$x^{2000} + \frac{1}{x^{2000}}$ in terms of $x + \frac 1x$. If $x + \frac{1}{x} = 1$, then what is
$$ x^{2000} + \frac{1}{x^{2000}} = ?$$

This is a not so elegant approach. 
Let $a_n=x^n+\frac1{x^n}$ then 
$$a_n\left(x+\frac1x\right)=a_{n+1}+a_{n-1}$$ or $$a_{n+1}=a_{n}-a_{n-1}$$
We have $a_1=1$ and $a_2=-1$, hence solving recursively - plugging - we obtain \begin{align}
a_{6n-4}=a_{6n-2}&=-1\\
a_{6n-3}&=-2\\
a_{6n}&=2\\
a_{6n-1}=a_{6n+1}&=1
\end{align}
Now since $2000=6\times 334 -4$, we have that $a_{2000}=-1$.
