2^(n-1) -1