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ganeshie8
 one year ago
show that \(\large \binom{2n}{n}\) is divisible by \(2\) for all \(n\ge 1\)
ganeshie8
 one year ago
show that \(\large \binom{2n}{n}\) is divisible by \(2\) for all \(n\ge 1\)

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Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0\[\left(\begin{matrix}2n \\ n\end{matrix}\right) = \frac{ (2n)! }{ n!n! }\]

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1dw:1436166061011:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Hmmm, I guess:\[ {2n\choose n} = {2n1\choose n1} + {2n1 \choose n} = 2 {2n1 \choose n} \]

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1wow! pascal's rule does the job is it

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0Lol that's actually pretty clever

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1I got excited about this problem because it has many solutions but what wio has is the most clever+short+simple one i think

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0So is that also skew symmetry

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1whats skew symmetry ? i do see the left and right ends are symmetrical dw:1436166600905:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0You could say: \[ f(k,m) ={k+m\choose k} = {m+k \choose m} = f(m,k) \]A function defined this way would be symmetric, kinda

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0It's not skew symmetry, but I do see symmetry obviously, skew symmetry is for a matrix (I actually learnt that yesterday from empty) haha...thought I could apply it at other places as well

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0\[aij = −aji\] xD

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Skew symmetry is more like subtraction, I suppose: \[ f(x,y) = xy = (yx) = f(y,x) \]

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0Yeah you're right

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0But it's better to say 'commutative' than 'symmetric' when talking about maps, and to say 'symmetric' when talking about relations.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1\[ f(k,m) ={k+m\choose k} = {m+k \choose m} = f(m,k) \] Again this follows from identity right \[\binom{n}{r} = \binom{n}{nr}\]

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1thats an interesting way to see it

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1dw:1436167013449:dw

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1another proof which is just algebra but looks kinda cool \[x+\binom{2n}{n}+x = 2^{2n} \implies \binom{2n}{n} = 2(x+2^{2n1})\]

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0I like that one lol

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[ x = \sum_{k=0}^{n1}{2n\choose k} = \sum_{k=0}^{n1}{2n\choose 2nk} \]

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1The odd rows have a nice closed form for the sum http://www.wolframalpha.com/input/?i=%5Csum%5Climits_%7Bk%3D0%7D%5E%7Bn%7D%28%282n%2B1%29+choose+k%29

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1\[ \sum_{k=0}^{n}{2n+1\choose k} = \sum_{k=0}^{2n}{2n\choose k} \] not sure if this is an obvious result
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