Prove identity $1 + {n \choose 3} + {n \choose 6} + \cdots = \frac{1}{3}\left(2^n + 2\cos{\frac{n\pi}{3}}\right)$ without using complex numbers

Question

How to prove the following identity without using complex numbers (and de Moivre's formula)? $$1 + {n \choose 3} + {n \choose 6} + {n \choose 9} + \cdots = \frac{1}{3}\left(2^n + 2\cos{\frac{n\pi}{3}}\right)$$

Answer 1

Let $a_n$ be your left-hand side, the number of subsets of $\{1,\ldots,n\}$ with a multiple of $3$ elements.

To make such a subset, select any subset of $\{4,5,\ldots,n\}$. If it happens to have a multiple of 3 elements, then you can either add nothing or all of $\{1,2,3\}$; if it is not a multiple of 3 elements, then you have 3 ways to complete it. Therefore $$ a_n = 3\cdot 2^{n-3} - a_{n-3} $$ You should be able to use this for an induction proof for each residue class of $n$ modulo $3$.

Actually a single induction proof (still with a step size of 3 and base cases $n=0,1,2$) will do if you write your goal in the form $$ a_n = \frac{2^n + f(n\bmod 3)\cdot(-1)^{\lfloor n/3\rfloor}}3 $$ where $f(n)=2\cos(n\frac\pi3) = \begin{cases} 2 & \text{for }n=0 \\ 1 &\text{for }n=1\\ -1 &\text{for }n=2 \end{cases} \qquad (= 3-2^n \text{ for }n\in\{0,1,2\})$