It is known that a (n + 1) - 2An = 3 * 2 ^ (n-1) Find an

It is known that a (n + 1) - 2An = 3 * 2 ^ (n-1) Find an

Divide both sides by 2 ^ (n + 1). Note an / 2 ^ n = BN
Then B (n + 1) - BN = 3 / 4
That is, BN is an arithmetic sequence with 3 / 4 tolerance
The following is easy

F (1) = 2, f (n + 1) = (f (n) + 1) / 2, (n belongs to n *), find the general term formula
I can't do it any more

Both sides of the recursive formula are multiplied by 2 ^ n at the same time
f(n+1)*2^(n+1)=f(n)*2^n+2^n
Let g (n) = f (n) * 2 ^ n
Then G (n + 1) = g (n) + 2 ^ n
In order, G (n) = 2 ^ (n-1) + 2 ^ (n-2) +... + 2 + G (1)
And G (1) = f (1) * 2 = 4
g(n)=2^n+2
So f (n) = 1 + 1 / 2 ^ (n-1)

How to find the general formula of F (n) = 2F (n-1) + 1

f(n)=2f(n-1)+1
f(n)+1=2[f(n-1)+1]
So f (n) + 1 is an equal ratio sequence with F (1) + 1 as the first term and 2 as the common ratio
f(n)+1=[f(1)+1]*2^(n-1)
f(n)=[f(1)+1]*2^(n-1) -1