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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 \(C_{n+1}=\left(\begin{matrix}2n\\n\end{matrix}\right)\left(\begin{matrix}2n\\n1\end{matrix}\right)\)
Please, help
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 \(C_{n+1}=\left(\begin{matrix}2n\\n\end{matrix}\right)\left(\begin{matrix}2n\\n1\end{matrix}\right)\) Please, help

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Loser66
 one year ago
Best ResponseYou've already chosen the best response.0Appreciate any help. I have to go now but will be back later

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.0I know that such proofs are rarely elegant, but you can just apply the factorial definition.

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0I 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
 one year ago
Best ResponseYou've already chosen the best response.0\(C_{n+1}= \dfrac{1}{n+1}\dfrac{2n!}{n!n!}*\dfrac{4n+2}{n+2}=\left(\begin{matrix}2n\\n\end{matrix}\right)\left(\dfrac{4n+2}{(n+2)(n+1)}\right)\)
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