chứng minh rằng neu n là sô nguyen dương thì (2n)! < 2^(2n) * (n!)^2

Giải thích các bước giải:

Với $n=1\to \begin{cases}\left(2\cdot 1\right)!=2!=2\\ 2^{2\cdot 1}\cdot \left(1!\right)^2=4\end{cases}$

$\to \left(2\cdot 1\right)!<2^{2\cdot 1}\cdot \left(1!\right)^2$ đúng

Giả sử $n=k\to \left(2k\right)!<2^{2k}\cdot \left(k!\right)^2\left(*\right)$ đúng

Ta cần chứng minh $n=k+1$ thì mệnh đề vẫn đúng

$\to 2^{2\left(k+1\right)}\cdot \left(\left(k+1\right)!\right)^2>\left(2\left(k+1\right)\right)!$

Thật vậy ta có:

$2^{2k+2}\left(\left(k+1\right)!\right)^2-\left(2k+2\right)!$

$=2^2\cdot 2^{2k}\left(\left(k+1\right)\cdot k!\right)^2-\left(2k+2\right)!$

$=4\cdot 2^{2k}\cdot\left(k+1\right)^2\cdot \left(k!\right)^2-\left(2k+2\right)!$

$=4\left(k+1\right)^2\cdot 2^{2k}\cdot \left(k!\right)^2-\left(2k+2\right)!$ 

$>4\left(k+1\right)^2\cdot\left(2k\right)!-\left(2k+2\right)!\left(*\right)$

$>\left(2k+2\right)^2\cdot\left(2k\right)!-\left(2k+2\right)!$

$>\left(2k+1\right)\cdot \left(2k+2\right)\cdot\left(2k\right)!-\left(2k+2\right)!$

$>\left(2k+2\right)!-\left(2k+2\right)!$

$>0$

$\to 2^{2k+2}\left(\left(k+1\right)!\right)^2>\left(2k+2\right)!$

$\to n=k+1$ đúng

$\to$Giả sử đúng