With meager loss of generality, let's first assume that $A,C$ are diagonalizable so that $A=RDR^{-1},C=SES^{-1}$ where $R,S$ are invertible and $D,E$ are diagonal.

Then $$\sum_{k=0}^{N-1}A^kBC^k=\sum_{k=0}^{N-1}RD^kR^{-1}BSE^kS^{-1}$$

$$=R(\sum_{k=0}^{N-1}D^kR^{-1}BSE^k)S^{-1}.$$

Suppose now that $R^{-1}BS=(\alpha_{i,j})_{i,j}$ and that the diagonal entries in $D$ are $(\delta_1,\dots,\delta_n)$ and the diagonal entries in $E$ are $(\epsilon_1,\dots,\epsilon_m)$.

Then $$\sum_{k=0}^{N-1}A^kBC^k=R(\sum_{k=0}^{N-1}\delta_i^k\alpha_{i,j}\epsilon_j^k)_{i,j}S^{-1}$$

$$=R\cdot\Big(\frac{(\delta_i\epsilon_j)^N-1}{\delta_i\epsilon_j-1}\Big)_{i,j}\cdot S^{-1}.$$

It looks like there is a not quite as nice expression for our sum when $A,C$ are not diagonalizable, and $D,E$ are in Jordan normal form.