Combinatorial proof of a Fibonacci identity: n F 1 + ( n − 1 )...

Recall that $F_{n+1}$ is the number of ways to tile a board of length 𝑛 with tiles of length $1$ and $2$. So $F_{n+4}$ is the number of ways to tile a board of length $n+3$ with tiles of length $1$ and $2$. Note that $n+3$ such tilings use at most one tile of length $2$, so $F_{n+4}-(n+3)$ such tilings use at least two tiles of length $2$.
Given such a tiling, look at where the second-to-last tile of length $2$ is used. The part after this tile is a tiling of some section of length $k+1$ where exactly one tile of length $2$ is used (which can be done in $k$ ways), and the part before this tile is a tiling of the remaining portion of length $n-k$ (which can be done in $F_{n-k+1}$ ways). Sum over $k$.