D Given that $\alpha^{2005}+\beta^{2005}$ can be expressed as a bivariate polynomial in terms of $\alpha+\beta$ and $\alpha \beta$, find the sum of the coefficients of this polynomial. (Supplied by Zhu Huawei)
Method One: In the expansion of $\alpha^{k}+\beta^{k}$, let $\alpha+\beta=1, \alpha \beta=1$, and the sum of the coefficients sought is $S_{k}$. From
$$
\begin{aligned}
& (\alpha+\beta)\left(\alpha^{k-1}+\beta^{k-1}\right) \\
= & \left(\alpha^{k}+\beta^{k}\right)+\alpha \beta\left(\alpha^{k-2}+\beta^{k-2}\right),
\end{aligned}
$$

we have
$$
S_{k}=S_{k-1}-S_{k-2},
$$

thus
$$
S_{k}=\left(S_{k-2}-S_{k-3}\right)-S_{k-2}=-S_{k-3} \text {, }
$$

similarly
$$
S_{k-3}=-S_{k-6} .
$$

Therefore,
$$
S_{k}=S_{k-6} .
$$

Hence, the sequence $\left\{S_{k}\right\}$ is a periodic sequence with a period of 6. Therefore, $S_{2005}=S_{1}=1$.
Method Two: In the expansion of $\alpha^{k}+\beta^{k}$, let $\alpha+\beta=1, \alpha \beta=1$, and the sum of the coefficients sought is $S_{k}$. Then $\alpha, \beta$ are the roots of the equation $x^{2}-x+1=0$. Solving, we get
$$
\alpha=\cos \frac{\pi}{3}+\mathrm{i} \sin \frac{\pi}{3}, \beta=\cos \frac{\pi}{3}-\mathrm{i} \sin \frac{\pi}{3},
$$

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

Taking $k=2005$, we get $S_{k}=1$.