Fibonacci Number as Sum of Binomial Coefficients

Theorem
Let $F_n$ denote the $n$th Fibonacci number.

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

where:
 * $\dbinom a b$ denotes a binomial coefficient
 * $\left\lfloor{x}\right\rfloor$ denotes the floor function, which is the greatest integer less than or equal to $x$.

Proof
By definition of Fibonacci numbers:
 * $F_1 = 1, F_2 = 1, F_3 = 2, F_4 = 3, \ldots$

Proof by induction:

For all $n \in \Z_{>0}$, let $P(n)$ 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 $P \left({k}\right)$ are true, where $k > 2$ is an even integer, then it logically follows that $P \left({k + 1}\right)$ and $P \left({k + 2}\right)$ are both 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:
 * $\displaystyle \forall n \in \Z_{>0}: F_n = \sum_{k \mathop = 0}^{\left\lfloor{\frac {n - 1} 2}\right\rfloor} \dbinom {n - k - 1} k$