How to compute $ \sum_{2k \le n} (-3)^k \binom{n}{2k} $

Question

How can we compute this sum : $$ \sum_{2k \le n} (-3)^k \binom{n}{2k} $$ using complex numbers ? My main idea is to use Newton binomial like $ (1+i)^n $ but it doesn't seem to work.

Answer 1

$$(1+x)^n+(1-x)^n=2\sum_{k=0}^{2k\le n}\binom n{2k}x^{2k}$$

we need $x^{2k}=(-3)^k\implies x^2=-3\implies x=\pm\sqrt3i$

Now as $\sin(\pm y)=\pm\sin y$

$1\pm\sqrt3i=2(\cos60^\circ+i\sin(\pm60^\circ))=2e^{\pm60^\circ\cdot i}$ using How to prove Euler's formula: $e^{i\varphi}=\cos(\varphi) +i\sin(\varphi)$?

So, $(1\pm\sqrt3i)^n=2^ne^{\pm60^\circ\cdot i}=2^n\left(\cos60^\circ n\pm i\sin(60^\circ n)\right)$

Answer 2

$\displaystyle (1+i\sqrt{3})^n=\sum_{k=0}^{\lfloor n/2\rfloor}{n\choose 2k}(i\sqrt{3})^{2k}+\sum_{k=1}^{\lfloor n/2\rfloor}{n\choose 2k-1}(i\sqrt{3})^{2k-1}$

$\displaystyle 2^n\left(\cos \frac{n\pi}6+i\sin\frac{n\pi}6\right)=\sum_{k=0}^{\lfloor n/2\rfloor}{n\choose 2k}(-3)^{k}+\sum_{k=1}^{\lfloor n/2\rfloor}{n\choose 2k-1}(i\sqrt{3})^{2k-1}$

Compare the real parts.