<s>[INST] If $x+\frac{1}{x}=\frac{1+\sqrt{5}}{2}$ then $x^{2000}+\frac{1}{x^{2000}}= $? If $x+\frac{1}{x}=\frac{1+\sqrt{5}}{2}$ then
$$x^{2000}+\frac{1}{x^{2000}}=?$$
My try:
$$\left(x^{1000}\right)^2+\left(\frac{1}{x^{1000}}\right)^2=\left(x^{1000}+\frac{1}{x^{1000}}\right)^2-2$$
Continuation ?
 [/INST] Nice try of yours. You can further continue...:
$$
x^{2000} + \frac{1}{x^{2000}} = \left(x^{1000} + \frac{1}{x^{1000}}\right)^2 - 2 \\
x^{1000} + \frac{1}{x^{1000}} = \left(x^{500} + \frac{1}{x^{500}}\right)^2 - 2 \\
x^{500} + \frac{1}{x^{500}} = \left(x^{250} + \frac{1}{x^{250}}\right)^2 - 2 \\
\vdots \\
x^{n} + \frac{1}{x^n} = \left(x^{n/2} + \frac{1}{x^{n/2}}\right)^2 - 2 \\
$$
In the case of $2000 = 2^4 \cdot 5^3$. That is, when you are done with all the $2$s, you will be left with $5$:
$$
x^{125} + \frac{1}{x^{125}} = 
\left[\left(x^{25}\right)^5 + \left(\frac{1}{x^{25}}\right)^5\right] = 
$$
We do basically the same you did:
$$
(a+b)^5 = {{5}\choose{0}}a^5 + {{5}\choose{1}}a^4 b + \cdots \\
(a+a^{-1})^5 = {{5}\choose{0}}a^5 + {{5}\choose{5}}b^{-5} + \cdots \\
$$
Notice the intermediate terms:
$$
{{5}\choose{1}}a^{4} a^{-1} + 
{{5}\choose{2}}a^{3} a^{-2} + 
{{5}\choose{3}}a^{2} a^{-3} + 
{{5}\choose{4}}a^{1} a^{-4}
= 
{{5}\choose{1}}a^{3} + 
{{5}\choose{2}}a + 
{{5}\choose{3}}a^{-1} + 
{{5}\choose{4}}a^{-3}
$$
That is:
$$
\left[x^{5} + \frac{1}{x^{5}}\right]^5 = 
\left[x^{5} + \frac{1}{x^{5}}\right] + 
10\left[\left(x^{5}\right)^3 + \left(\frac{1}{x^{5}}\right)^3\right] + 
5\left[\left(x^{5}\right)^5 + \left(\frac{1}{x^{5}}\right)^5\right]
$$

You can keep doing so and simplifying to smaller and smaller terms. You can do this because:
$$
{{n}\choose{k}} = {{n}\choose{n-k}} = \frac{n!}{k!(n-k)!}
$$
That is, $(a + a^{-1})^n$ will have the terms $a^{-k} a^{n-k}$ and $a^k a^{-(n-k)}$ with the same coeficients. And this term forms: 
$$
a^k a^{-n+k} = a^{k-n+k} = a^{2k-n} \\
a^{-k} a^{n-k} = a^{-k+n-k} = a^{-2k+n} = a^{-(2k-n)}
$$
So, the coeficients opposite terms will be equal, then you will always be able to build terms like $x^k + 1/x^k$. Thus you can keep doing it until you are left only with the $x + 1/x$. It might take a while..... =D.
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