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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
    • one year ago
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    \[\left(\begin{matrix}2n \\ n\end{matrix}\right) = \frac{ (2n)! }{ n!n! }\]

  2. Astrophysics
    • one year ago
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    nvm

  3. ganeshie8
    • one year ago
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    |dw:1436166061011:dw|

  4. anonymous
    • one year ago
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    Hmmm, I guess:\[ {2n\choose n} = {2n-1\choose n-1} + {2n-1 \choose n} = 2 {2n-1 \choose n} \]

  5. ganeshie8
    • one year ago
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    wow! pascal's rule does the job is it

  6. Astrophysics
    • one year ago
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    Lol that's actually pretty clever

  7. ganeshie8
    • one year ago
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    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
    • one year ago
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    So is that also skew symmetry

  9. anonymous
    • one year ago
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    No, it isn't

  10. ganeshie8
    • one year ago
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    whats skew symmetry ? i do see the left and right ends are symmetrical |dw:1436166600905:dw|

  11. anonymous
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    \[aij = −aji\] xD

  14. anonymous
    • one year ago
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    Skew symmetry is more like subtraction, I suppose: \[ f(x,y) = x-y = -(y-x) = -f(y,x) \]

  15. Astrophysics
    • one year ago
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    Yeah you're right

  16. anonymous
    • one year ago
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    But it's better to say 'commutative' than 'symmetric' when talking about maps, and to say 'symmetric' when talking about relations.

  17. ganeshie8
    • one year ago
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    \[ f(k,m) ={k+m\choose k} = {m+k \choose m} = f(m,k) \] Again this follows from identity right \[\binom{n}{r} = \binom{n}{n-r}\]

  18. ganeshie8
    • one year ago
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    thats an interesting way to see it

  19. ganeshie8
    • one year ago
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    |dw:1436167013449:dw|

  20. ganeshie8
    • one year ago
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    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
    • one year ago
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    I like that one lol

  22. anonymous
    • one year ago
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    \[ x = \sum_{k=0}^{n-1}{2n\choose k} = \sum_{k=0}^{n-1}{2n\choose 2n-k} \]

  23. ganeshie8
    • one year ago
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    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
    • one year ago
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    \[ \sum_{k=0}^{n}{2n+1\choose k} = \sum_{k=0}^{2n}{2n\choose k} \] not sure if this is an obvious result

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