Fibonacci Number as Sum of Binomial Coefficients

Theorem
Let $F_k$ be the $k$th Fibonacci number.

Then:
 * $\displaystyle \forall n : F_n = \sum_{k \mathop = 0}^{\left\lfloor{\frac{n-1}{2}}\right\rfloor} \dbinom {n-k-1} {k}$

where $\dbinom a b$ denotes a binomial coefficient, and $\lfloor x \rfloor$ denotes the floor function, which is the greatest integer less than or equal to $x$.

Proof
From the initial definition of Fibonacci numbers, we have:
 * $F_1 = 1, F_2 = 1, F_3 = 2, F_4 = 3$

Proof by induction:

For all $n \in \N_{>0}$, let $P \left({b}\right)$ be the proposition:
 * $\displaystyle F_n = \sum_{k \mathop = 0}^{\left\lfloor{\frac{n-1}{2}}\right\rfloor} \dbinom {n-k-1} {k}$

Basis for the Induction
$P(1)$ is the case $\displaystyle F_1 = \sum_{k \mathop = 0}^{0} \dbinom {1-k-1} {k}$, which holds.

$P(2)$ is the case $\displaystyle F_2 = \sum_{k \mathop = 0}^{0} \dbinom {2-k-1} {k}$, which also holds.

This is our basis for the induction.

Induction Hypothesis
Now we need to show that, if $P \left({k-1}\right)$ and $P \left({k}\right)$ is true, where $k > 2$ is an even number, then it logically follows that $P \left({k+1}\right)$ and $P \left({k+2}\right)$ is true.

So this is our induction hypothesis:
 * $\displaystyle F_{k-1} = \sum_{i \mathop = 0}^{\frac{k}{2}-1} \dbinom {k-i-2} {i}$
 * $\displaystyle F_{k} = \sum_{i \mathop = 0}^{\frac{k}{2}-1} \dbinom {k-i-1} {i}$

Then we need to show:
 * $\displaystyle F_{k+1} = \sum_{i \mathop = 0}^{\frac{k}{2}} \dbinom {k-i} {i}$
 * $\displaystyle F_{k+2} = \sum_{i \mathop = 0}^{\frac{k}{2}} \dbinom {k-i+1} {i}$

Induction Step
This is our induction step:

For the first part:

For the second part:

So $P\left({k-1}\right) \land P\left({k}\right) \implies P\left({k+1}\right) \land P\left({k+2}\right)$ and the result follows by the Principle of Mathematical Induction.

Therefore:
 * $\displaystyle \forall n : F_n = \sum_{k \mathop = 0}^{\left\lfloor{\frac{n-1}{2}}\right\rfloor} \dbinom {n-k-1} {k}$