Let $S$ be a set of size $n$. There is an easy way to count the number of subsets with an even number of elements. Algebraically, it comes from the fact that
$\displaystyle \sum_{k=0}^{n} {n \choose k} = (1 + 1)^n$
while
$\displaystyle \sum_{k=0}^{n} (-1)^k {n \choose k} = (1 - 1)^n$.
It follows that
$\displaystyle \sum_{k=0}^{n/2} {n \choose 2k} = 2^{n-1}$.
A direct combinatorial proof is as follows: fix an element $s \in S$. If a given subset has $s$ in it, add it in; otherwise, take it out. This defines a bijection between the number of subsets with an even number of elements and the number of subsets with an odd number of elements.
The analogous formulas for the subsets with a number of elements divisible by $3$ or $4$ are more complicated, and divide into cases depending on the residue of $n \bmod 6$ and $n \bmod 8$, respectively. The algebraic derivations of these formulas are as follows (with $\omega$ a primitive third root of unity): observe that
$\displaystyle \sum_{k=0}^{n} \omega^k {n \choose k} = (1 + \omega)^n = (-\omega^2)^n$
while
$\displaystyle \sum_{k=0}^{n} \omega^{2k} {n \choose k} = (1 + \omega^2)^n = (-\omega)^n$
and that $1 + \omega^k + \omega^{2k} = 0$ if $k$ is not divisible by $3$ and equals $3$ otherwise. (This is a special case of the discrete Fourier transform.) It follows that
$\displaystyle \sum_{k=0}^{n/3} {n \choose 3k} = \frac{2^n + (-\omega)^n + (-\omega)^{2n}}{3}.$
$-\omega$ and $-\omega^2$ are sixth roots of unity, so this formula splits into six cases (or maybe three). Similar observations about fourth roots of unity show that
$\displaystyle \sum_{k=0}^{n/4} {n \choose 4k} = \frac{2^n + (1+i)^n + (1-i)^n}{4}$
where $1+i = \sqrt{2} e^{ \frac{\pi i}{4} }$ is a scalar multiple of an eighth root of unity, so this formula splits into eight cases (or maybe four).
Question: Does anyone know a direct combinatorial proof of these identities?