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?


Why not develop directly wrt the first column? The subindex means the order of the square matrix:
$$\begin{vmatrix}2 & 1 &  &  &  & ...\\
1 & 2 & 1 &  &  & ...\\
 & 1 & 2 & 1 &  & ...\\
 &  & 1 & 2 & 1 & ...\\
 &  &  & 1 & ... & 1\\
... & ... & ... & ... & 1 &2\end{vmatrix}_n=2\begin{vmatrix}2 & 1 &  &  &  & ...\\
1 & 2 & 1 &  &  & ...\\
 & 1 & 2 & 1 &  & ...\\
 &  & 1 & 2 & 1 & ...\\
 &  &  & 1 & ... & 1\\
... & ... & ... & ... & 1 &2\end{vmatrix}_{n-1}-$$
$$-\begin{vmatrix}1 & 0 &  &  &  & ...\\
1 & 2 & 1 &  &  & ...\\
 & 1 & 2 & 1 &  & ...\\
 &  & 1 & 2 & 1 & ...\\
 &  &  & 1 & ... & 1\\
... & ... & ... & ... & 1 &2\end{vmatrix}_{n-1}\stackrel{\text{Ind. Hyp.}}=2\left[(n-1)+1\right]-\left[(n-2)+1\right]=$$
$$=2n-n+1=n+1$$
as the last determinant is easily developed wrt the first entry and we get a $\;(n-2)\times(n-2)\;$ copy of the original matrix!
