Suppose I'm practicing penalties. Initially my goalscoring probability is $p$. As I miss a chance, the probability increases by constant value $q$. So the probability that I score my first goal at $n$-th take is
\begin{align*} p_n &= [p+(n-1)q] \prod_{k=0}^{n-2}{(1-p-kq)} \\ &=nq^n\left[\frac{p}{q}+(n-1)\right]\prod_{k=0}^{n-2}{\left( \frac{1-p}{q}-k \right)} \end{align*}
And my first goal will arrive at
$$E(n)=\sum_{k=1}^{\lfloor(1-p)/q\rfloor} {kp_k}$$
But how can I sum this? Some other assumptions that may simplify the calculation without trivializing the problem could be made.