Fibonacci Number as Sum of Binomial Coefficients

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

Then:

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

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

The proof proceeds by induction.

For all $n \in \Z_{>0}$, let $P(n)$ be the proposition:
 * $\ds F_n = \sum_{k \mathop = 0}^{\floor {\frac {n - 1} 2} } \dbinom {n - k - 1} k$

Basis for the Induction
$\map P 1$ is the case:

So $\map P 1$ is seen to hold.

$\map P 2$ is the case:

So $\map P 2$ is also seen to hold.

This is our basis for the induction.

Induction Hypothesis
Now we need to show that, if $\map P {k - 1}$ and $\map P k$ are true, where $k > 2$ is an even integer, then it logically follows that $\map P {k + 1}$ and $\map P {k + 2}$ are both true.

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

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

Induction Step
This is our induction step:

For the first part:

For the second part:

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

Therefore:
 * $\ds \forall n \in \Z_{>0}: F_n = \sum_{k \mathop = 0}^{\floor {\frac {n - 1} 2} } \dbinom {n - k - 1} k$