Recurrence Problem involving multiple dependencies.

Question

I have 3 equations :-

1. $r_n=r_{n-1}+5m_{n-1}$
2. $m_n = r_{n-1} + 3m_{n-1}$
3. $p_n = 5m_{n-1}$

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 n^th term of the series without the recurrence?

A ready-made will be useful.

Answer 1

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 $r_n,m_n,p_n$.

Answer 2

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 $m_n.$