数列{an}满足a1=1,an^2=(2an+1)a(n+1),令bn=lg(1+1/an),求证{bn}为等比数列

an^2=(2an+1)a(n+1),
a(n+1)=an²/(2an+1) (1)
a(n+1)+1=an²/(2an+1)+1=(an+1)²/(2an+1) (2)
(2)÷(1) [a(n+1)+1]/a(n+1)=[(an+1)/an]²
依次顺推)(an+1)/an={[a(n-1)+1]/a(n-1)}²=.=[(a1+1)/a1]^2^(n-1)=2^[2^(n-1)]
即1+1/an=2^[2^(n-1)] (1)
所以bn=lg2^[2^(n-1)]=2^(n-1)*ln2 b(n-1)=2^(n-2)*ln2
bn/b(n-1)=2ln2
所以{bn}是公比为2ln2的等比数列
由(1) 得通项公式an=1/{1-2^[2^(n-1)]}
an/(an+1)=1/2^[2^(n-1)]
∑（ai/(1+ai)）=1/2+1/2^2+1/2^4+1/2^8+.+1/2^[2^(n-1)]
=3/4+(1/2^3)[1/2+1/2^3+.+1/2^[2^(n-1)-3]
再问： “