$$\text{Let } T_n=\begin{Bmatrix} a & b & \boldsymbol{0} \\ b & a & \ddots \\ \boldsymbol{0} & \ddots & \ddots \end{Bmatrix}\text{ a symmetric tridiagonal Toeplitz matrix of size }n\text{.}$$

I need to find its determinant in terms of $n$.

We know $\det T_1=a$ and $\det T_2=a^2-b^2$.

We can prove that, $\forall n\in\mathbb{N}^*$,
$$\det T_{n+1}=a\det T_n-b^2\det T_{n-1}$$

But I would like to get an absolute formula (with no recurrence relation).