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

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1. Astrophysics

$\left(\begin{matrix}2n \\ n\end{matrix}\right) = \frac{ (2n)! }{ n!n! }$

2. Astrophysics

nvm

3. ganeshie8

|dw:1436166061011:dw|

4. anonymous

Hmmm, I guess:${2n\choose n} = {2n-1\choose n-1} + {2n-1 \choose n} = 2 {2n-1 \choose n}$

5. ganeshie8

wow! pascal's rule does the job is it

6. Astrophysics

Lol that's actually pretty clever

7. 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

8. Astrophysics

So is that also skew symmetry

9. anonymous

No, it isn't

10. ganeshie8

whats skew symmetry ? i do see the left and right ends are symmetrical |dw:1436166600905:dw|

11. 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

12. 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

13. Astrophysics

$aij = −aji$ xD

14. anonymous

Skew symmetry is more like subtraction, I suppose: $f(x,y) = x-y = -(y-x) = -f(y,x)$

15. Astrophysics

Yeah you're right

16. anonymous

But it's better to say 'commutative' than 'symmetric' when talking about maps, and to say 'symmetric' when talking about relations.

17. ganeshie8

$f(k,m) ={k+m\choose k} = {m+k \choose m} = f(m,k)$ Again this follows from identity right $\binom{n}{r} = \binom{n}{n-r}$

18. ganeshie8

thats an interesting way to see it

19. ganeshie8

|dw:1436167013449:dw|

20. ganeshie8

another proof which is just algebra but looks kinda cool $x+\binom{2n}{n}+x = 2^{2n} \implies \binom{2n}{n} = 2(-x+2^{2n-1})$

21. Astrophysics

I like that one lol

22. anonymous

$x = \sum_{k=0}^{n-1}{2n\choose k} = \sum_{k=0}^{n-1}{2n\choose 2n-k}$

23. 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

24. 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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