Suppose we want to find a closed formula for the sum of the $\displaystyle N $ squares. Start with the formula $\displaystyle \sum_{n=0}^{N} x^n = \frac{x^{N+1}-1}{x-1} $. Then do we do the following:

$\displaystyle \sum_{n=0}^{N} (xD)^n = \frac{(xD)^{N+1}-1}{(xD)-1} $

e.g. apply the $\displaystyle (xD)^2 $ operator?