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 July 19th, 2013, 02:12 AM #1 Member   Joined: May 2013 From: Phnom Penh, Cambodia Posts: 77 Thanks: 0 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.
 July 19th, 2013, 02:32 AM #2 Senior Member   Joined: Jun 2013 From: London, England Posts: 1,316 Thanks: 116 Re: Newton binomial How do you define C(n-k, (n-k)/2) when (n-k)/2 is not a whole number?
 July 19th, 2013, 04:07 PM #3 Member   Joined: May 2013 From: Phnom Penh, Cambodia Posts: 77 Thanks: 0 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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