Complex

$\text{well.. let's brute force it and see if we get any insight}\\ z^2+z+1=0\\ z = \dfrac{-1 \pm i \sqrt{3}}{2} = e^{\pm i \frac{2\pi}{3}}\\ \text{so picking either root }\\ z^n + \dfrac{1}{z^n} = e^{i\frac{2n\pi}{3}}+e^{-i\frac{2n\pi}{3}} = 2 \cos\left(\dfrac{2n\pi}{3}\right)$

$\text{looking at }a_n = 2\cos\left(\dfrac{2n\pi}{3}\right) \text{ we find that}\\ a_n = \begin{cases} -1 &n \pmod{3} \in \{1,2\}\\ 2 &n \pmod{3} = 0 \end{cases}$

$\text{and thus the original sum becomes} \\ \sum \limits_{k=1}^{15}~(1 + 1 + 4) = 15 \cdot 6 = 90$