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:
$$\color{red}{\det(A_n) = (n+1) \bigl(\frac{a}{2}\bigr)^n}.$$
The matrix $\Sigma$ can be written as
$$
\Sigma=\lambda I+(1-\lambda)E,
$$
where $E=ee^T$, $e=[1,1,\ldots,1]^T$. If $V$ is an orthogonal matrix such that $V^TEV=D$ is diagonal, then
$$\tag{1}
V^T\Sigma V=\lambda I+(1-\lambda)D
$$
is the diagonalization of $\Sigma$.
The only thing which remains to find are the matrices $D$ and $V$.
Since $E$ has rank one, it has one nonzero eigenvalue and $n-1$ zero eigenvalues and we can chose an orthonormal set of eigenvectors since $E$ is symmetric.
We have
$$
Ee=ne,
$$
hence $n$ is the only nonzero eigenvalue of $E$ with the constant eigenvector $v_1=e/\sqrt{n}$ (normalized to $\|v_1\|=1$).
The other eigenvectors $v_2,\ldots,v_n$ of $E$ corresponding to the zero eigenvalues can be any set of orthonormal eigenvectors orthogonal to $e$.
Indeed, if $0\neq v\perp e$, then
$$
Ev=(ee^T)v=e(e^Tv)=0.
$$
Since the eigenvalues of $\Sigma$ are $\lambda$ plus $(1-\lambda)$ times the eigenvalues of $E$, the eigenvalues of $\Sigma$ are $\lambda$ with the multiplicity $n-1$ and the simple eigenvalue $\lambda+(1-\lambda)n$.
Note that there are infinitely many bases of the eigenspace of $E$ corresponding to the zero eigenvalues. You can chose any $n-1$ orthonormal vectors orthogonal to a constant vector.
Best Answer
Hint. In general, let $d_0=d_1=1$ and let $(a_k)_{k=1,2,\ldots}$ be any sequence of numbers. For every $n\ge2$, denote by $d_n$ the determinant of the $n\times n$ Toeplitz-Hessenberg matrix $$ \begin{pmatrix} a_1 &1 &0 &0 &\cdots &0\\ a_2 &a_1 &1 &0 &\cdots &0\\ a_3 &a_2 &a_1 &1 &\cdots &0\\ \vdots &\vdots &\vdots &\ddots &\ddots &\vdots\\ a_{n-1} &a_{n-2} &a_{n-3} &\cdots &a_1 &1\\ a_n &a_{n-1} &a_{n-2} &a_{n-3} &\cdots &a_1 \end{pmatrix}. $$ If one expands the determinant by the first column, one obtains $$ d_n=-\sum_{k=1}^n(-1)^ka_kd_{n-k}. $$