Simplifying the sum of powers of the golden ratio 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
 A: What's $\left(\frac{1+\sqrt{5}}{2}\right)^2$?
A: $$ \left (\frac{1+\sqrt{5}}{2} \right )^{k-2} + \left (\frac{1+\sqrt{5}}{2} \right )^{k-1} = \left ( \frac{1+ \sqrt{5}}{2}\right )^{k-2} \left ( 1+ \frac{1+ \sqrt{5}}{2}\right)$$
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:
$$ \left ( \frac{1+ \sqrt{5}}{2}\right )^{k-2} \left ( 1+ \frac{1+ \sqrt{5}}{2}\right)= \left ( 1+ \frac{\sqrt{5}}{2}\right)^k$$
A: 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.
