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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 nth term of the series without the recurrence?

A ready-made will be useful.

5xum
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Soumyadeep Roy
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2 Answers2

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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 $r_n,m_n,p_n$.

Community
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Surb
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  • By the way, note that $$\begin{pmatrix} r_n \ m_n \end{pmatrix} =\begin{pmatrix} 1 & 5 \ 1 & 3 \end{pmatrix} \begin{pmatrix} r_{n-1} \ m_{n-1} \end{pmatrix}$$ and $p_n = 5m_{n-1}$, but this don't significantly reduce the diagonalization work since it is straightforward to see that $0$ is an eigenvalue of $M$. – Surb Sep 10 '14 at 08:34
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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 $m_n.$

Bumblebee
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