Determinant involving recurrence

Question

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.

Answer 1

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

$$D_n=\frac{x^n-y^n}{x-y}(x+y)-xy\frac{x^{n-1}-y^{n-1}}{x-y}= \frac{(x^n-y^n)(x+y)-xy(x^{n-1}-y^{n-1})}{x-y}= \frac{x^{n+1}-y^{n+1}+x^ny -xy^n-x^{n}y+xy^{n}}{x-y}= \frac{x^{n+1}-y^{n+1}}{x-y}$$

Answer 2

Hint: compute $D_1 = x+y, D_2 = x^2 + xy + y^2$.

Then, prove that $D_n = x^n + x^{n-1}y + \dots + xy^{n-1} +y^n$ via induction.

Eventually, as $(x^n + x^{n-1}y + \dots + xy^{n-1} +y^n)(x-y) = x^{n+1} - y^{n+1}$, you are done.