Talk:Fibonacci Number as Sum of Binomial Coefficients
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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)