TITLE: Prove the following equality: $\sum_{k=0}^n\binom {n-k }{k} = F_n$
QUESTION [9 upvotes]: I need to prove that there is the following equality:
$$
\sum\limits_{k=0}^n {n-k \choose k} = F_{n}
$$
where $F_{n}$ is a n-th Fibonacci number.
The problem seems easy but I can't find the way to prove it. Could you give me any hints? I'm looking for a combinatorial proof.

REPLY [2 votes]: The sequences $F_{n}$ and $G_n=\sum_{i+j = n} {i \choose j}$ have the same initial conditions:
$$G_0 = \binom{0}{0}=0=F_0, G_1 = \binom{1}{0}+\binom{0}{1} = 1 = F_1$$
And they satisfy the same order-2 recursion:
$$G_{n+2} = \sum_{i+j = n+2} {i \choose j} = \sum_{i+j = n+2} ({i-1 \choose j}+{i-1 \choose j-1})= $$
$$\sum_{(i-1)+j = n+1} {i-1 \choose j} + \sum_{(i-1)+(j-1) = n} {i-1 \choose j-1}=G_n + G_{n+1}$$
Hence those 2 sequences coincide.