Question Number 176102 by MathsFan last updated on 12/Sep/22
$$\:\:\:\:\:\:\:\:\:\boldsymbol{\mathrm{Prove}}\:\boldsymbol{\mathrm{that}}\: \\ $$$$\:\:\:\:\:\mathrm{4}\boldsymbol{\mathrm{n}}<\mathrm{2}^{\boldsymbol{\mathrm{n}}} \:\boldsymbol{\mathrm{for}}\:\boldsymbol{\mathrm{all}}\:\boldsymbol{\mathrm{n}}\geqslant\mathrm{5} \\ $$
Answered by a.lgnaoui last updated on 12/Sep/22
![n=5 2^5 =32 4×5=20<32 n=6 2^6 =64 4×6=24<64 n=7 2^7 =128 4×7=28<128 ........................... n+1 2^(n+1) 4(n+1)=4n+4<2^n +4 n[4(n+1)]<n(2^n +4)<n(2×2^n )<n2^(n+1) donc 4(n+1)<2^(n+1) n est vrai pour tout n≥5 (n∈ N)](https://www.tinkutara.com/question/Q176109.png)
$${n}=\mathrm{5}\:\:\:\mathrm{2}^{\mathrm{5}} =\mathrm{32}\:\:\:\:\:\:\:\:\:\mathrm{4}×\mathrm{5}=\mathrm{20}<\mathrm{32} \\ $$$${n}=\mathrm{6}\:\:\:\mathrm{2}^{\mathrm{6}} =\mathrm{64}\:\:\:\:\:\:\:\:\:\mathrm{4}×\mathrm{6}=\mathrm{24}<\mathrm{64} \\ $$$${n}=\mathrm{7}\:\:\:\:\mathrm{2}^{\mathrm{7}} =\mathrm{128}\:\:\:\:\:\:\mathrm{4}×\mathrm{7}=\mathrm{28}<\mathrm{128} \\ $$$$……………………… \\ $$$${n}+\mathrm{1}\:\:\:\:\:\mathrm{2}^{{n}+\mathrm{1}} \:\:\:\:\:\:\:\:\:\:\mathrm{4}\left({n}+\mathrm{1}\right)=\mathrm{4}{n}+\mathrm{4}<\mathrm{2}^{{n}} +\mathrm{4} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{n}\left[\mathrm{4}\left({n}+\mathrm{1}\right)\right]<{n}\left(\mathrm{2}^{{n}} +\mathrm{4}\right)<{n}\left(\mathrm{2}×\mathrm{2}^{{n}} \right)<{n}\mathrm{2}^{{n}+\mathrm{1}} \\ $$$${donc}\:\:\:\mathrm{4}\left({n}+\mathrm{1}\right)<\mathrm{2}^{{n}+\mathrm{1}} \\ $$$${n}\:{est}\:{vrai}\:{pour}\:{tout}\:{n}\geqslant\mathrm{5}\:\left({n}\in\:\mathbb{N}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$
Commented by MathsFan last updated on 13/Sep/22
$$ \\ $$Merci Monsieur
Answered by mahdipoor last updated on 12/Sep/22
$${if}\:\mathrm{4}{a}<\mathrm{2}^{{a}} \:\Rightarrow \\ $$$$\mathrm{4}{a}+\mathrm{4}<\mathrm{2}^{{a}} +\mathrm{4}=\mathrm{2}^{{a}} ×\left(\mathrm{1}+\frac{\mathrm{4}}{\mathrm{2}^{{a}} }\right)<\mathrm{2}^{{a}} ×\mathrm{2}\:\:\:\left({a}>\mathrm{5}\right) \\ $$$$\Rightarrow\mathrm{4}\left({a}+\mathrm{1}\right)<\mathrm{2}^{{a}+\mathrm{1}} \:\: \\ $$$${and}\:\mathrm{4}×\mathrm{5}<\mathrm{2}^{\mathrm{5}} \\ $$$$\Rightarrow\Rightarrow\:\mathrm{4}{n}<\mathrm{2}^{{n}} \:{for}\:\forall{n}\geqslant\mathrm{5} \\ $$
Commented by MathsFan last updated on 13/Sep/22
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir} \\ $$