Solve for $a_{n}$ where $a_{n} = 4a_{n-1} + 2^{2n-1}$, $a_0=1$ and $a_1=6$ . $a_{n} = 4a_{n-1} + 2^{2n-1}$, $a_0=1$ and $a_1=6$.
So I am trying to find $a_n$ by using the generating function let's call it $A(x)$.
the equation then is written as (if I doing this correctly) :
$A(x)-ao = 4xA(x) + \sum_{k=1}2^{2n-1}*x^k$
now for  $\sum_{k=0}2^{2n-1}*x^k = \sum_{k=1}\frac{1}{2}4^{k}x^k \mapsto \frac{1}{2}\frac{1}{1-4x} - \frac{1}{2}$
Then we have: $(1-4x)A(x)= 1+\frac{1}{2}\frac{1}{1-4x} -\frac{1}{2}\Rightarrow A(x)=\frac{1}{2}\frac{1}{1-4x}+\frac{1}{2}\frac{1}{(1-4x)^2}\\ \Rightarrow a_n= ? $
$ (1-4x)^{-2} \mapsto 1 + \sum_{k=1}\binom{2+k-1}{k}*4^kx^{k}\\ \Rightarrow \frac{1}{(1-4x)^2}= 1+ \sum_{k=1}\binom{1+k}{k}*4^kx^{k} = 1+ \sum_{k=1}(k+1)4^kx^{k}= \sum_{k=0}(k+1)4^kx^{k}$
$And, \frac{1}{1-4x} \mapsto 4^n$
So $an = \frac{1}{2}4^n + \frac{1}{2}(n+1)4^n$  agrees with a1 = 6

Note, you didn't provide the boundary value for $a_1$.
Do you specifically want to use a GF? If you set
$$
b_n = \frac{a_n}{2^{2n-1}}
$$
to get (since $2(n-1)-1 = 2n -3$)
$$
b_n = b_{n-1} + 1
$$
and,
$$
a_n = 2^{n-2}a_1 + 2^{n-1}
$$
