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.
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 $$\lambda_2=\frac{a-\sqrt{a^2-4bc}}{2}$$ $...$