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

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 March 28th, 2018, 03:28 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, 04:19 PM #2 Senior Member     Joined: Sep 2015 From: USA Posts: 2,163 Thanks: 1135 \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*}

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