Solving a $n\times n$ determinant. $$
\left|
\begin{array}{cccccc}
3&2&0&0&0&\ldots\\
1&3&1&0&0&\ldots\\
0&2&3&2&0&\ldots\\
0&0&1&3&1&\ldots\\
\vdots&\vdots&\vdots&\vdots&\vdots&\ddots
\end{array}
\right|
$$
I solved the determinant by realising it's matrix is tridiagonal and using the continuant, and have gotten a recursive formula $D_n = 2^{n+1} -1$, if $D_n$ is a determinant with respect to $n$. I'm having trouble solving the determinat using expansion on rows/columns, since this is how we've done similar ones in class. Appreciate any responses.

If you develop along the first column, you get
$$\begin{align*} D_n &= 3\begin{vmatrix}3 & 1 & 0 & 0 & \cdots \\
2 & 3 & 2 & 0 & \cdots \\
0 & 1 & 3 & 1 & \dots \\
\vdots & \vdots & \vdots & \vdots & \ddots  \end{vmatrix} - 
\begin{vmatrix} 2 & 0 & 0 & 0 & \cdots \\
2 & 3 & 2 & 0 & \cdots \\
0 & 1 & 3 & 1 & \dots \\
\vdots & \vdots & \vdots & \vdots & \ddots 
 \end{vmatrix} \\
&= 3D_{n-1} - 2\begin{vmatrix} 
3 & 2 & 0 & \cdots \\
1 & 3 & 1 & \dots \\
\vdots & \vdots & \vdots & \ddots 
 \end{vmatrix} \\
&= 3D_{n-1} - 2D_{n-2}
\end{align*}$$
where the $D_{n-1}$ term comes from the fact that the determinant of a matrix is the same as the determinant of its transpose (you really get the transposed determinant of $D_{n-1}$, but that is just $D_{n-1}$).
Now this is just a linear recursion formula, which you can solve using your favourite method, or you can just observe that $D_n = 2^{n+1}-1$ satisfies this formula.
