We will generalize Calvin Lin's answer a bit. Let
$$A_n = \begin{bmatrix} a & b & 0 & 0 & \cdots & 0\\ c & a & b & 0 & \cdots & 0\\ 0 & c & a & b & \cdots & 0\\ \vdots & \vdots & \vdots & \ddots & \cdots & 0\\ \vdots & \vdots & \vdots & \vdots & a & b\\ 0 & 0 & 0 & \cdots & c & a \end{bmatrix}.$$
We then have, by using Laplace expansion twice,
$$\det(A_n) = a \det(A_{n-1}) - bc \det(A_{n-2}).$$
Calling $\det(A_n) = d_n$ we have the following linear homogeneous recurrence relation:
$$d_n = a d_{n-1} - bc d_{n-2}.$$
The characteristic equation is
$$\begin{align}
x^2 - ax + bc = 0 & \implies \left(x - \frac{a}2 \right)^2 - \left(\frac{a}2 \right)^2 + bc = 0 \\
& \implies x = \frac{a \pm \sqrt{a^2-4bc}}2.
\end{align}$$
(This assumes a square roots exist. It's always the case in $\mathbb{C}$.)
Case 1: $a^2 - 4bc \neq 0$
In this case the characteristic polynomial has two distinct roots, so we have (for some constants $k_1$, $k_2$): $$d_n = k_1 \left( \dfrac{a + \sqrt{a^2-4bc}}2\right)^n + k_2 \left( \dfrac{a - \sqrt{a^2-4bc}}2\right)^n.$$
We have $d_1 = a$ and $d_2 = a^2 - bc$. We then get that $d_0 = 1$. Hence,
$$k_1 + k_2 = 1.$$
$$a (k_1 + k_2) + (k_1 - k_2)\sqrt{a^2-4bc} = 2a \implies k_1 - k_2 = \dfrac{a}{\sqrt{a^2-4bc}}.$$
Hence,
$$\begin{align}
k_1 & = \dfrac{a + \sqrt{a^2-4bc}}{2\sqrt{a^2-4bc}}, & k_2 & = -\dfrac{a-\sqrt{a^2-4bc}}{2\sqrt{a^2-4bc}}
\end{align}$$
And finally:
$$\color{red}{\det(A_n) = \dfrac1{\sqrt{a^2-4bc}} \left( \left( \dfrac{a + \sqrt{a^2-4bc}}2\right)^{n+1} - \left( \dfrac{a - \sqrt{a^2-4bc}}2\right)^{n+1}\right)}.$$
Plug in $a = 5$ and $b=c=2$ ($a^2 - 4 bc \neq 0$), to get
$$\det(A_n) = \frac{1}{3} ( 4^{n+1} - 1)$$
Case 2: $a^2 - 4bc = 0$
If the characteristic polynomial has a double root $x = a/2$, there exist constants $k_1$, $k_2$ such that:
$$d_n = (k_1 + k_2 n) \bigl(\frac{a}{2}\bigr)^n.$$
The initial conditions are $d_0 = 1$ and $d_1 = a$, thus:
$$\begin{align}
k_1 & = 1 & (k_1 + k_2) a = 2a
\end{align}$$
If $a = 0$, then $4bc = a^2$ implies either $b$ or $c$ is zero, and $d_n = 0$ for $n \ge 1$. Otherwise $$(k_1 + k_2) a = 2a \implies k_1 + k_2 = 2 \implies k_2 = 1.$$
And finally:
$$\color{red}{\det(A_n) = (n+1) \bigl(\frac{a}{2}\bigr)^n}.$$
You have a tridiagonal matrix. A tridiagonal matrix has a nice form for the determinant. If the diagonal is $a_1,a_2, \ldots$, above diagonal $b_1,b_2,\ldots$ and below diagonal is $c_1,c_2,\ldots$, then the determinant of the $n$-th principal minor (i.e. the matrix formed by the top left $n \times n$ submatrix) is given by the following recursion:
$f_1 = |a_1|, f_0 = 1, f_{-1} =0$
$f_n = a_n f_{n-1} - c_{n-1} b_{n-1} f_{n-2}$
Best Answer
While I don't quite understand why you only want to use integers, you can prove this by induction. Let $d_n$ be the determinant of your matrix of size $n \times n$. Then $d_1 = 2$, $d_2 = 3$. So assume that $d_k = k+1$ for all $k < n$. By Laplace expansion and the induction hypothesis, we get $$d_n = 2d_{n-1} - d_{n-2} = 2n - (n-1) = n+1$$ and we are done.