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$(4)^n$> ${2n\choose n}$

I have attempted to prove this by placing it as $(4)^n$=$(1+1)^n$ with another square like $2^{2n},$ since I can't write it properly; then I used the formula for ${2n\choose n}$ which I separated into a sum, ${2n\choose 0}$ + ${2n\choose 1}$ + ... + ${2n\choose n}$, but I still came to the wrong conclusion that $(4)^n$>$(4)^n$, so could you please lend a hand?

PinkyWay
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Danny_007
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1 Answers1

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Your inequalities are strange... $(1+1)^n=4^n$ is of course wrong. But the idea is good : $$4^n=(1+1)^{2n}=\sum_{k=0}^{2n}\binom{2n}{k},$$ and thus $4^n>\binom{2n}{n}$. By the way, this also prove that $$4^n\geq \sum_{k=m}^{p}\binom{2n}{k},$$ for all $0\leq m\leq p\leq 2n.$

Surb
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  • Thanks for the quick reply, i meant to write the first expression with the power 2n but since this is my first post i couldn't figure it out. I don't seem to understand one part of the solution, i understand that if this is greater or equal to 4^n then it is true, but if not, if it is only 4^n> some x we found to be the same expression would that mean i am wrong, or does the sum decrease at some poing and 4^n is always bigger than that – Danny_007 Feb 08 '19 at 13:43
  • @Danny_007 : Your question is really unclear :"i understand that if this is greater or equal to 4^n then it is true, but if not, if it is only 4^n> some x we found to be the same expression"... what ? – Surb Feb 08 '19 at 13:47
  • Sorry for that, let me try to simplify. We have 4^n we know that if we express this first formula in a different way we get that sum, so in a logical sense that first formula is greater or equal to the second formula, now i am understanding that we have only expressed the first formula in a different manner and not shown that it is greater but only showed that is greater or equal, also we have only expressed it in a different manner , this means i believe that it is only true if it is greater or equal and not if it is greater, (unless the sum is smaller at some point idk) hope i did it better – Danny_007 Feb 08 '19 at 14:05
  • No, we have proved that $4^n>\binom{2n}{n}$ for all $n\geq 1$. – Surb Feb 08 '19 at 15:26
  • Thanks bro, i now have a better understanding of the proof process – Danny_007 Feb 09 '19 at 08:45