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Which is greater? $2^n$ or $\binom{2n}n$? Find with proof

$\frac{2n}{n}=2 $
$\frac{2n-1}{n-1} > 2$
$\frac{2n-2}{n-2} > 2$
.
.
$\frac{2n-(n-1)}{n-(n-1)} > 2$
multiplying all ....we have ,
$\frac{2n(2n-1)(2n-2)......(n+1)}{n(n-1)(n-2)......1}>2^{n}$
or, $\binom{2n}{n}>2^{n}$

it works for all n>1 . for n=1 both of them are equal

The following bound is stronger:

\[4^n=(1+1)^{2n}=\binom{2n}{0}+\cdots+\binom{2n}{2n}<(2n+1)\binom{2n}{n}\Rightarrow\binom{2n}{n}>\frac{4^n}{2n+1}\]

Or, you know, Cauchy–Schwarz -
\[(n+1)\binom{2n}n = (n+1)\left [ \binom{n}{0}^2 + \binom{n}{1}^2 + \cdots + \binom{n}{n}^2 \right ] \]\[ \geq \left [ \binom{n}{0} + \binom{n}{1} + \cdots + \binom{n}{n} \right ]^2 = 4^n \]