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Math Help - Sum of binomials

  1. #1
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    Sum of binomials

    Let p>3 be a prime. Show that p|\sum_{i=1}^p\binom {i\cdot p}{p}\cdot\binom {\left(p-i+1\right)p}{p}.
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  2. #2
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    Quote Originally Posted by james_bond View Post
    Let p>3 be a prime. Show that p|\sum_{i=1}^p\binom {i\cdot p}{p}\cdot\binom {\left(p-i+1\right)p}{p}.
    it's also true for p=2. anyway, this is just a simple application of Lucas' theorem. by this theorem we have:

    \sum_{i=1}^p \binom{ip}{p} \binom{(p-i+1)p}{p} \equiv \sum_{i=1}^p \binom{i}{1} \binom{p-i+1}{1}=\sum_{i=1}^p i(p-i+1) \equiv \sum_{i=1}^p i(i-1)

    =\frac{p(p^2 - 1)}{3} \equiv 0 \mod p, because every prime which is \neq 3 is in the form 3k \pm 1 and thus 3 \mid p^2-1. \ \Box



    Remark: you might say that i didn't apply Lucas' theorem correctly for i=1 and i=p. but actually what i did

    is perfectly fine because:

    if i=1, then by Lucas' theorem: \binom{(p-i+1)p}{p}=\binom{p^2}{p} \equiv \binom{1}{0} \binom{0}{1} = 0 \equiv \binom{p}{1}=\binom{p-i+1}{1} \mod p.

    similarly if i=p, then by Lucas' theorem: \binom{ip}{p}=\binom{p^2}{p} \equiv \binom{1}{0} \binom{0}{1} = 0 \equiv \binom{p}{1}=\binom{i}{1} \mod p.
    Last edited by NonCommAlg; March 1st 2010 at 12:11 AM.
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