Proof help; binomial coefficients

Question

I'm stuck on this and am not sure about the right way to proceed; what I did ended up being entirely wrong so I won't post it -at the time, our professor touched on it and it was entirely different from what I did. I really want to know how to correctly do this problem and I am not able to receive relevant help on it now outside of this site so I would be very appreciative of insight.

Prove that

$$(n+1)\mid{{2n}\choose{n}}$$ for every $n \in\mathbb N$.

There was a suggestion to write the quotient as a difference of binomial coefficients.

Thanks

Answer 1

$$\binom{2n}n=\frac{(2n)!}{n!n!}=\frac{(n+1)(n+2)\cdot\ldots\cdot2n}{n!} \ldots$$

So your problem in fact is to prove that $\;n!\,\mid\,(n+2)\cdot\ldots\cdot2n\;$ and you can try, say, induction here. And it is a pretty beautiful and interesting little exercise.

Answer 2

Note that:

$${{2n}\choose{n}}=\frac{(2n)!}{n!n!}= 2^n \frac{ 1 \cdot 3 \cdot 5 \cdots (2n-1)}{n!}$$

Indeed multiplying both sides by $(n!)^2$

$$(2n)!=2^n\cdot 1\cdot 3\cdot 5\cdots (2n-1)\cdot n!=\\1\cdot 3\cdot 5\cdots (2n-1)\cdot 2\cdot 4\cdot 6\cdots 2n=1\cdot 2\cdot 3\cdots (2n-1)(2n)=(2n)!$$

Answer 3

Following the hint, notice that $$\binom{2n}{n} - \binom{2n}{n+1} = \frac{1}{n+1} \binom{2n}{n}$$ This follows from $$\binom{2n}{n+1} = \frac{n}{n+1} \binom{2n}{n}$$