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March 28th, 2018, 03:28 PM   #1
Adm
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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 ...
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March 28th, 2018, 04:19 PM   #2
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$\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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