How to find the General expression of $\sum_{k=0}^ {\lfloor n/3\rfloor} {n \choose 3k}$

Question

Well as the title says I'm having problems trying to derive a general expression for this sum which involves cubic roots of unity

$$\sum_{k=0}^ {\lfloor \frac n 3\rfloor} {n \choose 3k}$$

Need help guys!

Answer 1

Here's a hint:

The binomial theorem gives $f(x) = (1+x)^n = \sum_{i = 0} \binom{n}{i} x^i$.

Now if $\zeta$ is a cubic root of unity, consider an appropriate linear combination of $f(1), f(\zeta)$ and $f(\zeta^2)$. Since $1 + \zeta + \zeta^2 = 0$, you should be able to make all but the threeven terms in the sum cancel.