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 March 28th, 2018, 02:28 PM #1 Newbie   Joined: May 2017 From: Neverland Posts: 8 Thanks: 0 why 2n above n (binomial) is, equal or less than, 4^n ? i mean i know that 4^n = (1+1)^(2n) and 2n above n (binomial) is equals to 2n!/n!(2n-n)! but how do i put those in words to prove the equation (Title) ...? sorry i haven't seen any method in the formulas of how to write the binomial expression ... March 28th, 2018, 03:19 PM #2 Senior Member   Joined: Sep 2015 From: USA Posts: 2,549 Thanks: 1399 \begin{align*} &4^n = \\ \\ &(1+1)^{2n} = \\ \\ &\sum \limits_{k=0}^{2n}~\dbinom{2n}{k}(1)^k(1)^{2n-k} = \\ \\ &\sum \limits_{k=0}^{2n}~\dbinom{2n}{k} = \\ \\ &\dbinom{2n}{n} + \sum \limits_{k=0,~k\neq n}^{2n}~\dbinom{2n}{k} \geq \\ \\ &\dbinom{2n}{n} \end{align*} Tags binomial, equal, prove 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 GIjoefan1976 Algebra 2 February 15th, 2016 02:56 PM beesee Probability and Statistics 5 September 18th, 2015 01:38 PM rnck Number Theory 9 May 24th, 2015 03:50 AM Shen Geometry 9 April 14th, 2014 01:21 PM honzik Applied Math 0 June 9th, 2009 10:38 AM

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