Prove that, $4^n<(n+1)(2n+1)\binom n{\left\lfloor \frac n2\right\rfloor}^2$.
Hint: see the title
Binomial and power of 4(or 2?)
One one thing is neutral in the universe, that is $0$.
Re: Binomial and power of 4(or 2?)
Typo?
$\binom{5}{2} \cdot 6 \cdot 11 = 660 < 4^5$
$\binom{5}{2} \cdot 6 \cdot 11 = 660 < 4^5$
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Nur Muhammad Shafiullah | Mahi
Use $L^AT_EX$, It makes our work a lot easier!
Nur Muhammad Shafiullah | Mahi
Re: Binomial and power of 4(or 2?)
Fixed. Now it's ok
One one thing is neutral in the universe, that is $0$.
Re: Binomial and power of 4(or 2?)
Same idea from the other thread gives a stronger bound:
"Everything should be made as simple as possible, but not simpler." - Albert Einstein