Can this equation simplify to the property of a sum of a geometric series, such as $ \frac{1}{1-r} $
$$\sum_{y=1}^{\infty}y^2q^{y}p$$
I understand that
$$\sum_{y=1}^{\infty}yq^{y} = q \sum_{y=0}^{\infty}(y-1)q^{y-1} = q \frac{d}{dq} \sum_{y=0}^{\infty}q^{y} = q \frac{d}{dq}\frac{1}{1-q} $$
\begin{align} \sum_y y^2 q^y p &= p \sum_y (y(y-1)+y) q^y \\ &= p q^2 \sum_y y(y-1) q^{y-2} + p q \sum_y y q^{y-1} \\ &= p q^2 \frac{d^2}{dq^2} \sum_y q^y + p q \frac{d}{dq} \sum_y q^y \\ &= p q^2 \frac{d^2}{dq^2} \frac{1}{1-q} + p q \frac{d}{dq} \frac{1}{1-q} \\ &= p q^2 \frac{2}{(1-q)^3} + p q \frac{1}{(1-q)^2} \\ &= \frac{pq(2 q +(1-q))}{(1-q)^3} \\ &= \frac{pq(1+ q)}{(1-q)^3} \\ \end{align}
