i'm trying to proof these two terms. I started with an induction, but I got stuck...
Can anybody help?
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1related : this – mathlove Jun 10 '15 at 13:26
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The first one is easy:
$\displaystyle {{2n}\choose{n}} = \frac{2n}{n} \frac{2n-1}{n-1} \cdots \frac{n+1}{1} > 2 \cdot 2 \cdots 2 = 2^n $
The second one needs Pascal's relation and induction:
$\displaystyle {{2n+1}\choose{n}} = {{2n}\choose{n}} + {{2n}\choose{n-1}} $
$\displaystyle\qquad\qquad = {{2n}\choose{n}} + {{2n-1}\choose{n-1}} + {{2n-1}\choose{n-2}} $
$\displaystyle\qquad\qquad > {{2n}\choose{n}} + {{2n-1}\choose{n-1}} $
$\displaystyle\qquad\qquad > 2^n + 2^n = 2^{n+1} $
lhf
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ah and ${{2(n-1)+1}\choose {n-1}}=2{{2(n-1)}\choose {n-1}}\gt 2(2^{n-1})=2^n$ – miniparser Jun 11 '15 at 15:33
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@underdog, no. I meant this: ${{2n-1}\choose{n-1}}={{2(n-1)+1}\choose {n-1}} > 2^{(n-1)+1}$. – lhf Jun 11 '15 at 16:20
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yes. i know. 6-1 half dozen-other. one twice the other per twice as many subsets with $+1$ element. actually i don't follow without claiming one twice the other. – miniparser Jun 11 '15 at 17:34
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Induction works pretty well, since: $$\frac{\binom{2n+2}{n+1}}{\binom{2n}{n}}=\frac{4n+2}{n+1}=2\cdot\frac{2n+1}{n+1}\geq 3 $$ and: $$\frac{\binom{2n+3}{n+1}}{\binom{2n+1}{n}}=2\cdot\frac{2n+3}{n+2}\geq 3$$ as soon as $n\geq 1$.
Jack D'Aurizio
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that's cool. never seen induction done like that. so base case and then assumption and then times $3+\gt$ times $2$ – miniparser Jun 11 '15 at 15:44