power of 2 or binomial?

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Masum
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power of 2 or binomial?

Unread post by Masum » Sun Feb 08, 2015 11:59 pm

Which is greater? $2^n$ or $\binom{2n}n$? Find with proof
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Nirjhor
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Re: power of 2 or binomial?

Unread post by Nirjhor » Mon Feb 09, 2015 1:34 am

\[\dbinom{2n}{n}=\dfrac{2n}{n}\dfrac{2n-1}{n-1}\dfrac{2n-2}{n-2}\cdots\dfrac{n+2}{2}\dfrac{n+1}{1}\]
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Tahmid
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Re: power of 2 or binomial?

Unread post by Tahmid » Wed Feb 18, 2015 7:54 pm

$\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

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nayel
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Re: power of 2 or binomial?

Unread post by nayel » Wed Feb 18, 2015 11:06 pm

The following bound is stronger:

\[4^n=(1+1)^{2n}=\binom{2n}{0}+\cdots+\binom{2n}{2n}<(2n+1)\binom{2n}{n}\Rightarrow\binom{2n}{n}>\frac{4^n}{2n+1}\]
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*Mahi*
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Re: power of 2 or binomial?

Unread post by *Mahi* » Thu Feb 19, 2015 1:01 pm

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