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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(nk,[(nk)/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(nk, (nk)/2) when (nk)/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. (nk)/2 is not an integer but [(nk)/2] is the integer part of (nk)/2 which is the whole number.


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