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 Probability and Statistics Basic Probability and Statistics Math Forum

 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. Tags binomial, newton Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post OriaG Probability and Statistics 1 February 13th, 2013 01:10 AM gelatine1 Probability and Statistics 2 December 23rd, 2012 12:50 PM julian21 Probability and Statistics 3 November 28th, 2010 01:29 PM DouglasM Probability and Statistics 4 February 25th, 2010 03:06 AM rooster Applied Math 1 October 27th, 2009 08:30 AM

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