Calculate f(n+1)-f(n-1) based on f(n)??? Being:
$$f(n) = \left(\frac{5+3\sqrt5}{10}\right)\cdot\left(\frac{1+\sqrt5}{2}\right)^n+\left(\frac{5-3\sqrt5}{10}\right)\cdot\left(\frac{1-\sqrt5}{2}\right)^n$$
Calculate:
$$f(n+1)-f(n-1)\\ \text{Based on } f(n)$$
I really can't see a simple way to solve this, It'd really help me if you could point it out.
 A: Here's the beginning of a solution that doesn't require too many rabbits-out-of-hats.  First, to save on writing, let
$$ \alpha=\frac{5+3\sqrt5}{10}
\text{ ,}\quad
\overline\alpha=\frac{5-3\sqrt5}{10}
\text{ ,}\quad
\varphi=\frac{1+\sqrt5}{2}
\text{ ,}\quad
\overline\varphi=\frac{1-\sqrt5}{2}
\text{ .}
$$
Then
$$ f(n+1)-f(n-1)
= \alpha\varphi^{n+1} + \overline\alpha\,\overline\varphi^{n+1}
- \alpha\varphi^{n-1} - \overline\alpha\,\overline\varphi^{n-1}
$$
Since we're trying to get something in terms of $f(n)$, we'll now try to get some $\varphi^n$ and $\overline\varphi^n$.  That suggests grouping the terms thus:
$$ f(n+1)-f(n-1)
= \alpha\varphi^n(\varphi-\varphi^{-1})
+ \overline\alpha\,\overline\varphi^n(\overline\varphi-\overline\varphi^{-1})
\tag{$\ast$}
$$
This looks a lot like $f(n)$, but it's got extra $(\varphi-\varphi^{-1})$ and $(\overline\varphi-\overline\varphi^{-1})$, so we'd like to get rid of those, or manipulate them into something else.  It'd be swell if they were equal, for example, since then we could factor them out, and we'd have $(\text{something})f(n)$.  So, the next step is to get a handle on these expressions; you could start by computing them.
A: Hint: the term $\phi=\frac{1+\sqrt{5}}{2}$ is an important mathematical constant known as the golden ratio. The other terms in the expression defining $f(n)$ can be conveniently written in terms of this single constant, e.g.,
$$\begin{cases}
\frac{1-\sqrt{5}}{2}=-\phi^{-1},\\
\frac{5+3\sqrt{5}}{10}=\frac{\phi^4-1}{5}.
\end{cases}$$
Can you take it from here?
