A sequence $\{a_n\}_{n \ge 1}$ is defined recursively by
$$a_0 = 1, a_1 = 1$$$$a_n = 5a_{n-1}-6a_{n-2}, \text{ for } n \ge 2 $$
Find an explicit formula for the general term $A_n$.
So, I want to let the vector
$ \begin{equation*} A_n = \begin{bmatrix} a_n\\ a_{n-1}\\ \end{bmatrix} \end{equation*} $
Then if I can find a relation between $A_n$ and $A_{n-1}$ using a $2 \times 2$ matrix I believe I'm close. Once the relation is found I could use eigenvector diagonalization to possibly find the explicit formula for $A_n$?
There should be a relation, I believe, between the characteristic polynomial of a $2 \times 2$ matrix to the coefficients of
$a_n - 5a_{n-1} + 6a_{n-2} = 0$
Thank you for the help.
$\endgroup$2 Answers
$\begingroup$Hint:$$ A_n = \begin{bmatrix} a_n \\ a_{n-1} \end{bmatrix} = \begin{bmatrix} 5a_{n-1} - 6a_{n-2} \\ a_{n-1} \end{bmatrix} = \begin{bmatrix} 5 & -6 \\ 1 & 0 \end{bmatrix} \begin{bmatrix} a_{n-1}\\ a_{n-2} \end{bmatrix} = \begin{bmatrix} 5 & -6 \\ 1 & 0 \end{bmatrix} A_{n-1}. $$
$\endgroup$ $\begingroup$To compare with the matrix method a classic method is given here. Let $a_{n} = r^{n}$ to obtain\begin{align} a_{n} &= 5 \, a_{n-1} - 6 \, a_{n-2} \\ r^{n} &= 5 \, r^{n-1} - 6 \, r^{n-2} \\ r^{2} &= 5 \, r - 6 \\ r &\in \{2, 3\}. \end{align}From this it is given that $a_{n} = c_{0} \, 2^{n} + c_{1} \, 3^{n}$. Using $a_{0} = a_{1} = 1$ then the solution becomes$$a_{n} = 2^{n+1} - 3^{n}.$$
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