Finding the closed form for a recurrence relation I'm having trouble finding a closed form for a geometric recurrence relation where the term being recursively multiplied is of the form (x+a) instead of just (x). 
Here's the recursive sequence:
$a_{n} = 4a_{n-1} + 5$ for $n \geq 1$ with the initial condition $a_{0} = 2$.
I know that in general the way to solve these problems is to start by writing out all of the arithmetic for the first few values of $a_{n}$, starting with the initial condition. Here's what I have:
$a_{0} = 2$
$a_{1} = 4 (2) + 5 \equiv ((2)(2)(2) + 5)$
$a_{2} = 4(4(2)+5)+5 \equiv (2)(2)((2)(2)(2)+5)+5$
$a_{3} = 4(4(4(2)+5)+5)+5 \equiv (2)(2)((2)(2)((2)(2)(2)+5)+5)+5$
$\ldots$ etc.
So at this point it's pretty clear to me that
$a_{n} = 2^{2n + 1} + something$
My problem is figuring out how to account for all of those 5's. Especially since the first 5 is being multiplied by $2^{3}$ and all of the other 5's are being multiplied by $2^{2}$. 
I guessed something like this:
$a_{n} = 2^{2n+1} + 5(4^{n-1}) + 5^{n-1}$
$\ldots$ and the results were close, but not exact.
Can anyone help me out with the correct method for solving these types of problems? Thanks very much for your time.

This sequence is an affine recursion of the form $a_{n+1} = \lambda a_n + \mu$, with $\lambda=4\neq 1$ and $\mu=5$. Its $n$th term is given by
$$a_n = \lambda^n (a_0 - \rho) +\rho \, ,$$
where $\rho= \frac{\mu}{1-\lambda}$. The formula for the $n$th term can be obtained by setting $b_n = a_{n+1}-a_n$, which is a geometric sequence with common ratio $\lambda$,
$$
b_{n+1} = \underbrace{\lambda a_{n+1} + \mu}_{a_{n+2}} - \underbrace{(\lambda a_{n} + \mu)}_{a_{n+1}} = \lambda b_{n} \, ,
$$
and initial condition $$b_0 = a_1 - a_0 = \left(\lambda - 1\right) a_{0} + \mu \, .$$
The $n$th term $a_n$ is deduced from the telescoping series
\begin{aligned}
\sum_{k=0}^{n-1} b_k  = a_n - a_0 = b_0 \frac{1-\lambda^n}{1-\lambda} \, .
\end{aligned}
