$n+1$ is a divisor of $\binom{2n} {n}$

Question

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

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

I would appreciate some help.

Answer 1

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.

Answer 2

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

Answer 3

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 ${2n\choose n}$

Answer 4

Theorem $\displaystyle \rm\,\ \ \color{#c00}{(n\!+\!1,\color{#0af}{k\!+\!1}) = 1}\:\Rightarrow\: \color{#c00}{n\!+\!1}\:\bigg|\:\color{#0a0}{{k\choose n}}\ \ $ [OP is case $\,\rm k = 2n$]. $\ \ $ Proof $\displaystyle\rm\Bbb Z\ni {k\!+\!1\choose n\!+\!1} = \frac{(\color{#0af}{k\!+\!1})\:\!\color{#0a0}k!}{\color{#c00}{(n\!+\!1)}\:\!\color{#0a0}{n!\:\!(k\!-\!n)!}}$ $=\rm \displaystyle \underbrace{\frac{\color{#0af}{(k\!+\!1)}\:\!\color{#0a0}{\large{k\choose n}^{\phantom{|^|}}}}{\color{#c00}{n\!+\!1}}\Rightarrow\, \color{#c00}{n\!+\!1}\:\bigg|\:\color{#0a0}{{k\choose n}}}_{\rm\textstyle\color{#c00}{(n\!+\!1,\color{#0af}{k\!+\!1})\!=\! 1}}\,$ by Euclid's Lemma.

Answer 5

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 $\binom{2n}{n}$.

Answer 6

This is equivalent to showing that $n!(n+1)!$ is a divisor of $2n!$ because, $$ \frac{1}{(n+1)}\binom{2n}{n} = \frac{1}{n+1}\frac{2n!}{n!n!} = \frac{2n!}{n!(n+1)!}. $$

Further, it is sufficient to show that any prime $p$ which occurs in $n!(n+1)!$ occurs to at least as high a power in $2n!$. We can use Legendre's formula to obtain the largest powers of any prime $p$, $\nu_p\left(n!(n+1)!\right)$ and $\nu_p(2n!)$, that divide $n!(n+1)!$ and $2n!$ respectively. Thus, we need to show

$$ \nu_p(2n!) = \sum_{i=1}^{\infty}\left\lfloor \frac{2n}{p^i} \right\rfloor \ge \nu_p\left(n!(n+1)!\right) = \sum_{i=1}^{\infty}\left\lfloor \frac{n}{p^i} \right\rfloor + \sum_{i=1}^{\infty}\left\lfloor \frac{n+1}{p^i} \right\rfloor \qquad (\because a^{m}a^{n} = a^{m+n}). $$

Observe that $$ \left\lfloor 2\frac{n}{p^i}\right\rfloor = \left\lfloor \frac{n}{p^i}\right\rfloor + \left\lfloor \frac{n}{p^i} + \frac{1}{2}\right\rfloor \qquad \left(\because [nx] = \sum_{i=0}^{n-1}\left\lfloor x + \frac{i}{n}\right\rfloor\right). $$

However, for $i \ge 1$, $$ \left\lfloor \frac{n}{p^i}\right\rfloor + \left\lfloor \frac{n}{p^i} + \frac{1}{2}\right\rfloor \ge \left\lfloor \frac{n}{p^i}\right\rfloor + \left\lfloor \frac{n}{p^i} + \frac{1}{p^i}\right\rfloor \qquad (\because p \ge 2) \\ =\left\lfloor \frac{n}{p^i}\right\rfloor + \left\lfloor \frac{n+1}{p^i} \right\rfloor \\ \implies \sum_{i=1}^{\infty}\left\lfloor \frac{2n}{p^i} \right\rfloor \ge \sum_{i=1}^{\infty}\left\lfloor \frac{n}{p^i} \right\rfloor + \sum_{i=1}^{\infty}\left\lfloor \frac{n+1}{p^i} \right\rfloor. $$

Therefore, $$ \nu_p(2n!) \ge \nu_p\left(n!(n+1)!\right). $$

Hence, $n+1$ divides $\binom{2n}{n}$.

Answer 7

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 $(n+1)$.