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


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).



Solving recurrence relation for symmetric Toeplitz matrices determinant
$$T\_n=\frac{1}{2^{n+1}} \left(c\_-^n+c\_+^n+\left(c\_+^n-c\_-^n\right)\frac{c\_++c\_- }{c\_+-c\_-}\right),\;\;\text{with}\;\; c\_\pm=a\pm\sqrt{a^2-4 b^2}.$$


