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?)

### 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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Use $L^AT_EX$, It makes our work a lot easier!

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