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How to prove 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 $n\ge 0$, $r\ge 0$, $n>r$, $w^r=1$ which means that $w$ was primitive $rth$ root of unity. .

For example, $$(1 + 1)^n + (1 + \omega)^n + (1 + \omega^2)^n) = 2^n + (-\omega^2)^n + (-\omega)^n)$$ $$(1 + 1)^n + (1 + \omega)^n= 2^n $$

How to prove 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 $n\ge 0$, $r\ge 0$, $n>r$ and $w^r=1$ is a primitive $r$-th root of unity. .

For example, $$(1 + 1)^n + (1 + \omega)^n + (1 + \omega^2)^n) = 2^n + (-\omega^2)^n + (-\omega)^n)$$ $$(1 + 1)^n + (1 + \omega)^n= 2^n $$

How to prove 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, $$(1 + 1)^n + (1 + \omega)^n + (1 + \omega^2)^n) = 2^n + (-\omega^2)^n + (-\omega)^n)$$ $$(1 + 1)^n + (1 + \omega)^n= 2^n $$

How to prove 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 $n\ge 0$, $r\ge 0$, $n>r$, $w^r=1$ which means that $w$ was primitive $rth$ root of unity. .

For example, $$(1 + 1)^n + (1 + \omega)^n + (1 + \omega^2)^n) = 2^n + (-\omega^2)^n + (-\omega)^n)$$ $$(1 + 1)^n + (1 + \omega)^n= 2^n $$