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For all n > 0 prove that ${2n}\choose{n}$ < $4^n$

It seems like a simple proof but I'm not sure how to continue it.

What I have so far:

Base Case:

Let n=1 ${2}\choose{1}$ = 2
$4^1$ = 4
2 < 4

Inductive Step: Assume ${2(x+1)}\choose{x+1}$ < $4^{x+1}$

Left side: $\frac{(2x+2)!}{(x+1)1(x+1)!}$

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    Do we have to use induction? – 2017-02-16
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    No, I think I can prove any way. – 2017-02-16

1 Answers 1

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For the inductive step:

$$ \begin{align*} \binom{2n+2}{n+1} &= \frac{(2n+2)!}{(n+1)!\,(n+1)!} \\ &= \frac{(2n+2)(2n+1)}{(n+1)^2} \times \frac{(2n)!}{n!\,n!} \\ &= \frac{(2n+2)(2n+1)}{(n+1)^2} \binom{2n}{n} \\ &= \frac{2(2n+1)}{n+1}\binom{2n}{n} \\ &= \frac{4n + 4 - 2}{n+1}\binom{2n}{n} \\ &= \left(4 - \frac{2}{n+1}\right)\binom{2n}{n} \\ &< 4\binom{2n}{n} \\ &< 4^{n+1}. \end{align*} $$

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    Why are you multiplying by the (2n choose n) – 2017-02-16
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    I'm confused why you're multiplying everything by ${2n}\choose{n}$ – 2017-02-16
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    @JanoyCresva: I edited my answer. Is it more understandable ? – 2017-02-16
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    It's still not clear to me, I don't know if I'm missing something. Where is the ${(2n)!}\choose{n!n!}$ coming from? I meant that as a fraction – 2017-02-16
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    $$(2n+2)! = (2n+2)(2n+1)(2n)(2n-1)...3\times2\times1= (2n+2)(2n+1)(2n)!$$ and $$(n+1)! = (n+1)(n)(n-1)...3*2*1=(n+1)n!$$ $$\therefore (n+1)!(n+1)!= (n+1)n!(n+1)n!=(n+1)^2 n!$$ using those two you can change $\frac{(2n+2)!}{(n+1)!(n+1)!} = \frac{(2n+2)(2n+1)}{(n+1)^2} \times \frac{(2n)!}{n!n!}$ – 2017-02-16
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    Ok I think I got it. I'm still trying to get used to the notation for these. Thanks – 2017-02-16
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    @JanoyCresva No Problem – 2017-02-16