Does the geometric series have any special result for $r = \phi$? I am wondering whether partial sums of the geometric series 
$$S_n = \sum_{k=0}^n r^k = \frac{1 - r^{n+1}}{1-r},$$
and especially,
$$S'_n = \sum_{k=\color{red}{1}}^n r^k = r \cdot \frac{1 - r^n}{1-r}$$
have any particular result (or meaning, or interpretation) for $r = (1 + \sqrt5)/2$.
Any hint would be nice. This is not a homework topic, but an expression that I encountered while fitting curve models to some data.
Note: an early version of this question had an incorrect expression for $S_n'$, which I have corrected. The current expressions can be validated in MATLAB:
N = 12; % or whatever you use to test
r = (1 + sqrt(5)) / 2;

sum(r.^(0 : N))
(1 - r^(N+1)) / (1 - r)

sum(r.^(1 : N))
r * (1-r^N) / (1 - r)

 A: $$\sum_{k=0}^{n} \phi^{k}=\phi^{n+2}-\phi$$
$$\sum_{k=1}^{n} \phi^{k}=\phi^{n+2}-\phi^2$$
You can prove both of these formulas by induction on $n.$
A: Let $r=\phi$, the golden ratio, then $r^2=r+1$. By induction you can show that $r^n=F_nr+F_{n-1}$ where $F_n$ denotes the $n$-th Fibonacci number. Hence 
$$\sum_{i=0}^nr^i=1+r+\sum_{i=2}^n(F_ir+F_{i-1}).$$
Now this is not really an interpretation of this sum, but a rather a nice way of rewriting it. The bottom line is that since $r^2=r+1$, you can rewrite higher powers of $r$ in terms of $r$ and some constant. You can also use this to rewrite $\frac{1-r^{n+1}}{1-r}$. Maybe this is useful?
A: As an addendum to @MitchellSpector's answer, and since the title of my original question involved $\Phi$ instead of $\phi$, I'll give the expressions for $\Phi = 1 / \phi$ as well:
$$\sum_{k=0}^{n} \Phi^{k} = \sum_{k=0}^{n} \phi^{-k} = \phi^2 - \phi^{1-n}$$
$$\sum_{k=1}^{n} \Phi^{k} = \sum_{k=1}^{n} \phi^{-k} = \phi - \phi^{1-n}$$
This can be shown from @MitchellSpector's answer by dividing everything by $\phi^n$ (for $\sum_{k=0}^{n}$) or $\phi^{n+1}$, (for $\sum_{k=1}^{n}$), respectively. In the partial sums, you can then rearrange the terms (inverse order), change the summation index, and there you are.
In MATLAB:
N = 4;
phi = (1 + sqrt(5)) / 2;

sum(phi.^-(0 : N))
phi^2 - phi^(1 - N)

sum(phi.^-(1 : N))
phi - phi^(1 - N)

