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Question Number 176102 by MathsFan last updated on 12/Sep/22

$$\mathbf{Prove}\mathbf{that}$$ $$4\mathbf{n}<2^{\mathbf{n}}\mathbf{for}\mathbf{all}\mathbf{n}⩾5$$

Answered by a.lgnaoui last updated on 12/Sep/22

$$n=52^5=324\times 5=20<32$$ $$n=62^6=644\times 6=24<64$$ $$n=72^7=1284\times 7=28<128$$ $$...........................$$ $$n+12^{n+1}4(n+1)=4n+4<2^n+4$$ $$n\left[4(n+1)\right]<n(2^n+4)<n\left(2\times 2^n\right)<n{2}^{n+1}$$ $$donc4(n+1)<2^{n+1}$$ $$nestvraipourtoutn⩾5(n\in \mathbb{N})$$ $$$$

Commented byMathsFan last updated on 13/Sep/22

$$$$ Merci Monsieur \n

Answered by mahdipoor last updated on 12/Sep/22

$$if4a<2^a\Rightarrow$$ $$4a+4<2^a+4=2^a\times (1+\frac{4}{2^a})<2^a\times 2(a>5)$$ $$\Rightarrow 4(a+1)<2^{a+1}$$ $$and4\times 5<2^5$$ $$\Rightarrow \Rightarrow 4n<2^{n}for\mathrm{\forall }n⩾5$$

Commented byMathsFan last updated on 13/Sep/22

$$\mathrm{thank}\mathrm{you}\mathrm{sir}$$

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