Seeking combinatorial proof for $F_{n+1} -1=\sum\limits_{k=0}^{n-1} F_k$

Question

In order to give a combinatorial proof for this equation, we need to find what these two count for.

But I don't know what they count for and how I can pivot the RHS to show that it actually counts the same thing as LHS.

Answer 1

Presumably $F_n$ is the $n$'th Fibonacci number.

Note that $F_n$ is the number of binary sequences of length $n$ with no consecutive $0$'s. Condition on the position of the last $0$ to get your identity.

Answer 2

Let $T_n$ be the number of ways to tile a $1\times n$ strip with $1\times 1$ and $1\times 2$ tiles, which I’ll call squares and dominoes, respectively; $T_n=F_{n+1}$. On the lefthand side $F_{n+1}-1=T_n-1$ is $1$ less than the number of ways to tile a board of length $n$.

Let $T_n'$ be the number of such tilings in which the last tile is a domino; clearly $T_n'=T_{n-2}=F_{n-1}$. Thus,

$$\sum_{k=0}^{n-1}F_k=\sum_{k=1}^nT_k'\tag{1}$$

is the number of ways to tile a board of length at most $n$ so that the last tile is a domino. Each such tiling can be extended with squares to a tiling of the board of length $n$, and every such tiling will include at least one domino. Thus, $(1)$ counts every tiling of the $n$-board except ... ?