I need help solving this task, if anyone had a similar problem it would help me.
The task is:
Calculate: $$\sum\limits_{k=0}^{n} { 3n \choose 3k } $$
I tried something, with $$2^n= \sum\limits_{k=0}^{n} { n \choose k }$$
But, have no idea how to get the correct solution which is:
$$ \frac{2(-1)^n+8^n}{3}$$
We know that $$(1+x)^{3n}=\sum_{r=0}^{3n}{3n \choose r}x^r$$
Put $x=1,\omega ,\omega^2$ and evaluate the sums as, $$\begin{aligned} (1+1)^{3n}&=\sum_{r=0}^{3n}{3n \choose r} \\ (1+\omega)^{3n}&=\sum_{r=0}^{3n}{3n \choose r}\omega^r \\ (1+\omega^2)^{3n}&=\sum_{r=0}^{3n}{3n \choose r}\omega^{2r} \end{aligned}$$
Here $\omega$ is the cube root of unity.
Add all the equations, we get $$2^{3n}+(1+\omega)^{3n}+(1+\omega^2)^{3n}=3\sum_{k=0}^n{3n \choose 3k}$$
Using the well known property of $$\omega^2+\omega+1=0 {\tag 1}$$ $$ \omega^3=1 {\tag 2}$$
Hence the required sum $\sum_{k=0}^n{3n \choose 3k}=\dfrac{2^{3n}+(1+\omega)^{3n}+(1+\omega^2)^{3n}}{3}$.
The RHS can be simplified by substituting (1), i.e. $$\dfrac{2^{3n}+(-\omega^2)^{3n}+(-\omega)^{3n}}{3}$$ and using (2) we finally get
$$\frac{2(-1)^n+8^n}{3}$$
I think that this is way too complicated for this problem.
So here is a hint:
$$\sum_{k=0}^{3n}\binom{3n}{k}=2^{3n}=8^n$$
Moreover, observer that $$\sum_{k=0}^{3n}\binom{3n}{k}=\sum_{k=0,k\equiv 0\pmod{3}}^{3n}\binom{3n}{k}+\sum_{k=0,k\equiv 1\pmod{3}}^{3n}\binom{3n}{k}+\sum_{k=0,k \equiv 2\pmod{3}}^{3n}\binom{3n}{k}$$
And another very useful thing is that $$\sum_{k=0,k\equiv 1\pmod{3}}^{3n}\binom{3n}{k}=\sum_{k=0,k \equiv 2\pmod{3}}^{3n}\binom{3n}{k}$$
So you have $$8^n=\sum_{k=0,k\equiv 0\pmod{3}}^{3n}\binom{3n}{k}+2\sum_{k=0,k\equiv 1\pmod{3}}^{3n}\binom{3n}{k}$$
Thus, if you prove that $$\sum_{k=0,k\equiv 1\pmod{3}}^{3n}\binom{3n}{k}=\sum_{k=0,k\equiv 0\pmod{3}}^{3n}\binom{3n}{k}+(-1)^n$$
You can substitute in $$8^n=\sum_{k=0,k\equiv 0\pmod{3}}^{3n}\binom{3n}{k}+2\sum_{k=0,k\equiv 1\pmod{3}}^{3n}\binom{3n}{k}$$
And you get what you want. Now lets prove
$$\sum_{k=0,k\equiv 1\pmod{3}}^{3n}\binom{3n}{k}=\sum_{k=0,k\equiv 0\pmod{3}}^{3n}\binom{3n}{k}+(-1)^n$$
I hope this helped you. Try to prove this yourself, you can reach out in the comments section.
I tried to generalise this as a result :
$\displaystyle \sum_{k = 0}^{n} \binom {\phi n}{\phi k} = \frac {1}{\phi} \cdot \sum_{k=0}^{ \phi -1} \left ( 1 + e^{\displaystyle \frac {\iota 2k \pi}{\phi} } \right )^{\phi n} $
Also,
$\displaystyle \sum_{k=0}^{ \phi -1} \left ( 1 + e^{\displaystyle \frac {\iota 2k \pi}{\phi} } \right ) = \sum_{k=0}^{ \phi -1} \left ( 2\iota \sin \left ( \frac{k \pi }{\phi} \right ) \right )^{ \phi n} e^{ \iota k\cdot n\pi} $
