$F_q$ (explanation): There are also other finite fields $F_q$, constructed in a somewhat more elaborate way, where the total number of elements is some power of a prime: $q=p^m$. Let me just give the simplest example, namely the case $q=4=2^2$. Here we can label the different elements as $0,1,\omega,\omega^2$ where $\omega^3=1$ and where each element $x$ is subject to $x+x=0$. This slightly extends the multiplicative group of complex numbers 1, o, o2 that are cube roots of unity (described in §5.4 and mentioned in §5.5 as describing the ‘quarkiness’ of strongly interacting particles). To get $F_4$, we just adjoin a zero ‘0’ and supply an ‘addition’ operation for which $x+x=0$. In the general case $F_{p^m}$, we would have $x+x+x...=0$, where the number of xs in the sum is p
.... later,
We find that $P^n(F_q)$ has exactly $\sum_i^n q^i = \frac{q^{n+1}-1}{q-1}$ points.
Sect 16.2, Road to Reality by Roger Penrose
Could it be explained how the number of points of the $n$th projective space of $F_q$ was found? Where did the geometric series come from?
p.s: I understand how projective spaces work, if that's relevant at all.