What is the difference between closed form and opened form in z transform
One possible difference occurs for a pole outside of unit circle. Note that the original system is a truncated IIR filter. As such, it's an FIR filter and have no stability issue for any finite $a$.

However, the closed form representation includes a pole, and implements a pole zero cancellation. Eventhough theoretically (with infinite precision arithmetic) this would not pose a problem, for finite length arithmetic and for large values of $N$, and for $|a| &gt; 1$, you could end up in numerical issues.

Specifcally, the instantaneous dynamic range (the mantissa length of the floating point representation) would be large enough to be able to accurately represent the arithmetic difference indicated in the numerator. Indeed this could also be aproblem with the finite length implementation when $a^N$ is too large to be added to $ax[n-1]$, and that could happen when $|a|&gt;1$. So which one would be more sensitive to this numerical issue? You better test by yourself. Furthermore, the closed-from version could impose an additional problem for those $ a \approx 1$ which would not be the case for the FIR filter.

As an advantage of te closed form, note that the FIR form and IIR form requires the same amount of memory (to hold past values of input up to a delay of $N$ samples), however the closed form (truncated IIR) requires much less computation MACS compared to FIR version for large N.