How do I prove that for all positive integer n, the inequality $2n\choose n$$<4^n$ holds?
Thank you!
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4$\begingroup$ See also math.stackexchange.com/questions/448861/… and math.stackexchange.com/questions/931306/… $\endgroup$– Martin SleziakOct 4, 2016 at 9:13
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1 Answer
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Hint: The LHS is the number of $n$-element subsets of $[2n]$, while the RHS is the number of all subsets of $[2n]$.
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$\begingroup$ Thanks you! But how do I set the equations up? $\endgroup$– EmileMay 16, 2013 at 16:30
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2$\begingroup$ @Emile You can use this hint the following way: $4^n=(1+1)^{2n} =\sum_{k=0}^{2n} \binom{2n}{k} > \binom{2n}{n}$. $\endgroup$– N. S.May 16, 2013 at 16:32
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$\begingroup$ @emile N.S. said it, I was just about to type it, oh by the way, what have you tried? $\endgroup$– user67258May 16, 2013 at 16:33