While playing with Arithmetico-Geometric progression formula(i.e
$$\sum_{k=1}^{n}(a+(k-1)d)y^{k-1} = \frac{a-[a+(n-1)d]y^n}{1-y} +\frac{1-y^{n-1}}{(1-y)^2}yd$$
I realized it could be generalized as:
If $x \ne y$,
$$\sum_{k=0}^{n}(a+kd)x^{n-k}y^k = \frac{ax^{n+1}-(a+nd)y^{n+1}}{x-y} +\frac{x^n-y^n}{(x-y)^2}xyd$$
Though, I can't prove this yet.
QUESTION: If per chance the claim is correct, Is it safe to say, if $-1\le x \le 1$, $-1\le y \le 1$, that $$\sum_{k=0}^{\infty}(a+kd)x^{n-k}y^k = 0$$