= Solution
Let $A_n$ denote the given <tridiagonal matrix> and put $D_n=\det A_n$, including $D_0=1$ for the empty <determinant>. The starting values are
$$
D_0=1,\qquad D_1=1,\qquad D_2=\det\begin{pmatrix}1&1\\1&1\end{pmatrix}=0.
$$
For $n\ge3$, apply the following <elementary row operations> in the indicated order, always using the updated rows:
$$
R_2\leftarrow R_2-R_1,\qquad R_3\leftarrow R_3-R_2,\qquad R_2\leftrightarrow R_3.
$$
The first three rows now begin
$$
\begin{pmatrix}
1&1&0&0&\cdots\\
0&1&0&1&\cdots\\
0&0&1&0&\cdots
\end{pmatrix},
$$
with the fourth column omitted if $n=3$. When $n\ge4$, also use $R_4\leftarrow R_4-R_3$. The resulting <block upper triangular matrix> has a $3\times3$ <upper triangular matrix> with diagonal entries all one in its top left corner, and $A_{n-3}$ in its bottom right corner. All <elementary row operations> preserved the <determinant> except the single row interchange. Therefore
$$
\boxed{D_n=-D_{n-3}\quad(n\ge3).}
$$
Applying this <recurrence relation> to the three starting values proves by <mathematical induction> that
$$
\boxed{D_n=\begin{cases}
(-1)^k,&n=3k,\\
(-1)^k,&n=3k+1,\\
0,&n=3k+2.
\end{cases}}
$$
Equivalently, the sequence is periodic with period six, with values $1,1,0,-1,-1,0$ starting at $n=0$. A compact closed form is $D_n=\sin((n+1)\pi/3)/\sin(\pi/3)$.
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