Let prove the theorem. Suppose the determinant of tri-diagonal matrix as $\Delta_{n}$, and operate the following calculation.
$$
\begin{align}
\Delta_{n}=& \det
\begin{bmatrix}
a_{1} & b_{1} & 0 & \cdots & 0 & 0 & 0 \\
c_{1} & a_{2} & b_{2} & \ddots & \vdots & \vdots & \vdots \\
0 & c_{2} & a_{3} & \ddots & a_{n-2} & b_{n-2} & 0 \\
\vdots & \vdots & \vdots & \ddots & c_{n-2} & a_{n-1} & b_{n-1} \\
0 & 0 & 0 & \cdots & 0 & c_{n-1} & a_{n}
\end{bmatrix}
\\ \\ =& \det
\begin{bmatrix}
a_{1} & b_{1} & 0 & \cdots & 0 & 0 & 0 \\
c_{1} & a_{2} & b_{2} & \ddots & \vdots & \vdots & \vdots \\
0 & c_{2} & a_{3} & \ddots & a_{n-2} & b_{n-2}-c_{n-1}\frac{b_{n-1}}{a_{n-1}} & 0 \\
\vdots & \vdots & \vdots & \ddots & c_{n-2} & a_{n-1}-c_{n-1}\frac{b_{n-1}}{a_{n-1}} & 0 \\
0 & 0 & 0 & \cdots & 0 & c_{n-1} & a_{n}
\end{bmatrix}
\\ \\ =& a_{n} \Delta_{n-1} - b_{n-1}c_{n-1} \Delta_{n-2}
\end{align}
$$
Based on this formula, it can be described as below using matrix and vector product. Then, we try to estimate $\Delta_{n}$.
$$
\begin{align}
\begin{bmatrix}
\Delta_{n} \\
\Delta_{n-1}
\end{bmatrix}
=&
\begin{bmatrix}
a_{n} & -b_{n-1}c_{n-1} \\
1 & 0
\end{bmatrix}
\begin{bmatrix}
\Delta_{n-1} \\
\Delta_{n-2}
\end{bmatrix}
\\ =&
\prod_{k=4}^n
\begin{bmatrix}
a_{n+4-k} & -b_{n-k+3}c_{n-k+3} \\
1 & 0
\end{bmatrix}
\begin{bmatrix}
\Delta_{3} \\
\Delta_{2}
\end{bmatrix}
\end{align}
$$
This problem's case, these elements are identity each diagonal factors like $a_{i}=5$ $b_{i}=3$, $c_{i}=2$. Therefore this equation can be simplified as follows.
$$
\begin{bmatrix}
\Delta_{n} \\
\Delta_{n-1}
\end{bmatrix}
=
\begin{bmatrix}
5 & -6 \\
1 & 0
\end{bmatrix}
^{n-3}
\begin{bmatrix}
65 \\
19
\end{bmatrix}
$$
After that, we get the eigenvalues, eigenvectors and diagonalization of the matrix.
$$
\begin{align}
\begin{bmatrix}
\Delta_{n} \\
\Delta_{n-1}
\end{bmatrix}
=&
\begin{bmatrix}
3 & 2 \\
1 & 1
\end{bmatrix}
\begin{bmatrix}
3^{n-1} & 0 \\
0 & 2^{n-1}
\end{bmatrix}
\begin{bmatrix}
3 & 2 \\
1 & 1
\end{bmatrix}
^{-1}
\begin{bmatrix}
65 \\
19
\end{bmatrix}
\\ =&
\begin{bmatrix}
3^n-2^n & 3 \cdot 2^{n-2} -2 \cdot 3^{n-2} \\
3^{n-3} - 2^{n-3} & 3 \cdot 2^{n-3} - 2 \cdot 3^{n-3}
\end{bmatrix}
\begin{bmatrix}
65 \\
19
\end{bmatrix}
\end{align}
$$
Eventually, $\Delta_{n}$ is
$$
\begin{align}
\Delta_{n}=&
65(3^{n-2}-2^{n-2})+19(3 \cdot 2^{n-2} -2 \cdot 3^{n-2}) \\ =&
3^{n+1} - 2^{n+1}
\end{align}
$$
