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July 19th, 2013, 02:12 AM   #1
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Newton binomial

Prove that sum(k=0 to k=n) of (2^k)C(n,k)C(n-k,[(n-k)/2]) =C(2n+1,n)
Where n is a positive integer.
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July 19th, 2013, 02:32 AM   #2
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Re: Newton binomial

How do you define C(n-k, (n-k)/2) when (n-k)/2 is not a whole number?
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July 19th, 2013, 04:07 PM   #3
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Re: Newton binomial

Yes, you're right. (n-k)/2 is not an integer but [(n-k)/2] is the integer part of (n-k)/2 which is the whole number.
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