<s>[INST] Using determinants to find a recursive sequence I am trying to compute a three diagonal determinant in order to find the recursive relation.
Let $\Delta_{n}$=$\begin{vmatrix}
11 & 3 & 0 & 0 & \dots & 0\\
13 & 11 & 3 & 0 & \dots & 0\\
0 & 13 & 11 & 3 & \dots & 0\\
\vdots & \vdots & \vdots & \vdots & \ddots & \vdots\\
0 & 0 & 0 & 0 & \dots 11 & 3\\
0 & 0 & 0 & 0 & \dots 13 & 11\\
\end{vmatrix}$
Once I have expanded $\Delta_{n+2}$ along the first row I obtain:
$\Delta_{n+2}$=$11\Delta_{n+1}$ - 3$\begin{vmatrix}
13 & 3 & 0 & 0 & \dots & 0\\
0 & 11 & 3 & 0 & \dots & 0\\
0 & 0 & 11 & 3 & \dots & 0\\
\vdots & \vdots & \vdots & \vdots & \ddots & \vdots\\
0 & 0 & 0 & 0 & \dots 11 & 3\\
0 & 0 & 0 & 0 & \dots 13 & 11\\
\end{vmatrix}$
Expanding the above matrix along the first row I obtain:
$\Delta_{n+2}$=$11\Delta_{n+1}$ - 3(13)$\Delta{n} + 0 + 0 + \dots$
I hope that I am on the right track with my matrix expansions. Any comments are appreciated. 
 [/INST] Set 
$$A_n(a,b,c)=\left( \begin{matrix}
   a & b & 0 & \cdots  & \cdots  & 0  \\
   c & a & b & \cdots  & \cdots  & 0  \\
   0 & c & a & b & \cdots  & 0  \\
   \vdots  & \ddots  & \ddots  & \ddots  & \ddots  & \vdots  \\
   0 & \cdots  & \cdots  & c & a & b  \\
   0 & 0 & \cdots  & 0 & c & a  \\
\end{matrix} \right)_{n\times n}
$$
then
$$|A_{n+2}(a,b,c)|=a|A_{n+1}(a,b,c)|-bc\,|A_{n}(a,b,c)|$$
Indeed we can show
$$|A_n(a,b,c)|=\frac{1}{\sqrt{a^2-4bc}}\left[\left(\frac{a+\sqrt{a^2-4bc}}{2}\right)^{n+1}-\left(\frac{a-\sqrt{a^2-4bc}}{2}\right)^{n+1}\right]$$
Hint
set $x_{n}=|A_n(a,b,c)|$ then
$$x_{n+2}=a\,{x_{n+1}}-bc\,{x_{n}}$$
where
$x_1=a$ and $x_0=1$
we have 
$$\lambda_1=\frac{a+\sqrt{a^2-4bc}}{2}$$
and
$$\lambda_2=\frac{a-\sqrt{a^2-4bc}}{2}$$
$...$
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