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