I'm trying to find a closed form for $\sum_{i=0}^{n}r^{ai^2+bi+c}$.
I'm thinking I might be able to use the geometric sum for a polynomial of one less degree, $\sum_{i=0}^{n}r^{ai+b} = \frac{r^{b}}{1-r^{a}}(1-(r^{a})^{n+1})$, because the second degree polynomial contains the first degree polynomial $r^{ai^2+bi+c}=r^{i(ai+b)}r^{c}$.
I've also tried completing the square:
$\sum_{i=0}^{n}r^{ai^2+bi+c} = r^{c-\frac{b^{2}}{4a}}\sum_{i=0}^{n}r^{a(i+\frac{b}{2a})^{2}}$
But then it seems like I need to solve a sum similar to $\sum_{j=0}^{n}r^{j^2}$