Recursive equation with limit Find $\alpha, \beta, \gamma$ for recursive equation:
$$ \alpha a_{n+3}-3a_{n+1}+\beta a_n = 18n$$  $$a_0=0,a_1=\gamma, a_2=3 $$
$$\lim_{n\rightarrow\infty}\frac{3a_n+(-2)^{n}}{n^3}=3$$
Hey guys, 
normally I know my way around recursive equations, but when it comes to exercise with limits like this one I don't know how to use the information from the limit condition. I looked in to answers and found out, that first you have to figure out that $a_n$ is formed like $n^3-\frac{1}{3}(-2)^n+ ...$
How can I come to this and move on from there?
My attempt was something like this:
$$\frac{3a_n+(-2)^{n}}{n^3}=3/*n^3$$
$$3a_n+(-2)^{n}=3n^3/-(-2)^{n}$$
$$3a_n=3n^3-(-2)^{n}/*\frac{1}{3}$$
$$a_n=n^3-\frac{1}{3}(-2)^{n}$$
but I don't know if this is correct and which parts are missing in $+...$ Please help
 A: First, convert the recursion to a linear recursion:
$$\alpha~a_{n+3}-3~a_{n+1}+\beta~ a_{n} = 18~n \tag{1}$$
$$\alpha~a_{n+4}-3~a_{n+2}+\beta~ a_{n+1} = 18~n +18 \tag{2}$$
$$\alpha~a_{n+5}-3~a_{n+3}+\beta~ a_{n+2} = 18~n + 36\tag{3}$$
Eliminate the nonlinear terms with linear algebra:
$$a_{n+5} = 2~a_{n+4} + 
\left(\frac{3}{\alpha} - 1\right)~a_{n+3} +
\left(-\frac{\beta}{\alpha} - \frac{6}{\alpha}\right)~a_{n+2} +
\left(\frac{2\beta}{\alpha} + \frac{3}{\alpha}\right)~a_{n+1} -
\left(\frac{\beta}{\alpha} \right )~a_{n} \tag{4}$$
This is a linear recursion represented by the matrix:
$$\begin{bmatrix} a_{n+4} \\ a_{n+3} \\ a_{n+2} \\ a_{n+1} \\ a_{n+0} \end{bmatrix} = 
\underbrace{\begin{bmatrix}
2 & 
\frac{3}{\alpha} - 1 &
-\frac{\beta}{\alpha} - \frac{6}{\alpha} &
\frac{2\beta}{\alpha} + \frac{3}{\alpha} &
-\frac{\beta}{\alpha} \\
1 & 0 & 0 & 0 & 0\\
0 & 1 & 0 & 0 & 0\\
0 & 0 & 1 & 0 & 0\\
0 & 0 & 0 & 1 & 0\\
\end{bmatrix}^n}_M
\begin{bmatrix}a_{4} \\ a_{3} \\ a_{2} \\ a_{1} \\ a_{0}\end{bmatrix}
\tag{5}$$
For the recurrence to have a cubic $k^nn^3$ and an exponential $r^n$ term, then Jordan form of the matrix must contain subblocks:
$$\begin{bmatrix} 
k & 1 & 0 & 0 \\
0 & k & 1 & 0 \\
0 & 0 & k & 1 \\
0 & 0 & 0 & k \\
\end{bmatrix} \text{ and } \begin{bmatrix} r \end{bmatrix}$$
Since the matrix is 5 by 5, with $k = 1$ and $r = -2$ the Jordan form can only be:
$${\rm Jordan}(M) = 
\begin{bmatrix} 
1 & 1 & 0 & 0 & 0\\
0 & 1 & 1 & 0 & 0 \\
0 & 0 & 1 & 1 & 0 \\
0 & 0 & 0 & 1 & 0 \\
0 & 0 & 0 & 0 & -2
\end{bmatrix}$$
So $M$ has an eigenvalue $1$ with multiplicity $4$ and eigenvalue $-2$ with multiplicity $1$.  Expanding $\det (\lambda I - M) = 0$  you get:
$$\begin{cases}\lambda^5
 - 2~\lambda^4
 - \left(\frac 3{\alpha} - 1\right)~\lambda^3
 + \frac{\beta+6}{~\alpha}~\lambda^2
 - \frac{2~\beta+3}{\alpha}\lambda + \frac \beta{\alpha} = 0 \\
(\lambda - 1)^4(\lambda + 2) = 0
\end{cases}$$
Solving gives $~\alpha=1$, $\beta=2$.  
