ganeshie8
  • ganeshie8
show that \(\large \binom{2n}{n}\) is divisible by \(2\) for all \(n\ge 1\)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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Astrophysics
  • Astrophysics
\[\left(\begin{matrix}2n \\ n\end{matrix}\right) = \frac{ (2n)! }{ n!n! }\]
Astrophysics
  • Astrophysics
nvm
ganeshie8
  • ganeshie8
|dw:1436166061011:dw|

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anonymous
  • anonymous
Hmmm, I guess:\[ {2n\choose n} = {2n-1\choose n-1} + {2n-1 \choose n} = 2 {2n-1 \choose n} \]
ganeshie8
  • ganeshie8
wow! pascal's rule does the job is it
Astrophysics
  • Astrophysics
Lol that's actually pretty clever
ganeshie8
  • ganeshie8
I got excited about this problem because it has many solutions but what wio has is the most clever+short+simple one i think
Astrophysics
  • Astrophysics
So is that also skew symmetry
anonymous
  • anonymous
No, it isn't
ganeshie8
  • ganeshie8
whats skew symmetry ? i do see the left and right ends are symmetrical |dw:1436166600905:dw|
anonymous
  • anonymous
You 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
  • Astrophysics
It'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
  • Astrophysics
\[aij = −aji\] xD
anonymous
  • anonymous
Skew symmetry is more like subtraction, I suppose: \[ f(x,y) = x-y = -(y-x) = -f(y,x) \]
Astrophysics
  • Astrophysics
Yeah you're right
anonymous
  • anonymous
But it's better to say 'commutative' than 'symmetric' when talking about maps, and to say 'symmetric' when talking about relations.
ganeshie8
  • ganeshie8
\[ f(k,m) ={k+m\choose k} = {m+k \choose m} = f(m,k) \] Again this follows from identity right \[\binom{n}{r} = \binom{n}{n-r}\]
ganeshie8
  • ganeshie8
thats an interesting way to see it
ganeshie8
  • ganeshie8
|dw:1436167013449:dw|
ganeshie8
  • ganeshie8
another proof which is just algebra but looks kinda cool \[x+\binom{2n}{n}+x = 2^{2n} \implies \binom{2n}{n} = 2(-x+2^{2n-1})\]
Astrophysics
  • Astrophysics
I like that one lol
anonymous
  • anonymous
\[ x = \sum_{k=0}^{n-1}{2n\choose k} = \sum_{k=0}^{n-1}{2n\choose 2n-k} \]
ganeshie8
  • ganeshie8
The 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
  • ganeshie8
\[ \sum_{k=0}^{n}{2n+1\choose k} = \sum_{k=0}^{2n}{2n\choose k} \] not sure if this is an obvious result

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