Prove
$\sum_{k\geq 0}^{\infty }\binom{n}{3k}=(2^n+m)/3$,
where m=where m depends on n and is equal to $2, 1, −1, −2, −1, 1,$ when n is congruent to $0,1,2,3,4,5$ ($\mod 6$) respectively. and
$\sum_{k\geq 0}^{\infty }\binom{n}{4k}=\frac{2^n+m2^{\left \lceil \frac{n}{2} \right \rceil}}{4}$
where $m = 2,1,0,−1,−2,−1,0,1,$ when $n\ge1$ is congruent, respectively,
to $0, 1, 2, 3, 4, 5, 6, 7$ ($\mod 8$).
And if possible, prove
$\sum_{k\geq 0}^{\infty }\binom{n}{rk}= \frac{1}{r}\sum_{j=0}^{r-1}(1+w^j)^n$
where $w=e^{2\pi i/r}$. This is an assigment that our teacher tell us to do. However the only clue that he gives us is to use binomial theorem... Don't really know how to do that. I can only use direct counting to prove$\sum_{k\geq 0}^{\infty }\binom{n}{3k}=(2^n+m)/3$ but starting form $k=4$ the case gets a lot complicated. Any help will be appreciated!
Let $\omega$ be a third root of unity, then $1^k + \omega^k + \omega^{2k}$ is $3$ if $k$ is divisible by 3, and $0$ otherwise. Hence:
$3 \sum \binom{n}{3k} = \sum \binom{n}{k} 1^k + \sum \binom{n}{k} \omega^k + \sum \binom{n}{k} (\omega^2)^k$
Now use the binomial theorem to evaluate those three sums. The other exercise can be done in the same way (this time, use 4 sums, with $1^k$, $i^k$, $(-1)^k$, and $(-i)^k$).