Menu Close

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




Question Number 73755 by ~blr237~ last updated on 15/Nov/19
show that   for all integer  n ,  n+1 divides  (((2n)),(n) )
$${show}\:{that}\:\:\:{for}\:{all}\:{integer}\:\:{n}\:,\:\:{n}+\mathrm{1}\:{divides}\:\begin{pmatrix}{\mathrm{2}{n}}\\{{n}}\end{pmatrix} \\ $$
Answered by mind is power last updated on 15/Nov/19
 (((2n)),(n) )=(((2n)!)/(n!n!))∈N  (n/(n+1)) (((2n)),(n) )=((2n!)/(n!.(n)!)).(n/(n+1))=((2n!)/((n−1)!.(n+1)!))= (((2n)),((n+1)) )  ⇔n (((2n)),(n) )=(n+1) (((2n)),((n+1)) )  since n and n+1 are coprim  we have n+1∣(n+1) (((2n)),((n+1)) )=n (((2n)),(n) )⇒(n+1)∣n. (((2n)),(n) )  n+1 and n are coprime⇒(n+1)∣ (((2n)),(n) )
$$\begin{pmatrix}{\mathrm{2}{n}}\\{{n}}\end{pmatrix}=\frac{\left(\mathrm{2}{n}\right)!}{{n}!{n}!}\in\mathbb{N} \\ $$$$\frac{{n}}{{n}+\mathrm{1}}\begin{pmatrix}{\mathrm{2}{n}}\\{{n}}\end{pmatrix}=\frac{\mathrm{2}{n}!}{{n}!.\left({n}\right)!}.\frac{{n}}{{n}+\mathrm{1}}=\frac{\mathrm{2}{n}!}{\left({n}−\mathrm{1}\right)!.\left({n}+\mathrm{1}\right)!}=\begin{pmatrix}{\mathrm{2}{n}}\\{{n}+\mathrm{1}}\end{pmatrix} \\ $$$$\Leftrightarrow{n}\begin{pmatrix}{\mathrm{2}{n}}\\{{n}}\end{pmatrix}=\left({n}+\mathrm{1}\right)\begin{pmatrix}{\mathrm{2}{n}}\\{{n}+\mathrm{1}}\end{pmatrix} \\ $$$${since}\:{n}\:{and}\:{n}+\mathrm{1}\:{are}\:{coprim} \\ $$$${we}\:{have}\:{n}+\mathrm{1}\mid\left({n}+\mathrm{1}\right)\begin{pmatrix}{\mathrm{2}{n}}\\{{n}+\mathrm{1}}\end{pmatrix}={n}\begin{pmatrix}{\mathrm{2}{n}}\\{{n}}\end{pmatrix}\Rightarrow\left({n}+\mathrm{1}\right)\mid{n}.\begin{pmatrix}{\mathrm{2}{n}}\\{{n}}\end{pmatrix} \\ $$$${n}+\mathrm{1}\:{and}\:{n}\:{are}\:{coprime}\Rightarrow\left({n}+\mathrm{1}\right)\mid\begin{pmatrix}{\mathrm{2}{n}}\\{{n}}\end{pmatrix} \\ $$$$ \\ $$
Commented by MJS last updated on 15/Nov/19
great!
$$\mathrm{great}! \\ $$
Commented by mind is power last updated on 15/Nov/19
thanx sir mjs
$${thanx}\:{sir}\:{mjs} \\ $$

Leave a Reply

Your email address will not be published. Required fields are marked *