Matrix eigenvalues Consider the matrix $$A_n=\begin{bmatrix}
 a & b & 0 & 0 & 0 & \dots & 0 & 0 & 0 \\
 c & a & b & 0 & 0 & \dots & 0 & 0 & 0 \\
 0 & c & a & b & 0 & \dots & 0 & 0 & 0 \\
 0 & 0 & c & a & b & \dots & 0 & 0 & 0 \\
 0 & 0 & 0 & c & a & \dots & 0 & 0 & 0 \\
 \vdots & \vdots & \vdots & \vdots & \vdots & \ddots & \vdots & \vdots & \vdots \\
 0 & 0 & 0 & 0 & 0  & \dots & a & b & 0 \\
 0 & 0 & 0 & 0 & 0  & \dots & c & a & b \\
 0 & 0 & 0 & 0 & 0  & \dots & 0 & c & a  
 \end{bmatrix}_{n\times n}$$
The matrix with $a=2$ and $b=c=-1$ is encountered in finite difference discretization of $u_{xx}.$
(a) If $D_n = \det(A_n),$ show that $D_n = aD_{n-1}-bcD_{n-2}.$
(b) Solve the recurrence analytically to obtain $D_n$ as a function of $n.$ (and ofcourse $D_n$ will also depend on $a, b, c.$)
(c) Obtain the eigenvalues of $A_n.$ (Hint: Replace $a$ by $a-\lambda$)$$ $$
$$ $$(a)Part can be shown easily by just simple Laplace expansion.
(b)We see that $D_0=1, D_1=a$.
Let $D_n=r^n$ be a solution of the recurrence relation \begin{equation}
D_n=aD_{n-1}-bcD_{n-2}
\end{equation}
Then characteristic equation corresponding to (1) \begin{alignat*}{3}
&\quad & r^n-ar^{n-1}+bcr^{n-2} &=0
\\&\implies &r^2-ar+bc &=0
\\&\implies &r_1=\tfrac{a-\sqrt{a^2-4bc}}{2}, r_2 &=\tfrac{a+\sqrt{a^2-4bc}}{2}
\end{alignat*}$ $
Case 1: $a^2-4bc=0$
$r_1=r_2=\frac{a}{2}$
General solution of (1) :
$D_n=(C_1+nC_2)(\frac{a}{2})^n$, where $C_1$ and $C_2$ are arbitrary constants. 
For $n=0$, we get $C_1=D_0=1$.
For $n=1$, we get $(C_1+C_2)\frac{a}{2}=D_1=a\implies C_2=1$  
Hence $D_n=(1+n)(\frac{a}{2})^n$ 
$$ $$Case 2: $a^2-4bc\neq0$
General solution of (1) :
$D_n=C_1r_1^n+C_2r_2^n$, with where $C_1$ and $C_2$ are arbitrary constants.
For $n=0$, we get $C_1+C_2=D_0=1$
For $n=1$, we get $C_1r_1+C_2r_2=D_1=a\implies (C_1+C_2)\frac{a}{2}+(C_2-C_1)\frac{\sqrt{a^2-4bc}}{2}=a  
\implies 2C_2-1=\frac{a}{\sqrt{a^2-4bc}}  
\implies C_2=\frac{r_2}{\sqrt{a^2-4bc}}$
$\therefore C_1=\frac{-r_1}{\sqrt{a^2-4bc}}$  
Hence $D_n=\frac{r_2^{n+1}-r_1^{n+1}}{\sqrt{a^2-4bc}}=\frac{1}{2^{n+1}\sqrt{a^2-4bc}}[(a+\sqrt{a^2-4bc})^{n+1}-(a-\sqrt{a^2-4bc})^{n+1}]$ $$------------------------------------$$
I have done this far, but I'm stuck now.
Is there any simpler expression for $D_n$?
How to obtain eigenvalues, if we consider replacing $a$ by $a-\lambda$?

You got questions (a) and (b) already. For (c) the eigenvalues, you need the characteristic equation $\det (A_n - \lambda I) = 0$. This is the same as $D_n = \det (A_n) = 0$, if in there $a$ is replaced by $a-\lambda$. From your result,
$$
0 = D_n({\rm a  \; replaced}) =\frac{1}{2^{n+1}\sqrt{(a-\lambda)^2-4bc}}[(a-\lambda+\sqrt{(a-\lambda)^2-4bc})^{n+1}-(a-\lambda-\sqrt{(a-\lambda)^2-4bc})^{n+1}]$$
i.e. for $(a-\lambda)^2-4bc \ne 0$ (denominator $\ne 0$) we have
$$
(a-\lambda+\sqrt{(a-\lambda)^2-4bc})^{n+1}=(a-\lambda-\sqrt{(a-\lambda)^2-4bc})^{n+1}$$
or (be careful to obtain all the roots in $\sqrt[n+1]{1}$)
$$
a-\lambda+\sqrt{(a-\lambda)^2-4bc}=(a-\lambda-\sqrt{(a-\lambda)^2-4bc})\exp{(2\pi i k/(n+1))}$$
for $k = 0,1,\cdots,n$. Indexing the $\lambda_k$ with $k$, you get the results.
E.g.
$\lambda_0 = a \pm 2 \sqrt{bc}$ but that contradicts the above condition $(a-\lambda)^2-4bc \ne 0$. 
Since $k=0$ is excluded, the general result is $\lambda_k = a \pm 2 \sqrt{bc} \cos(\frac{\pi k}{n+1})$ for $k = 1,2,\cdots,n$. Since $\cos(x) = -\cos(\pi -x)$, one of the two signs in $\pm$ actually suffices:  $\lambda_k = a - 2 \sqrt{bc} \cos(\frac{\pi k}{n+1}) = a + 2 \sqrt{bc} \cos(\pi - \frac{\pi k}{n+1})  \\= a + 2 \sqrt{bc} \cos(\frac{\pi (n+1-k)}{n+1})  =a + 2 \sqrt{bc} \cos(\frac{\pi m}{n+1}) $ 
where $1 \le m = n+1-k \le n$, so the results with the positive sign are reproduced with the same range of the counting variable $m$. 
We show the general result by plugging  $\lambda_k = a + 2 \sqrt{bc} \cos(\frac{\pi k}{n+1})$  (plugging $\lambda_k = a - 2 \sqrt{bc} \cos(\frac{\pi k}{n+1})$  works as well )  in the determining equation for the eigenvalues. Indeed
$$
2 \sqrt{bc} [\cos(\frac{\pi k}{n+1})+\sqrt{\cos^2(\frac{\pi k}{n+1})-1}]=2 \sqrt{bc} [\cos(\frac{\pi k}{n+1})-\sqrt{\cos^2(\frac{\pi k}{n+1})-1}]\exp{(2\pi i k/(n+1))}$$
or
$$
\cos(\frac{\pi k}{n+1}) + i \sin(\frac{\pi k}{n+1})=[\cos(\frac{\pi k}{n+1}) - i \sin(\frac{\pi k}{n+1})]\exp{(2\pi i k/(n+1))}$$
or
$$
\exp{(\pi i k/(n+1))}=\exp{(-\pi i k/(n+1))}\exp{(2\pi i k/(n+1))}$$
which is an identity.
By the way, technically, what you have here is a tridiagonal Toeplitz matrix, where references can be found easily. 
