Given function, f (x) = 2x + 1, G (x0 = x), (x belongs to R, sequence, {an}, {BN} satisfies A1 = 1, an = f (BN) = g (BN + 1) to find the general term formula of sequence {an}

Given function, f (x) = 2x + 1, G (x0 = x), (x belongs to R, sequence, {an}, {BN} satisfies A1 = 1, an = f (BN) = g (BN + 1) to find the general term formula of sequence {an}

Because A1 = f (B1), 1 = 1 + 2B1, B1 = 0;
Because f (BN) = g (BN + 1), 1 + 2bn = B (n + 1), 2 + 2bn = 1 + B (n + 1),
Let 1 + BN = BN, then B (n + 1) / BN = 2, B1 = 1 + B1 = 1,
The sequence {BN} is an equal ratio sequence with 1 as the first term and 2 as the common ratio, that is, BN = 2 ^ (n-1),
bn=Bn-1=2^(n-1)-1,
an=f(bn)=1+2bn=1+（2^n-2）=2^n-1.

Given the function f (x) = 2x / (x + 1), the sequence {an} satisfies A1 = 4 / 5, a (n + 1) = f (an), BN = 1 / an-1
(1) The general term formula of the sequence {BN}
(2) Let CN = [an * a (n + 1)] / 2 ^ (n + 2) and tn be the sum of the first n terms of the sequence CN

(1) From BN = 1 / an-1,
An = 1 / (BN + 1), substituting a (n + 1) = f (an)
1/(b(n+1) + 1) = 2/(bn + 1) / [1/(bn + 1) + 1]
B (n + 1) = BN / 2
And B1 = 1 / A1 - 1 = 1 / 4
So BN = 1 / 2 ^ (n + 1)
(2) By (1), an = 1 / (BN + 1) = 2 ^ (n + 1) / [2 ^ (n + 1) + 1],
So CN = 2 ^ (n + 1) / [(2 ^ (n + 1) + 1) (2 ^ (n + 2) + 1)] = 1 / [2 ^ (n + 1) + 1] - 1 / [2 ^ (n + 2) + 1]
So TN = C1 +... + CN = 1 / 5 - 1 / [2 ^ (n + 2) + 1]