## Loser66 one year ago $$\forall n\in\mathbb N$$ $$C_n= \dfrac{1}{n+1}\left(\begin{matrix}2n\\n\end{matrix}\right)$$ Prove that $$C_{n+1}=\left(\begin{matrix}2n\\n\end{matrix}\right)-\left(\begin{matrix}2n\\n-1\end{matrix}\right)$$ Please, help

1. Loser66

Appreciate any help. I have to go now but will be back later

2. ParthKohli

I know that such proofs are rarely elegant, but you can just apply the factorial definition.

3. Loser66

I did, but get nothing $$C_{n+1}= \dfrac{1}{n+2}\left(\begin{matrix}2(n+1)\\n+1\end{matrix}\right)=\dfrac{1}{n+2} \left(\begin{matrix}2n+2\\n+1\end{matrix}\right)\\=\dfrac{1}{n+2}*\dfrac{ (2n+2)!}{(n+1)!(n+1)!}$$ $$= \dfrac{1}{n+2}*\dfrac{(2n+2)(2n+1)2n!}{(n+1)(n+1) n!n!}=\dfrac{1}{n+2}*\dfrac{2(2n+1)2n!}{(n+1)n!n!}$$

4. Loser66

$$C_{n+1}= \dfrac{1}{n+1}\dfrac{2n!}{n!n!}*\dfrac{4n+2}{n+2}=\left(\begin{matrix}2n\\n\end{matrix}\right)\left(\dfrac{4n+2}{(n+2)(n+1)}\right)$$

5. Loser66

@mathstudent55

Find more explanations on OpenStudy