27. Find the value of $\left\lfloor\left(\frac{3+\sqrt{17}}{2}\right)^{6}\right\rfloor$.
27. Answer: 2040.
Let $\alpha=\frac{3+\sqrt{17}}{2}$ and $\beta=\frac{3-\sqrt{17}}{2}$. Then $\alpha \beta=-2$ and $\alpha+\beta=3$.
Set $S_{n}=\alpha^{n}+\beta^{n}$. Then
$$
\begin{aligned}
3 S_{n+1}+2 S_{n} & =(\alpha+\beta)\left(\alpha^{n+1}+\beta^{n+2}\right)-\alpha \beta\left(\alpha^{n}+\beta^{n}\right) \\
& =\alpha^{n+2}-\beta^{n+2}=S_{n+2} .
\end{aligned}
$$

Note that $|\beta|<1$. Then for even positive integer $n,\left\lfloor\alpha^{n}\right\rfloor=S_{n}+\left\lfloor-\beta^{n}\right\rfloor=S_{n}-1$.
Since $S_{0}=2$ and $S_{1}=3$, we can proceed to evaluate that $S_{6}=2041$.