Proving the determinant of a tridiagonal matrix with $-1, 2, -1$ on diagonal. Let $A_n$ denote an $n \times n$ tridiagonal matrix. 
$$A_n=\begin{pmatrix}2 & -1 & & & 0 \\ -1 & 2 & -1 & & \\ & \ddots & \ddots & \ddots & \\ & & -1 & 2 & -1 \\ 0 & & & -1 & 2 \end{pmatrix} \quad\text{for }n \ge 2$$
Set $D_n = \det(A_n)$
Prove that $D_n = 2D_{n-1} - D_{n-2}$ for $n \ge 4$. 

Expand with respect to the first column: you get
$$
D_n
=2 \times
\begin{vmatrix}2 & -1 & & & 0 \\ -1 & 2 & -1 & & \\ & \ddots & \ddots & \ddots & \\ & & -1 & 2 & -1 \\ 0 & & & -1 & 2 \end{vmatrix}_{(n-1)}
- (-1) \times
\begin{vmatrix}-1 & 0 & & & 0 \\
-1 & 2 & -1 & & 0 \\
 & \ddots & \ddots & \ddots & \\ & & -1 & 2 & -1 \\ 0 & & & -1 & 2 \end{vmatrix}_{(n-1)}
$$
Now expand according to the first line the second determinant: you get
$$
=2 \times
\begin{vmatrix}2 & -1 & & & 0 \\ -1 & 2 & -1 & & \\ & \ddots & \ddots & \ddots & \\ & & -1 & 2 & -1 \\ 0 & & & -1 & 2 \end{vmatrix}_{(n-1)}
- (-1)^2\times
\begin{vmatrix}2 & -1 & & & 0 \\ -1 & 2 & -1 & & \\ & \ddots & \ddots & \ddots & \\ & & -1 & 2 & -1 \\ 0 & & & -1 & 2 \end{vmatrix}_{(n-2)}
$$
Conclusion:
$$
D_n = 2D_{n-1} - D_{n-2}
$$
