Recurrence Problem involving multiple dependencies. I have 3 equations :-


*

*$r_n=r_{n-1}+5m_{n-1}$

*$m_n = r_{n-1} + 3m_{n-1}$

*$p_n = 5m_{n-1}$


The initial values of the sequences are
$$r_0=3, m_0=1, p_0=0$$
How can I get the formula to get the nth term of the series without the recurrence?
A ready-made  will be useful.
 A: Hint Define the matrix $$M = \begin{pmatrix}1 & 5 & 0 \\ 1 & 3 & 0 \\ 0 & 5 & 0 \end{pmatrix},$$
Then you have $$\begin{pmatrix} r_{n} \\ m_n \\ p_n \end{pmatrix} = M \begin{pmatrix} r_{n-1} \\ m_{n-1} \\ p_{n-1} \end{pmatrix} = M^{n} \begin{pmatrix} r_{0} \\ m_{0} \\ p_{0} \end{pmatrix}.$$
So you have to compute $M^n$ for $n \in \Bbb N$. You can diagonalize $M$ for this purpose in order to have a relation of the form $M= SJS^{-1}$. You'll then have $M^n = SJ^nS^{-1}$ and $J^n$ is easy to compute since it is diagonal. Finally computing
$$\begin{pmatrix} r_{n} \\ m_n \\ p_n \end{pmatrix} = SJ^nS^{-1} \begin{pmatrix} r_{0} \\ m_{0} \\ p_{0} \end{pmatrix},$$
will give you a close form for $r_n,m_n,p_n$.
A: It is given that,
$$r_n=r_{n-1}+5m_{n-1}$$ 
$$m_n = r_{n-1} + 3m_{n-1}.$$ 
By subtracting we can obtain that,
$$r_n=m_n+2m_{n-1}$$
$$r_{n-1}=m_{n-1}+2m_{n-2}.$$
Hence
$$m_n=4m_{n-1}+2m_{n-2}.$$
This second order liner recurrence with $m_0=1$ and $m_1=6.$
I think you can solve this one and try to find $r_n$ and $p_n$ via substituting $m_n.$
