China Western Mathematical Olympiad

In the expansion of ${\alpha }^k+{\beta }^k$, let $\alpha +\beta =1$ and $\alpha \beta =1$. We get the sum of coefficients $S_k={\alpha }^k+{\beta }^k$. Since

$$(\alpha +\beta )({\alpha }^{k-1}+{\beta }^{k-1})=({\alpha }^k+{\beta }^k)+\alpha \beta ({\alpha }^{k-2}+{\beta }^{k-2}),$$

we get $S_k=S_{k-1}-S_{k-2}$. Thus $S_k=S_{k-6}$ and $\{S_k\}$ is a periodic sequence with period 6 and $S_{2005}=S_1=1$.

---

Alternative solution.

Set $\alpha +\beta =1$ and $\alpha \beta =1$ in the expansion of ${\alpha }^k+{\beta }^k$. The sum of the coefficients is $S_k={\alpha }^k+{\beta }^k$. Since $\alpha ,\beta$ are solutions of the equation $x^2-x+1=0$, $\alpha =\cos \frac{\pi }{3}+i\sin \frac{\pi }{3}$, $\beta =\cos \frac{\pi }{3}-i\sin \frac{\pi }{3}$. Therefore

$$\begin{array}{rl}{\alpha }^k+{\beta }^k& ={\left(\cos \frac{\pi }{3}+i\sin \frac{\pi }{3}\right)}^k+{\left(\cos \frac{\pi }{3}-i\sin \frac{\pi }{3}\right)}^k\\ & =\left(\cos \frac{k\pi }{3}+i\sin \frac{k\pi }{3}\right)+\left(\cos \frac{k\pi }{3}-i\sin \frac{k\pi }{3}\right)\\ & =2\cos \frac{k\pi }{3}.\end{array}$$

Let $k=2005$, we have $S_k=1$.