show-that-for-all-integer-n-n-1-divides-2n-n-

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

s h o w t h a t f o r a l l i n t e g e r n, n + 1 d i v i d e s (\begin{matrix} 2 n \\ n \end{matrix})

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

(\begin{matrix} 2 n \\ n \end{matrix}) = \frac{(2 n)!}{n! n!} \in N

\frac{n}{n + 1} (\begin{matrix} 2 n \\ n \end{matrix}) = \frac{2 n!}{n! . (n)!} . \frac{n}{n + 1} = \frac{2 n!}{(n - 1)! . (n + 1)!} = (\begin{matrix} 2 n \\ n + 1 \end{matrix})

\Leftrightarrow n (\begin{matrix} 2 n \\ n \end{matrix}) = (n + 1) (\begin{matrix} 2 n \\ n + 1 \end{matrix})

s i n c e n a n d n + 1 a r e c o p r i m

w e h a v e n + 1 ∣ (n + 1) (\begin{matrix} 2 n \\ n + 1 \end{matrix}) = n (\begin{matrix} 2 n \\ n \end{matrix}) \Rightarrow (n + 1) ∣ n . (\begin{matrix} 2 n \\ n \end{matrix})

n + 1 a n d n a r e c o p r i m e \Rightarrow (n + 1) ∣ (\begin{matrix} 2 n \\ n \end{matrix})

Commented by MJS last updated on 15/Nov/19

great!

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

t h a n x s i r m j s