We've already know that $$\sum_{k\ge 0} \binom{n}{rk} =\frac{1}{r}\sum_{j=0}^{r-1}(1+w^j)^n$$ can be approximated as $\frac{2^n}{r}$, where $n\ge 0$, $r\ge 0$, $n>r$, $w^r=1$.
For example, $$(1 + 1)^n + (1 + \omega)^n + (1 + \omega^2)^n) = 2^n + (-\omega^2)^n + (-\omega)^n)$$ $$(1 + 1)^n + (1 + \omega)^n= 2^n $$
However, how to derive a close formula for $$\sum_{k\ge 0} \binom{n}{a+rk} $$
For $0\le a<r$, letting $w=\exp(2\pi i/r)$ consider $$\sum_{j=0}^{k-1}w^{-aj}(1+w^j)^n.$$