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[Гост]

Бернули?

[Добромир Глухаров]

https://en.wikipedia.org/wiki/Stirling%27s_approximation

$n\to\infty\Rightarrow n!\sim\sqrt{2\pi n}n^n e^{-n}$

${4n+1\choose2n}=\frac{(4n+1)!}{(2n)!(2n+1)!}\sim\frac{\sqrt{2\pi(4n+1)}(4n+1)^{4n+1}e^{-4n-1}}{\sqrt{4\pi n}(2n)^{2n}e^{-2n}\sqrt{4\pi n+2\pi}(2n+1)^{2n+1}e^{-2n-1}}\sim$

$\sim\frac{\sqrt{\pi}\sqrt{8n+2}(4n+1)^{4n+1}e^{-4n-1}}{\pi\sqrt{4n}\sqrt{4n+2}(2n)^{2n}(2n+1)^{2n+1}e^{-4n-1}}\sim\frac{1}{\sqrt{\pi}}\sqrt{\frac{8n+2}{16n^2+8n}}\cdot2^{4n+1}\cdot\frac{(2n+0,5)^{4n+1}}{(2n)^{2n}(2n+1)^{2n+1}}\sim$

$\sim\frac{1}{\sqrt{\pi}}\cdot\frac{1}{\sqrt{2n}}\cdot2^{4n+1}\cdot\frac{2n+0,5}{2n+1}\cdot\left[\frac{(2n+0,5)^2}{(2n)(2n+1)}\right]^{2n}\sim\frac{1}{\sqrt{2\pi n}}\cdot2^{4n+1}.1.\left(\frac{4n^2+2n+0,25}{4n^2+2n}\right)^{2n}\sim$

$\sim\frac{1}{\sqrt{2\pi n}}\cdot2^{4n+1}.\left(1+\frac{1}{16n^2+8n}\right)^{2n}\sim\frac{1}{\sqrt{2\pi n}}\cdot2^{4n+1}.\left(1+\frac{1}{16n^2+8n}\right)^{\frac{16n^2+8n}{8n}-1}\sim$

$\sim\frac{1}{\sqrt{2\pi n}}\cdot2^{4n+1}.e^{\frac{1}{8n}}\left(1+\frac{1}{16n^2+8n}\right)^{-1}\sim\frac{1}{\sqrt{2\pi n}}\cdot2^{4n+1}$

$\sqrt[n]{{4n+1\choose2n}}\sim\frac{1}{\sqrt[2n]{2\pi n}}\cdot2^{4+\frac{1}{n}}\sim2^4=16$

[Сийка]

В успоредника ABCD точката M е средата на страната AD . Ако MB = n , MC = m и ъгъл BAC = x , x е по - малко от 90 градуса , то намерете лицето на успоредника .

[Добромир Глухаров]

Не толкова строго в смисъл на доказателство, може да се изведе така:

${4n+1\choose2n}=\frac{(4n+1)!}{(2n)!(2n+1)!}=\frac{2^{4n+1}(2n+0,5)(2n)(2n-0,5)(2n-1)...1,5.1.0,5}{(2n+1)(2n)^2(2n-1)^2...2^2.1^2}=$

$=2^{4n+1}\cdot\frac{2n+0,5}{2n+1}\cdot\frac{2n}{2n}\cdot\frac{2n-0,5}{2n}\cdot\frac{2n-1}{2n-1}\cdot\frac{2n-1,5}{2n-1}\cdots\frac{3}{3}\cdot\frac{2,5}{3}\cdot\frac{2}{2}\cdot\frac{1,5}{2}\cdot\frac{1}{1}\cdot\frac{0,5}{1}=$

$=2^{4n+1}\cdot\frac{2n+0,5}{2n+1}\cdot\prod_{k=0}^{2n-1}\frac{2n-k}{2n-k}\cdot\frac{2n-k-0,5}{2n-k}$

$n\to\infty\Rightarrow{4n+1\choose2n}\sim2^{4n+1}\cdot\frac{1+\frac{0,5}{2n}}{1+\frac{1}{2n}}\cdot\prod_{k=0}^{2n-1}\left(1-\frac{0,5}{2n-k}\right)\sim2^{4n+1}.1.O\left(\frac{1}{n^{\alpha}}\right)\sim\frac{2.16^n}{n^{\alpha}}$

$\Rightarrow\lim_{n\to\infty}\sqrt[n]{{4n+1\choose2n}}=\lim_{n\to\infty}\sqrt[n]{\frac{2.16^n}{n^{\alpha}}}=16$