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: :$\displaystyle F_n = \sum_{k \mathop = 0}^{\floor {\frac {n - 1} 2} } \dbinom {n - k - 1} k$ === Basis for the Induction === $\map P 1$ is the case: {{begin-eqn}} {{eqn | l = F_1       | r = 1       | c =  }} {{eqn | r = \dbinom 0 0       | c = Zero Choose Zero }} {{eqn | r = \dbinom {1 - 0 - 1} 0       | c =  }} {{eqn | r = \sum_{k \mathop = 0}^{\floor {\frac {1 - 1} 2} } \dbinom {1 - k - 1} k       | c =  }} {{end-eqn}} So $\map P 1$ is seen to hold. $\map P 2$ is the case: {{begin-eqn}} {{eqn | l = F_2       | r = 1 + 0       | c =  }} {{eqn | r = \dbinom 1 0       | c = Binomial Coefficient with Zero }} {{eqn | r = \dbinom {2 - 0 - 1} 0       | c =  }} {{eqn | r = \sum_{k \mathop = 0}^{\floor {\frac {2 - 1} 2} } \dbinom {2 - k - 1} k       | c =  }} {{end-eqn}} 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: :$\displaystyle F_{k - 1} = \sum_{i \mathop = 0}^{\frac k 2 - 1} \dbinom {k - i - 2} i$ :$\displaystyle F_k = \sum_{i \mathop = 0}^{\frac k 2 - 1} \dbinom {k - i - 1} i$ 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: {{begin-eqn}} {{eqn | l = \sum_{i \mathop = 0}^{\frac k 2} \dbinom {k - i} i       | r = \dbinom k 0 + \sum_{i \mathop = 1}^{\frac k 2 - 1} \dbinom {k - i} i + \dbinom {k - \frac k 2} {\frac k 2}       | c =  }} {{eqn | r = 1 + \sum_{i \mathop = 1}^{\frac k 2 - 1} \dbinom {k - i} i + \dbinom {\frac k 2} {\frac k 2}       | c = Binomial Coefficient with Zero }} {{eqn | r = 1 + \sum_{i \mathop = 1}^{\frac k 2 - 1} \dbinom {k - i} i + 1       | c = Binomial Coefficient with Self }} {{eqn | r = 1 + \sum_{i \mathop = 1}^{\frac k 2 - 1} \left({ \dbinom {k - i - 1} i + \dbinom {k - i - 1} {i - 1} }\right) + 1       | c = Pascal's Rule }} {{eqn | r = 1 + \sum_{i \mathop = 1}^{\frac k 2 - 1} \dbinom {k - i - 1} i + \sum_{i \mathop = 1}^{\frac k 2 - 1} \dbinom {k - i - 1} {i - 1} + 1       | c = Summation is Linear }} {{eqn | r = 1 + \sum_{i \mathop = 1}^{\frac k 2 - 1} \dbinom {k - i - 1} i + \sum_{i \mathop = 0}^{\frac k 2 - 2} \dbinom {k - i - 2} i + 1       | c =  }} {{eqn | r = \dbinom {k - 2} 0 + \sum_{i \mathop = 1}^{\frac k 2 - 1} \dbinom {k - i - 1} i + \sum_{i \mathop = 0}^{\frac k 2 - 2} \dbinom {k - i - 2} i + 1       | c = Binomial Coefficient with Zero }} {{eqn | r = \dbinom {k - 2} 0 + \sum_{i \mathop = 1}^{\frac k 2 - 1} \dbinom {k - i - 1} i + \sum_{i \mathop = 0}^{\frac k 2 - 2} \dbinom {k - i - 2} i + \dbinom {k - \paren {\frac k 2 - 1} - 2} {\frac k 2 - 1}       | c = Binomial Coefficient with Self }} {{eqn | r = \sum_{i \mathop = 0}^{\frac k 2 - 1} \dbinom {k - i - 1} i + \sum_{i \mathop = 0}^{\frac k 2 - 1} \dbinom {k - i - 2} i       | c =  }} {{eqn | r = F_k + F_{k - 1}       | c = Induction hypothesis }} {{eqn | r = F_{k + 1}       | c = {{Defof|Fibonacci Number}} }} {{end-eqn}} For the second part: {{begin-eqn}} {{eqn | l = \sum_{i \mathop = 0}^{\frac k 2} \dbinom {k - i + 1} i       | r = \dbinom k 0 + \sum_{i \mathop = 1}^{\frac k 2} \dbinom {k - i + 1} i       | c =  }} {{eqn | r = 1 + \sum_{i \mathop = 1}^{\frac k 2} \dbinom {k - i + 1} i       | c = Binomial Coefficient with Zero }} {{eqn | r = 1 + \sum_{i \mathop = 1}^{\frac k 2} \paren {\dbinom {k - i} i + \dbinom {k - i} {i - 1} }       | c = Pascal's Rule }} {{eqn | r = 1 + \sum_{i \mathop = 1}^{\frac k 2} \dbinom {k - i} i + \sum_{i \mathop = 1}^{\frac k 2} \dbinom {k - i} {i - 1}       | c = Summation is Linear }} {{eqn | r = 1 + \sum_{i \mathop = 1}^{\frac k 2} \dbinom {k - i} i + \sum_{i \mathop = 0}^{\frac k 2 - 1} \dbinom {k - i - 1} i       | c =  }} {{eqn | r = \dbinom {k - 2} 0 + \sum_{i \mathop = 1}^{\frac k 2} \dbinom {k - i} i + \sum_{i \mathop = 0}^{\frac k 2 - 1} \dbinom {k - i - 1} i       | c = Binomial Coefficient with Zero }} {{eqn | r = \sum_{i \mathop = 0}^{\frac k 2} \dbinom {k - i} i + \sum_{i \mathop = 0}^{\frac k 2 - 1} \dbinom {k - i - 1} i       | c =  }} {{eqn | r = F_{k + 1} + F_k       | c = Induction hypothesis }} {{eqn | r = F_{k + 2}       | c = {{Defof|Fibonacci Number}} }} {{end-eqn}} 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: :$\displaystyle \forall n \in \Z_{>0}: F_n = \sum_{k \mathop = 0}^{\floor {\frac {n - 1} 2} } \dbinom {n - k - 1} k$ {{qed}}