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I seem to have forgotten some fundamental algebra. I know that:

$(\frac{1+\sqrt{5}}{2})^{k-2} + (\frac{1+\sqrt{5}}{2})^{k-1} = (\frac{1+\sqrt{5}}{2})^{k}$

But I don't remember how to show it algebraicly

factoring out the biggest term on the LHS gives

$(\frac{1+\sqrt{5}}{2})^{k-2}(1+(\frac{1+\sqrt{5}}{2}))$ which doesn't really help

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    $\begingroup$ $x^{k-2}+x^{k-1}=x^k$ is true if you have $x+1=x^2$. Can you solve for $x$ in the quadratic equation $x^2-x-1=0$? Is $(1+\sqrt{5})/2$ one of the solutions? $\endgroup$
    – Kim Jong Un
    Commented Sep 1, 2014 at 1:09

3 Answers 3

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$$ \left (\frac{1+\sqrt{5}}{2} \right )^{k-2} + \left (\frac{1+\sqrt{5}}{2} \right )^{k-1} = \left ( \frac{1+ \sqrt{5}}{2}\right )^{k-2} \left ( 1+ \frac{1+ \sqrt{5}}{2}\right)$$

It is known that the Greek letter phi (φ) represents the golden ratio,which value is:

$$\phi=\frac{1+ \sqrt{5}}{2}$$

One of its identities is:

$$\phi^2=\phi+1$$

Therefore:

$$ \left ( 1+ \frac{1+ \sqrt{5}}{2}\right)= \left ( 1+ \frac{\sqrt{5}}{2}\right)^2$$

So:

$$ \left ( \frac{1+ \sqrt{5}}{2}\right )^{k-2} \left ( 1+ \frac{1+ \sqrt{5}}{2}\right)= \left ( 1+ \frac{\sqrt{5}}{2}\right)^k$$

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What's $\left(\frac{1+\sqrt{5}}{2}\right)^2$?

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You are almost done. You have already found that $$ ( \frac{1 + \sqrt{5}}{2} )^{k-2} + ( \frac{1 + \sqrt{5}}{2} )^{k-1} = ( \frac{1 + \sqrt{5}}{2} )^{k-2} (1 + \frac{1 + \sqrt{5}}{2} ) $$

You want to show that this quantity can be expressed as $ ( \frac {1 + \sqrt{5} }{2} )^k$.

Comparing what you have to what you need, you should be able to see that it would be sufficient to prove that $ 1 + \frac{1 + \sqrt{5}}{2} = ( \frac {1 + \sqrt{5} }{2} )^2 $. This can be verified directly by simplifying both sides.

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