Binomial and power of 4(or 2?)

For discussing Olympiad Level Algebra (and Inequality) problems
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Masum
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Binomial and power of 4(or 2?)

Unread post by Masum » Wed Feb 18, 2015 9:37 pm

Prove that, $4^n<(n+1)(2n+1)\binom n{\left\lfloor \frac n2\right\rfloor}^2$.
Hint: see the title
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*Mahi*
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Re: Binomial and power of 4(or 2?)

Unread post by *Mahi* » Thu Feb 19, 2015 1:09 pm

Typo? :?
$\binom{5}{2} \cdot 6 \cdot 11 = 660 < 4^5$
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Masum
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Re: Binomial and power of 4(or 2?)

Unread post by Masum » Fri Feb 20, 2015 12:38 am

Fixed. Now it's ok
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nayel
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Re: Binomial and power of 4(or 2?)

Unread post by nayel » Fri Feb 20, 2015 3:11 am

Same idea from the other thread gives a stronger bound:
\[
\begin{align*}
&2^n=(1+1)^n=\binom n0+\cdots+\binom nn\le (n+1)\binom{n}{\lfloor n/2\rfloor}\\
\Rightarrow &4^n\le (n+1)^2\binom{n}{\lfloor n/2\rfloor}^2<(n+1)(2n+1)\binom{n}{\lfloor n/2\rfloor}^2
\end{align*}
\]
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