General term: $s_n = \displaystyle \sum_{0}^{\lfloor\frac{n-1}{2}\rfloor}\binom{n-1-r}{r} \text{ for } n \in \mathbb{N} \tag{1}$
I think you struggled because the binomial sum $s_{n+2}$ has a decomposition that involves both its even and odd numbered predecessors, and not just any one of them, as you were trying in the inductive step.
Decomposition: $s_{n+2} = s_{n+1} + s_{n} \tag{2}$
This is easy to prove by using the combinatorial identity $\binom{n}{k}=\binom{n-1}{k}+\binom{n-1}{k-1}$. See here and here.
We can then use this property along with strong induction to prove:
$P(n): s_n = \frac{1}{\sqrt{5}}\left(\frac{1 + \sqrt{5}}{2}\right)^n - \frac{1}{\sqrt{5}}\left(\frac{1 - \sqrt{5}}{2}\right)^n \text{ true } \forall n \in \mathbb{N}\tag{3}$
Proof-sketch:
Base case: Show $P(1)$ and $P(2)$ to be true (by evaluating both sides).
Inductive step: Assume $P(n)$ and $P(n+1)$ true.
Then $\begin{aligned}s_{n+2} & = s_{n+1} + s_{n} \\ & = \frac{1}{\sqrt{5}}\left(\frac{1 + \sqrt{5}}{2}\right)^{n+1} - \frac{1}{\sqrt{5}}\left(\frac{1 - \sqrt{5}}{2}\right)^{n+1} + \frac{1}{\sqrt{5}}\left(\frac{1 + \sqrt{5}}{2}\right)^n - \frac{1}{\sqrt{5}}\left(\frac{1 - \sqrt{5}}{2}\right)^n \\ & = \frac{1}{\sqrt{5}}\left(\frac{1 + \sqrt{5}}{2}\right)^n\left(\frac{3 + \sqrt{5}}{2}\right) - \frac{1}{\sqrt{5}}\left(\frac{1 - \sqrt{5}}{2}\right)^n\left(\frac{3 - \sqrt{5}}{2}\right) \\ & = \frac{1}{\sqrt{5}}\left(\frac{1 + \sqrt{5}}{2}\right)^{n+2} - \frac{1}{\sqrt{5}}\left(\frac{1 - \sqrt{5}}{2}\right)^{n+2} \\ & \implies P(n+2) \text{ true }\end{aligned}$
This completes our proof. $ \blacksquare$
Note: The formulas for odd $n$ and even $n$ follow automatically from $(3)$ by the removal of the floor function in $s_n$.
First of all, we rewrite
$$F_n=\frac{\phi^n−(1−\phi)^n}{\sqrt5}$$
Now we see
\begin{align}
F_n&=F_{n-1}+F_{n-2}\\
&=\frac{\phi^{n-1}−(1−\phi)^{n-1}}{\sqrt5}+\frac{\phi^{n-2}−(1−\phi)^{n-2}}{\sqrt5}\\
&=\frac{\phi^{n-1}−(1−\phi)^{n-1}+\phi^{n-2}−(1−\phi)^{n-2}}{\sqrt5}\\
&=\frac{\phi^{n-2}(\phi+1)−(1−\phi)^{n-2}((1-\phi)+1)}{\sqrt5}\\
&=\frac{\phi^{n-2}(\phi^2)−(1−\phi)^{n-2}((1-\phi)^2)}{\sqrt5}\\
&=\frac{\phi^n−(1−\phi)^n}{\sqrt5}\\
\end{align}
Where we use $\phi^2=\phi+1$ and $(1-\phi)^2=2-\phi$. Now check the two base cases and we're done!
Turns out we don't need all the values below $n$ to prove it for $n$, but just $n-1$ and $n-2$ (this does mean that we need base case $n=0$ and $n=1$).
Best Answer
Let $\phi=\dfrac{\sqrt{5}+1}2$ and note that $\phi^{-1} =\dfrac 1\phi= \dfrac{\sqrt{5}-1}2$.
Note also that $1+\dfrac 1\phi=\phi$ and $1-\phi=-\dfrac 1\phi$.
From your formula,
$$F_n = \frac 1{\sqrt{5}}\left[\phi^n-(-\frac 1\phi)^n \right]$$
For $n=k$ and $n=k-1$, $$\begin{align} F_k &= \frac 1{\sqrt{5}}\left[\phi^k-(-\frac 1\phi)^k \right]\\ F_{k-1} &= \frac 1{\sqrt{5}}\left[\phi^{k-1}-(-\frac 1\phi)^{k-1} \right]\\ &=\frac 1{\sqrt{5}} \left[\phi^k \cdot \frac 1\phi -(-\frac 1\phi)^k \cdot (-\phi)\right]\\ \end{align}$$ Hence,
$$\begin{align} F_{k+1}&=F_{k}+F_{k-1}\\ &=\frac 1{\sqrt{5}} \left[\phi^k \cdot \left( 1+\frac 1\phi \right) -(-\frac 1\phi)^k \cdot \left( 1-\phi \right)\right]\\ &=\frac 1{\sqrt{5}} \left[\phi^k \cdot \phi -(-\frac 1\phi)^k \cdot \left( -\frac 1\phi \right)\right]\\ &=\frac 1{\sqrt{5}} \left[\phi^{k+1}-(-\frac 1\phi)^{k+1} \right] \end{align}$$
i.e. if formula is true for $n=k-1$ and $n=k$, it is also true for $n=k+1$.
For $n=0$ and $n=1$, $F_0=0$ and $F_1=1$ respectively. Hence $F_2=F_0+F_1=1$. It can easily be shown that the formula is true for $n=2$.
Hence, by induction, formula is true for all positive integer $n\geq 2$.