big determinant calculation I have found this exercise in a book, and having troubles solving it:
How to calculate this determinant?
$$\det\begin{pmatrix}
5 & 6 & 0 & 0      & 0      & \cdots & 0 \\
4 & 5 & 2 & 0      & 0      & \cdots & 0 \\
0 & 1 & 3 & 2      & 0      & \cdots & 0 \\
0 & 0 & 1 & 3      & 2      & \cdots & \vdots \\
0 & 0 & 0 & 1      & \ddots & \ddots & 0 \\
  &   &   &        & \ddots & 3      & 2 \\
0 & 0 & 0 & \cdots & 0      & 1      & 3 \\
\end{pmatrix}
_{n\times n}$$
Thanks!!

In general, when calculating a determinant of tridiagonal matrix of this form
$D_n=
\begin{vmatrix}
c& b & 0 & 0 & \ldots & 0 \\
a& c & b & 0 & \ldots & 0 \\
0& a & c & b & \ldots &  0 \\
\vdots & \ddots & \ddots & \ddots & \vdots \\
0 & \ldots & 0 & a & c & b \\
0 & \ldots & 0 & 0 & a & c
\end{vmatrix}$
you can use Laplace expansion twice to obtain recurrent relation.
$D_n=
\begin{vmatrix}
c& b & 0 & 0 & \ldots & 0 \\
a& c & b & 0 & \ldots & 0 \\
0& a & c & b & \ldots &  0 \\
\vdots & \ddots & \ddots & \ddots & \vdots \\
0 & \ldots & 0 & a & c & ab \\
0 & \ldots & 0 & 0 & a & c
\end{vmatrix}
\overset{(1)}= cD_{n-1} - b
\begin{vmatrix}
a & 0 & 0 & \ldots & 0 \\
a & c & b & \ldots & 0 \\
\vdots & \ddots & \ddots & \ddots & \vdots \\
0 & \ldots & a & c & ab \\
0 & \ldots & 0 & a & c
\end{vmatrix}
\overset{(2)}= cD_{n-1} - ab D_{n-2}$
The equality (1) is Laplace for the first row and the equality (2) is Laplace for the first column.

This is not exactly the situation from the original post, but we can modify it to a very similar situation  using reflection w.r.t vertical and horizontal axis. This can be done by reverting the order of rows and then the order of columns. Since this is the same as doing even number of exchanges of rows/columns, this does not change the determinant. So to solve the original question you simply have to solve the following recurrence:
$D_n=
\begin{vmatrix}
3& 1 & 0 & 0 & \ldots & 0 \\
2& 3 & 1 & 0 & \ldots & 0 \\
0& 2 & 3 & 1 & \ldots &  0 \\
\vdots & \ddots & \ddots & \ddots & \vdots \\
0 & \ldots & 0 & 2 & 5 & 4 \\
0 & \ldots & 0 & 0 & 6 & 5
\end{vmatrix}$
$D_3=
\begin{bmatrix}
  3 & 1 & 0 \\
  2 & 5 & 4 \\
  0 & 6 & 5
\end{bmatrix}
$
$D_4=
\begin{bmatrix}
  3 & 1 & 0 & 0\\
  2 & 3 & 1 & 0 \\
  0 & 2 & 5 & 4 \\
  0 & 0 & 6 & 5
\end{bmatrix}
$
$D_n=3D_{n-1}-2D_{n-2}$
