Determinant of a n x n Matrix - Main Diagonal = 2, Sub- & Super-Diagonal = 1 I'm stuck with this one - Any tips?
The Problem:

Let $n \in \mathbb{N}.$ The following $n \times n$ matrix:
$$A =  \left( \begin{array}{ccc}
2 & 1 &  &  &  & ...\\
1 & 2 & 1 &  &  & ...\\
 & 1 & 2 & 1 &  & ...\\
 &  & 1 & 2 & 1 & ...\\
 &  &  & 1 & ... & 1\\
... & ... & ... & ... & 1 &2
\end{array} \right) $$
e.g. for the main diagonal = 2, the sub- and superdiagonal = 1 . 
Show with Induction that $\det(A) = n + 1$.
My solution approach: Laplace Expansion starting with the 2 in the bottom right corner $(a_{n+1,n+1})$. But how can I tell wether its positive or negative? After that I'm stuck with the 1 $(a_{n,n+1})$(the sub matrix matrix becomes ugly and I get a recursively solution). How can I formalize this in a proper manner?


$A_n=2A_{n-1}-A_{n-2}$, therefore $A_n=a\cdot n\cdot1^n+b$. The coefficients $a=1,b=1$ can be computed from $A_2=2$ and $A_3=3$.
