TITLE: Find the value of special tridiagonal determinant
QUESTION [5 upvotes]: 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
 $$\begin{bmatrix}
 a_{n-1}+a_{n}& -a^{2}_{n-1}\\
 1& 0
\end{bmatrix}\begin{bmatrix}
 a_{n-2}+a_{n-1}& -a^{2}_{n-2}\\
 1& 0
\end{bmatrix}\cdots\begin{bmatrix}
 a_{2}+a_{3}& -a^{2}_{2}\\
 1& 0
\end{bmatrix}$$

REPLY [2 votes]: If we compute this determinant for $n=2$ and $n=3$ we get
\begin{align*}
D_2(a_0,a_1,a_2)&=a_0a_1+a_0a_2+a_0a_3\\
D_3(a_0,a_1,a_2,a_3)&=a_0a_1a_2+a_0a_1a_3+a_0a_2a_3+a_1a_2a_3
\end{align*}
This leads us to the conjecture that
$$D_n(a_0,a_1,\dots,a_n)=a_0a_1\cdots a_n\left(\frac1{a_0}+\frac1{a_1}+\dots+\frac1{a_n}\right).$$
We can prove this this by induction.
Inductive step. We assume that this is true for $n-1$. We have (using multilinearity)
$$
\begin{vmatrix}
 a_{0}+a_{1}&  a_{1}&  0&  0& \cdots&  0& \quad0\\
 a_{1}&  a_{1}+a_{2}&  a_{2}&  0&  \cdots&  0& \quad0\\
 0&  a_{2}&  a_{2}+a_{3}&  a_{3}&  \cdots&  0& \quad0\\
 \vdots&  \vdots&  \vdots&  \vdots&  &  \vdots& \quad\vdots\\
 0&  0&  0&  0&  \cdots&  a_{n-1}&   \quad a_{n-1}+a_{n}
\end{vmatrix} =
\begin{vmatrix}
 a_{0} &  a_{1}&  0&  0& \cdots&  0& \quad0\\
 0 &  a_{1}+a_{2}&  a_{2}&  0&  \cdots&  0& \quad0\\
 0&  a_{2}&  a_{2}+a_{3}&  a_{3}&  \cdots&  0& \quad0\\
 \vdots&  \vdots&  \vdots&  \vdots&  &  \vdots& \quad\vdots\\
 0&  0&  0&  0&  \cdots&  a_{n-1}&   \quad a_{n-1}+a_{n}
\end{vmatrix}+
\begin{vmatrix}
 a_{1}&  a_{1}&  0&  0& \cdots&  0& \quad0\\
 a_{1}&  a_{1}+a_{2}&  a_{2}&  0&  \cdots&  0& \quad0\\
 0&  a_{2}&  a_{2}+a_{3}&  a_{3}&  \cdots&  0& \quad0\\
 \vdots&  \vdots&  \vdots&  \vdots&  &  \vdots& \quad\vdots\\
 0&  0&  0&  0&  \cdots&  a_{n-1}&   \quad a_{n-1}+a_{n}
\end{vmatrix}
$$
The first determinant is equal to $a_0D_{n-1}(a_1,a_2,\dots,a_n)$.
The second determinant is equal to
$$\begin{vmatrix}
 a_{1}&  a_{1}&  0&  0& \cdots&  0&  0\\
 a_{1}&  a_{1}+a_{2}&  a_{2}&  0&  \cdots&  0&  0\\
 0&  a_{2}&  a_{2}+a_{3}&  a_{3}&  \cdots&  0&  0\\
 \vdots&  \vdots&  \vdots&  \vdots&  &  \vdots&  \vdots\\
 0&  0&  0&  0&  \cdots&  a_{n-1}&     a_{n-1}+a_{n}
\end{vmatrix}=
\begin{vmatrix}
 a_{1}&  a_{1}&  0&  0& \cdots&  0&  0\\
 0 & a_{2}&  a_{2}&  0&  \cdots&  0&  0\\
 0&  a_{2}&  a_{2}+a_{3}&  a_{3}&  \cdots&  0&  0\\
 \vdots&  \vdots&  \vdots&  \vdots&  &  \vdots&  \vdots\\
 0&  0&  0&  0&  \cdots&  a_{n-1}&     a_{n-1}+a_{n}
\end{vmatrix} = \dots =
\begin{vmatrix}
 a_{1}&  a_{1}&  0&  0& \cdots&  0&  0\\
 0 & a_{2}&  a_{2}&  0&  \cdots&  0&  0\\
 0 &  0 &  a_{3}&  a_{3}&  \cdots&  0&  0\\
 \vdots&  \vdots&  \vdots&  \vdots&  &  \vdots&  \vdots\\
 0&  0&  0&  0&  \cdots&  0 &   a_{n}
\end{vmatrix} = a_1a_2a_3\cdots a_n
$$ 
(In each step we subtracted a row from the following one. This does not change the determinant).
We can also get the value of the second determinant if we rewrite this matrix as a product
$$\begin{pmatrix}
a_1 &  a_1  & 0      &  \ldots & 0 \\
a_1 &a_1+a_2& a_2    &  \ldots & 0 \\
 0  &  a_2  & a_2+a_3&  \ldots & 0 \\
\vdots &    &        &   \ddots& \vdots\\
0 & \ldots &  \ldots       &   a_{n-1} & a_{n-1}+a_n
\end{pmatrix}=
\begin{pmatrix}
 1 & 0 & 0 & 0 & \ldots & 0 \\
 1 & 1 & 0 & 0 &  \ldots & 0 \\
 0 & 1 & 1 & 0 &  \ldots & 0 \\
\vdots &    &  \ddots  &  \ddots  &   & \vdots\\
0 & \ldots &   \ldots  & 1  &   1 & 0\\
0 & \ldots &   \ldots  & 0  &   1 & 1
\end{pmatrix}
\begin{pmatrix}
a_1 &  a_1  & 0  & 0 & \ldots & 0 \\
 0  &  a_2 & a_2 & 0 &  \ldots & 0 \\
 0  &   0  & a_3 & a_3 & \ldots & 0 \\
\vdots &    &    &    \ddots& \ddots &   \vdots\\
0 & \ldots & \ldots & 0   &   a_{n-1} & a_{n-1}\\
0 & \ldots &   \ldots  &  \ldots  &   0 & a_n
\end{pmatrix}$$
Now it remains to check that we indeed have 
$$\begin{multline}
a_0D_{n-1}(a_1,a_2,\dots,a_n) + a_1a_2a_3\cdots a_n =\\= 
a_0a_1a_2\cdots a_n\left(\frac1{a_1}+\frac1{a_2}+\dots+\frac1{a_n}\right) + a_1a_2a_3\cdots a_n =\\=
a_0a_1\cdots a_n\left(\frac1{a_0}+\frac1{a_1}+\dots+\frac1{a_n}\right).
\end{multline}
$$