$F_n$ has a nice combinatorial interpretation; it is the number of ways to tile a $1\times n$ rectangle with any number of squares and dominoes. This assumes the convention $F_0=F_1=1$.
For example, $F_4=5$, because of the five tilings below:
$$
\begin{array}{|c|c|c|c|}\hline \;\;&\;\;&\;\;&\;\;\\\hline\end{array}\qquad
\\
\begin{array}{|cc|c|c|}\hline \;\;&\;\;&\;\;&\;\;\\\hline\end{array}\qquad
\begin{array}{|c|cc|c|}\hline \;\;&\;\;&\;\;&\;\;\\\hline\end{array}\qquad
\begin{array}{|c|c|cc|}\hline \;\;&\;\;&\;\;&\;\;\\\hline\end{array}\qquad
\\
\begin{array}{|cc|cc|}\hline \;\;&\;\;&\;\;&\;\;\\\hline\end{array}\qquad\\
$$
You can prove this combinatorial interpretation for $F_n$ by induction on $n$.
To prove $F_n=\sum_k \binom{n-k}{k}$, note that $\binom{n-k}{k}$ is the number of tilings which use exactly $k$ dominos. Above, we see that $\binom{4}0=1$ tiling uses zero dominos, $\binom31=3$ tilings use one domino, and $\binom22=1$ tiling uses two dominoes.