Is $(2+i)^n + (2-i)^n $ a real number ($\in \Bbb R$)? The question:
$$\forall n \in \Bbb N$$
is the number
$$(2+i)^n+(2-i)^n$$
in the real numbers ($\Bbb R$)?

My try for solution using Newton binom:
$$(2+i)^n = \sum_{k=0}^{n}\binom{n}{k}2^{n-k}i^{k}$$
$$(2-i)^n = \sum_{k=0}^{n}\binom{n}{k}2^{n-k}(-i)^{k}$$
$$(2+i)^n + (2-i)^n$$
$$\Downarrow$$
$$\forall k - odd: \sum_{k=0}^{n}\binom{n}{k}2^{n-k}i^{k} + \sum_{k=0}^{n}\binom{n}{k}2^{n-k}(-i)^{k} = 0$$
$$\forall k - even: \sum_{k=0}^{n}\binom{n}{k}2^{n-k}i^{k} + \sum_{k=0}^{n}\binom{n}{k}2^{n-k}(-i)^{k} = Real \ number$$
So the answer is YES, the sum $(2+i)^n + (2-i)^n \in \Bbb R.$
Correct? any better/smarter/more efficient way? 

*I tried to solve it trigonometricaly, didnt realy work, stuck...:
$$(2+i) \Rightarrow r = \sqrt{2^2+1^2} =\sqrt{5}, \tan \theta = \frac{1}{2} \Rightarrow \theta = 26.565$$
$$(2+i) = \sqrt{5}(\cos 26.565 + i \sin 26.565)$$
$$(2+i)^n = \sqrt{5}^n(\cos 26.565n + i \sin 26.565n)$$
$$(2-i)^n = \sqrt{5}^n(\cos 26.565n - i \sin 26.565n)$$
$$(2+i)^n + (2-i)^n = ?$$

More efficient way:
A complex number plus its complex conjugate is real, so
$(2+i)^n+(2-i)^n=(2+i)^n+(\overline{2+i})^n=(2+i)^n+\overline{(2+i)^n}\in\mathbb R.$
