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I want to show that $n+1$ is a divisor of $\displaystyle\binom{2n} {n}$, for all $n\in\mathbb{N}.$

I have tried to show it by induction and Pascal's rule but it did not worked out.

I would appreciate some help.

Iuli
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  • It might be worth looking at Catalan number: http://en.wikipedia.org/wiki/Catalan_number – Thomas Andrews Aug 31 '12 at 17:09
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    $\binom{2n}{n} = \frac{2n(2n-1)\cdots(n+1)}{n!}$. You can further factor the numerator and cancel out $n!$ leaving a product of terms which include $(n+1)$. –  Aug 31 '12 at 17:13
  • @JenniferDylan Does that mean that in general $\binom n k$ is divisible by $n-k+1$? Because you haven't said anything that makes this true for this particular case. – Thomas Andrews Aug 31 '12 at 17:17
  • @spohreis regarding my comment above, I might be wrong. The numerator has a factor $2n(2n-2)(2n-4)\cdots$ but the product doesn't end with $2$. I have to write it down. –  Aug 31 '12 at 17:21

6 Answers6

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From the Wikipedia page for Catalan numbers: http://en.wikipedia.org/wiki/Catalan_number

Note that $$\binom {2n}{n+1} = \frac{(2n)!}{(n-1)!(n+1)!} = \frac{n}{n+1} \frac{(2n)!}{n!n!} = \frac{n}{n+1}\binom{2n}{n}$$

So $$\frac{1}{n+1}\binom{2n}{n} = \binom{2n}{n} - \binom{2n}{n+1}$$ is an integer.

Thomas Andrews
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$$\frac{n}{n+1}\binom{2n} {n} = \binom{2n} {n+1} \textrm{ is an integer.}$$

TonyK
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  • I also upvoted Merosity's answer and prefer it a bit over yours since it only requires elementary number theory... – CopyPasteIt Sep 26 '23 at 13:38
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Let $$x=\frac{1}{n+1}{2n\choose n}\implies x=\frac{(n+1)-n}{n+1}{2n\choose n}$$ $$ \implies x={2n\choose n}-{2n\choose n+1}$$.

Since $n\choose k$ is always an integer $\implies x$ is an integer $\implies (n+1)$ divides ${2n\choose n}$

Aang
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Special case $\rm\,k = 2n\,$ of the following

Theorem $\displaystyle \rm\ \ \color{#0af}{(n\!+\!1,k\!+\!1) = 1}\:\Rightarrow\: \color{#c00}{n\!+\!1}\:\bigg|\:\color{#0a0}{{k\choose n}}$

Proof $\displaystyle\rm\ \ \ \color{#c00}{(n\!+\!1)}{k\choose n\!+\!1}\, =\, \frac{(n\!+\!1)\,k!}{(n\!+\!1)! \,(k\!-\!n\!-\!1)!}\, =\, \frac{(k\!-\!n)\,k!}{n!\, (k\!-\!n)!} \,=\, \color{#c00}{(k\!-\!n)}\color{#0a0}{{k\choose n}}$

so the result follows from Euclid's Lemma, since $\rm\:\color{#c00}{(n\!+\!1,k\!-\!n)} = \color{#0af}{(n\!+\!1,k\!+\!1) = 1}.\ \ \bf\small QED$

Bill Dubuque
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Using the property $\binom{a}{b}=\frac{a}{b}\binom{a-1}{b-1}$ which comes by simply looking at the factorials in the binomial coefficients: $$\binom{2n+1}{n+1}=\frac{2n+1}{n+1}\binom{2n}{n}$$ Because $\gcd(2n+1,n+1)=\gcd(n,n+1)=1$ then $n+1$ must divide $\binom{2n}{n}$.

Merosity
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0

Here is another proof.

Lemma:

$$C^{2n}_{n} \text{ is even, }\forall n \in \mathbb{N}.$$

Proofs can be found here.

Proof:

Let $C^{2n+2}_{n+1} = 2k$.

$$2k = C^{2n+2}_{n+1} = \frac{(2n+2)(2n+1)(2n)...(n+2)}{(n+1)(n)(n-1)...(2)(1)}$$ $$= \frac{(2n+2)(2n+1)}{(n+1)^2} \times C^{2n}_{n}$$ $$= \frac{2(2n+1)}{n+1} \times C^{2n}_{n}$$ $$\therefore (2n+1)C^{2n}_{n} = k(n+1)$$ Note that $$2n+1 = 2(n+1)-1$$ That means $(2n+1)$ and $(n+1)$ are co-prime, provided that $n+1\geq 2$. Thus, $C^{2n}_{n}$ is a multiple of $(n+1)$.

Hirofumi Ryo
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