Determinant involving recurrence Evaluate
$$\left| A \right| = \left| {\matrix{
   {x + y} & {xy} & 0 &  \cdots  &  \cdots  & 0  \cr 
   1 & {x + y} & {xy} &  \cdots  &  \cdots  & 0  \cr 
   0 & 1 & {x + y} &  \cdots  &  \cdots  & 0  \cr 
    \cdots  &  \cdots  &  \cdots  &  \cdots  &  \cdots  &  \vdots   \cr 
   0 &  \cdots  & 0 & 1 & {x + y} & {xy}  \cr 
   0 &  \cdots  & 0 & 0 & 1 & {x + y}  \cr 
 } } \right|$$
And show that $\det(A) = \frac{x^{n+1}-y^{n+1}}{x-y}$ if $x\ne y$ and $\det(A) = (n+1)x^n$ if $x=y$.
I actually was able to get this recurrence formula:  
$$D_n = (x+y)\cdot D_{n-1} + xy\cdot D_{n-2}$$
I tried to prove it by induciton, but the algebric calculation didn't bring me to the desired result.

The problem is that the recurrence formula should be 
$$D_n=(x+y)D_{n-1}-xyD_{n-2}$$ then  
$$D_n=\frac{x^n-y^n}{x-y}(x+y)-xy\frac{x^{n-1}-y^{n-1}}{x-y}=
\frac{(x^n-y^n)(x+y)-xy(x^{n-1}-y^{n-1})}{x-y}=
\frac{x^{n+1}-y^{n+1}+x^ny -xy^n-x^{n}y+xy^{n}}{x-y}=
\frac{x^{n+1}-y^{n+1}}{x-y}$$
