Let $n$, $r$, and $m$ all be positive integers, $r\le m$, and $\omega_k=e^{\frac{2k\pi}{m}i}$ be a complex root to the equation $x^m=1$. Show $$\sum_{k=0}^{\lfloor{\frac{n-r}{m}}\rfloor}\binom{n}{r+km}x^{r+km}=\frac{1}{m}\sum_{k=0}^{m-1}\omega^{-r}(1+x\omega_k)^n$$
where function $\lfloor{x}\rfloor$ returns the largest integer not exceeding real number $x$.
Using binomial expansion can lead to $$(1+x\omega_k)^n=\sum_{l=0}^{n}\binom{n}{l}x^l\omega_k^l$$
Setting this to the right side of the to-be-proved equation gives $$\frac{1}{m}\sum_{k=0}^{m-1}\omega_k^{-r}(1+x\omega_k)^n = \frac{1}{m}\sum_{k=0}^{m-1}\omega_k^{-r}\sum_{l=0}^{n}\binom{n}{l}x^l\omega_k^l = \frac{1}{m}\sum_{l=0}^{n}\binom{n}{l}x^l\sum_{k-0}^{m-1}\omega_k^{l-r}$$
Because $\omega_0+\omega_1+\cdots+\omega_{m-1}=0$ and $\omega_{k}^{m}=1$, the term $\displaystyle\sum_{k-0}^{m-1}\omega_k^{l-r}$ will be non-zero if and only if $(l-r)$ is a multiple of $m$. When $l$ loops through $0$ to $n$, $\frac{l-r}{m}$ will loop through $0$ to $\lfloor{\frac{n-r}{m}}\rfloor$. This will lead to the to-be-proved relation.