Determinant of tridiagonal (banded) matrix Struggling with homework. I know that this is a banded matrix ( bandwidth = 3) but I don't know how to approach computing the determinant. I tried to compute it with matlab for n= 3,..,7 but I didn't find a specific pattern. 

 A: There are couple of things you can try. One posible induction is to make an induction on $n$. Let then the desired determinant be $d_n$ (where I explicitly put the index $n$). Then your matrix has general form
$$
\begin{array}{ccc}
1&1&0&\dots&0 \\
1&2&2&\dots &0\\
\vdots &\vdots&\vdots &\ddots&\vdots\\
0&\dots&0&n-1&n-1\\
0&\dots&0&n-1&n\\
\end{array}
$$
Then taking the determinant (expanding from the last row) we get $n\times d_{n-1}$ from the $n$ at low right corner and we get $-1\times(n-1)^2\times d_{n-2}$ from the other non-zero entry in such row.
Then $d_n=n d_{n-1}-(n-1)^2d_{n-2}$.
Lastly $d_1=1,d_2=1$. hence you can compute it recursively. I am trying to find some explicit expression for $d_n$ if I succeed I will post it.
Adition
As I was saying in the comment with $b_n=\frac{d_n}{nd_{n-1}}$ we get
$$
b_{n}=1-\frac{n-1}{n}\frac{1}{b_{n-1}}
$$
This sequence is a little simpler as it only involves the previous term (not the two previous as for $d_n$) and it has a chance of having some fix points as $\frac{n-1}{n}\to 1$. When you plot it you get something like 

In which you basically see it has three different behaviours. Ploting coloring $b_n$ by the value of $n\mod 3$ you get each of the branches, as you can see in the next figure.

Therefore one may think that the the series one could say something about is $b_{3n},b_{3n+1},b_{3n+2}$. 
I am not sure if that helps because the behavior of the series is really weird as you can see. By the way the fixed points for $b_n$ are $e^{\pm i\pi/3}$ I think the three branches stuff has to with the $3$ you have there, but not quite sure. I am still looking into it, so maybe I will get something else
