Given matrix $$ A=\begin{bmatrix} x+y&xy&0& .&.&. &0\\ 1&x+y&xy&0& .&.&0 \\ 0&1&x+y&xy&.&.&. \\ .&.&.&.&.&.&. \\ .&.&.&.&.&.&0 \\ .&.&.&.&.&.&xy \\ 0&.&.&.&0&1&x+y \end{bmatrix} $$
prove by induction that $$|A|=\frac{x^{n+1}-y^{n+1}}{x-y}$$ $x \neq y$, $A_{n \times n}$.
The determinant expression appears to be Dickson polynomial of second kind.
Let $D_n$ be the determinant of $A_n$. We can see that the appropriate recurrence relation is $$D_n=(x+y)D_{n-1}-xyD_{n-2}$$
Base cases: $$D_1=x+y=\frac{x^2-y^2}{x-y}$$
$$ D_2=(x+y)^2-xy=x^2+xy+y^2=\frac{x^3-y^3}{x-y} $$
Suppose that $$D_n=(x+y)D_{n-1}-xyD_{n-2}$$ Then we need to prove that $$D_{n+1}=(x+y)D_{n}-xyD_{n-1}$$
Which can be developed as: $$ D_{n+1}=(x+y)((x+y)D_{n-1}-xyD_{n-2})-xyD_{n-1}= $$ $$ =(x+y)^2D_{n-1}-xy(x+y)D_{n-2}-xyD_{n-1}= $$ $$ =(x^2+xy+y^2)D_{n-1}-xy(x+y)D_{n-2}= $$ $$ =\frac{x^3-y^3}{x-y}D_{n-1}-xy(x+y)D_{n-2} $$
I tried doing this up to $D_{n-6}$ in order to get any insights into possible simplification but I'm pretty stuck.