Talk:Fibonacci Number as Sum of Binomial Coefficients

Possible simplification
The statement of the result can be simplified to:
 * $\ds \forall n \in \Z_{>0}: F_n = \sum_{k \mathop \in \N} \dbinom {n - k - 1} k$

or even:
 * $\ds \forall n \in \Z_{>0}: F_n = \sum_{k \mathop \in \Z} \dbinom {n - k - 1} k$

because when $k > \floor {\dfrac {n - 1} 2}$ it happens that $k > n - k - 1$ and so $\dbinom {n - k - 1} k = 0$, and of course $\dbinom {n - k - 1} k = 0$ for all negative $k$.

Is it worth taking that step? --prime mover (talk) 08:29, 19 November 2016 (EST)