Question Number 194140 by MM42 last updated on 28/Jun/23
$${prove}\:{it}\:: \\ $$$$\begin{pmatrix}{{n}}\\{{k}}\end{pmatrix}^{\:\mathrm{2}} \geqslant\:\begin{pmatrix}{\:\:\:\:{n}}\\{{k}−\mathrm{1}}\end{pmatrix}\:×\begin{pmatrix}{\:\:{n}}\\{{k}+\mathrm{1}}\end{pmatrix}\:\:\:\:\:;\:\:\:\mathrm{1}\leqslant{k}\leqslant{n}−\mathrm{1} \\ $$$$ \\ $$
Answered by witcher3 last updated on 28/Jun/23
$$\left(\frac{\mathrm{n}!}{\mathrm{k}!.\left(\mathrm{n}−\mathrm{k}\right)!}\right)^{\mathrm{2}} \geqslant\frac{\mathrm{n}!}{\left(\mathrm{n}−\mathrm{k}+\mathrm{1}\right)!.\left(\mathrm{k}−\mathrm{1}\right)!}.\frac{\mathrm{n}!}{\left(\mathrm{k}+\mathrm{1}\right)!.\left(\mathrm{n}−\mathrm{k}−\mathrm{1}\right)!} \\ $$$$\Leftrightarrow\frac{\mathrm{n}−\mathrm{k}}{\mathrm{k}}.\frac{\mathrm{k}+\mathrm{1}}{\left(\mathrm{n}−\mathrm{k}\right)}\geqslant\mathrm{1}\Leftrightarrow\frac{\mathrm{k}+\mathrm{1}}{\mathrm{k}}\geqslant\mathrm{1}\:\mathrm{True} \\ $$
Commented by York12 last updated on 28/Jun/23
$${bro}\:{are}\:{using} \\ $$$${discord}\: \\ $$