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Are there combinatorial identities for the sum of "equal step sampled binomial coefficients" ?

Here "equal step sampled binomial coefficients", I referred to something like:

${\displaystyle {\binom {6}{0}} + {\binom {6}{2}}+{\binom {6}{4}} +{\binom {6}{6}} }$

or

${\displaystyle {\binom {6}{1}} + {\binom {6}{3}}+{\binom {6}{5}} }$

or in general, something like:

${\displaystyle {\binom {n}{1}} + {\binom {n}{1 + k}}+ {\binom {n}{1 + 2k}}+...+{\binom {n}{1 + jk}} }$

for any integer ${j, k}$ as long as they satisfy: ${1 + jk \leq n}$,

or another version:

${\displaystyle {\binom {n}{0}} + {\binom {n}{k}}+ {\binom {n}{2k}}+...+{\binom {n}{jk}} }$

for any integer ${j, k}$ as long as they satisfy: ${jk \leq n}$

My question is, for the above "summation of equally sampled subsets of binomial coefficients", are there known combinatorial identities ?

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This isn't necessarily combinatorial, but we can use roots of unity to determine a closed form for $\sum \limits_{k=0} ^ \infty \binom{n}{kr+a}$, for fixed $k, a, r$, and letting $\binom{n}{k}=0$ when $k>n$. In general, $\ \frac{1}{r} \sum \limits _{j=0}^{r-1} \omega ^{-ja} (1+\omega^j)^n=\sum \limits_{k=0} ^{\infty} \binom{n}{a+rk}$, where $\omega=e^{2\pi i/r}$

The above result can be proved with something called a root of unity filter.

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