Question

$Demonstrati~ca~n!\geq 2^{n}, n\geq 4.$

Answer 1

Demonstram prin inductie matematica :P(n):n!≥2ⁿ,∀n≥4

1.Verificarea:P(4):4!≥2⁴<=>1*2*3*4≥2*2*2*2|:8<=>3≥2(A)

2.Demonstratia:P(k)->P(k+1),∀k≥4

fie v(P(k))=1∀k≥4

$P(k):k!\geq 2^k, \forall k \geq 4\\P(k+1):(k+1)!\geq 2^{k+1},\forall k \geq 4$

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$(k+1)!\geq 2^{k+1},\forall k \geq 4<=>k!(k+1)\geq 2*2^k,\forall k \geq 4<=>k!\geq \frac{2*2^k}{k+1} (A),\forall k \geq 4\\=>P(k)_>P(k+1),\forall k \geq 4\\=>v(P(n))=1.\forall n \geq 4$

Answer 2

$n!\geq 2^n,\quad n\geq 4\\ \\ \displaystyle \prod\limits_{k=1}^{n}k\geq \prod\limits_{k=1}^{n}2 \,\Leftrightarrow\\ \\ \Leftrightarrow\,1\cdot 2\cdot 3\cdot \prod\limits_{k=4}^{n}k\geq 2\cdot 2\cdot 2\cdot \prod\limits_{k=4}^{n}2\\ \Leftrightarrow \, \dfrac{3}{4}\prod\limits_{k=4}^{n}k\geq \prod\limits_{k=4}^{n}2$

$\displaystyle \\\diamond\,\,\textbf{Pentru }\,n = 4:\\ \\\dfrac{3}{4}\cdot 4\geq 2 \quad (A)\\ \\ \diamond\,\,\textbf{Pentru }\,n\geq 5:\\ \\ \dfrac{3}{4}\cdot 4\cdot \prod\limits_{k=5}^{n}k \geq 2\cdot \prod\limits_{k=5}^{n}2 \,\Leftrightarrow\,3\prod\limits_{k=5}^{n}k \geq 2 \prod\limits_{k=5}^{n}2\\\\\\ {\!}\left\{\begin{aligned}3&\geq 2 \\ k \geq 2\Rightarrow \prod\limits_{k=5}^n k&\geq \prod\limits_{k=5}^n 2\end{aligned}\right|\,\Rightarrow\,3\prod\limits_{k=5}^{n}k \geq 2 \prod\limits_{k=5}^{n}2\quad (A)$

$\Rightarrow\, n!\geq 2^n,\quad \forall n\geq 4$