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
$S_{2005} = S_1 = 1$.

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

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

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