My Math Forum > Math why 2n above n (binomial) is, equal or less than, 4^n ?
 User Name Remember Me? Password

 Math General Math Forum - For general math related discussion and news

 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 Display Modes Linear 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

 Contact - Home - Forums - Cryptocurrency Forum - Top