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Question Number 73755 by ~blr237~ last updated on 15/Nov/19

$$showthatforallintegern,n+1divides\left(\begin{array}{c}2n\\ n\end{array}\right)$$

Answered by mind is power last updated on 15/Nov/19

$$\left(\begin{array}{c}2n\\ n\end{array}\right)=\frac{\left(2n\right)!}{n!n!}\in \mathbb{N}$$$$\frac{n}{n+1}\left(\begin{array}{c}2n\\ n\end{array}\right)=\frac{2n!}{n!.\left(n\right)!}.\frac{n}{n+1}=\frac{2n!}{(n-1)!.(n+1)!}=\left(\begin{array}{c}2n\\ n+1\end{array}\right)$$$$\iff n\left(\begin{array}{c}2n\\ n\end{array}\right)=(n+1)\left(\begin{array}{c}2n\\ n+1\end{array}\right)$$$$sincenandn+1arecoprim$$$$wehaven+1\mid (n+1)\left(\begin{array}{c}2n\\ n+1\end{array}\right)=n\left(\begin{array}{c}2n\\ n\end{array}\right)\Rightarrow (n+1)\mid n.\left(\begin{array}{c}2n\\ n\end{array}\right)$$$$n+1andnarecoprime\Rightarrow (n+1)\mid \left(\begin{array}{c}2n\\ n\end{array}\right)$$$$$$

Commented by MJS last updated on 15/Nov/19

$$\mathrm{great}!$$

Commented by mind is power last updated on 15/Nov/19

$$thanxsirmjs$$

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