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

Question

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

loser66 · Answer

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

parthkohli · Answer

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

loser66 · Answer

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)!} $ 

$= \dfrac{1}{n+2}*\dfrac{(2n+2)(2n+1)2n!}{(n+1)(n+1) n!n!}=\dfrac{1}{n+2}*\dfrac{2(2n+1)2n!}{(n+1)n!n!}$

loser66 · Answer

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

loser66 · Answer

@mathstudent55