Timeline for Binomial coefficient identity

9 events

when toggle format	what		by	license	comment
Jan 11, 2015 at 3:27	comment	added	GH from MO		@GjergjiZaimi: You are right. I checked this in my head, but stopped half-way :-)
Jan 11, 2015 at 3:17	comment	added	Gjergji Zaimi		@GHfromMO, It is actually correct as it is. If you notice the $\binom{n+m}{n}$, became $\binom{n+m}{n+1}$, and that's where the ${n+1}$ in the numerator came from.
Jan 11, 2015 at 3:11	comment	added	GH from MO		I think, if we want to be pedantic, in your 4th display the numerator $(n+1)(-1)^r$ should be $(r+1)(-1)^r$. And then we need to detach a zero sum similarly as before, if I am not mistaken.
Jan 11, 2015 at 2:20	history	edited	Gjergji Zaimi	CC BY-SA 3.0	added 78 characters in body
Jan 11, 2015 at 2:16	comment	added	Gjergji Zaimi		@KConrad, yes, the whole point of the proof is that we are reducing to the case $m=0$. I guess I can edit this to clarify.
Jan 11, 2015 at 2:05	comment	added	KConrad		Then you should say before you get to the step that you're taking $m \geq 1$ when doing the simplification, since the result itself is true at $m=0$.
Jan 11, 2015 at 1:58	comment	added	Gjergji Zaimi		True, but I'm also only using it whenever $m\geq 1$. :)
Jan 11, 2015 at 1:55	comment	added	KConrad		$\sum_{k=0}^m \binom{m}{k}(-1)^k$ is not $0$ if $m = 0$.
Jan 11, 2015 at 1:51	history	answered	Gjergji Zaimi	CC BY-SA 3.0