<s>[INST] Find the value of special tridiagonal determinant Let $A_{n}$ be the following  tridiagonal determinant  of order $n:$
\begin{vmatrix}
 a_{0}+a_{1}&  a_{1}&  0&  0& \cdots&  0& \quad0\\ 
 a_{1}&  a_{1}+a_{2}&  a_{2}&  0&  \cdots&  0& \quad0\\ 
 0&  a_{2}&  a_{2}+a_{3}&  a_{3}&  \cdots&  0& \quad0\\ 
 \vdots&  \vdots&  \vdots&  \vdots&  &  \vdots& \quad\vdots\\ 
 0&  0&  0&  0&  \cdots&  a_{n-1}&   \quad a_{n-1}+a_{n}
\end{vmatrix}
Find the value of $A_{n}.$
As we know,$$A_{n}=(a_{n-1}+a_{n})A_{n-1}-a^{2}_{n-1}A_{n-2}\Longrightarrow$$ $$\begin{bmatrix}
 A_{n}\\
 A_{n-1}
\end{bmatrix}=\begin{bmatrix}
 a_{n-1}+a_{n}& -a^{2}_{n-1}\\
 1& 0
\end{bmatrix}\begin{bmatrix}
 A_{n-1}\\
 A_{n-2}
\end{bmatrix}=$$$$\begin{bmatrix}
 a_{n-1}+a_{n}& -a^{2}_{n-1}\\
 1& 0
\end{bmatrix}\begin{bmatrix}
 a_{n-2}+a_{n-1}& -a^{2}_{n-2}\\
 1& 0
\end{bmatrix}\cdots\begin{bmatrix}
 a_{2}+a_{3}& -a^{2}_{2}\\
 1& 0
\end{bmatrix}\begin{bmatrix}
 A_{2}\\
 A_{1}
\end{bmatrix}.$$
But it is  not easy to deal with
 $$\begin{bmatrix}
 a_{n-1}+a_{n}& -a^{2}_{n-1}\\
 1& 0
\end{bmatrix}\begin{bmatrix}
 a_{n-2}+a_{n-1}& -a^{2}_{n-2}\\
 1& 0
\end{bmatrix}\cdots\begin{bmatrix}
 a_{2}+a_{3}& -a^{2}_{2}\\
 1& 0
\end{bmatrix}$$
 [/INST] Here is an observation that may save some labor later on. At first glance, the above recurrence relation only makes sense for $n>2$. But for $n=2$, the recurrence requires $A_2=(a_2+a_1)A_1-a_{1}^2 A_0$; since But $A_2=(a_0+a_1)(a_1+a_2)-a_1^2$ and $A_1=(a_1+a_0)$, the recurrence will be valid at $n=2$ if $A_0=1$. Similarly, for $n=1$ the recurrence would give $A_1 = (a_1+a_0)A_0-a_{-1}^2 A_{-1}$ which is valid if $A_{-1}=0$. Introducing $T_n=\begin{bmatrix} a_{n-1}+a_n & -a_{n-1}^2 \\ 1 & 0 \end{bmatrix}$ for convenience, one then has
$$\begin{bmatrix} A_n \\ A_{n-1} \end{bmatrix}
=T_n
\begin{bmatrix} A_{n-1} \\ A_{n-2} \end{bmatrix}
=\cdots =e_1^T (T_n T_{n-1}\cdots T_1)\begin{bmatrix} A_{0} \\ A_{-1} \end{bmatrix}=(T_n T_{n-1}\cdots T_1)e_1.$$ So obtaining $A_n$ amounts to finding the upper-left element of $T_n T_{n-1}\cdots T_1$.
</s>