Question Number 2750 by RasheedAhmad last updated on 26/Nov/15
$${Prove}\:{that}: \\ $$$$\left({i}\right)\:\:\mathrm{2}^{{n}} >{n}^{\mathrm{2}} \:\:{for}\:{all}\:{integral}\:{values} \\ $$$${of}\:\:{n}\geqslant\mathrm{5} \\ $$$$\left({ii}\right)\:{n}!>\mathrm{3}^{{n}−\mathrm{1}} ,{for}\:{all}\:{integral}\:{values} \\ $$$${of}\:{n}\geqslant\mathrm{5} \\ $$
Answered by Yozzi last updated on 26/Nov/15
![(Skeletons of proofs) (i) Let P(n): 2^n >n^2 , ∀n≥5,n∈Z. P(5): 2^5 =32>25=5^2 ⇒P(n) true for n=5. Supppose P(n) true for n=k (k≥5): 2^k >k^2 P(k+1): 2^k >k^2 ×2: 2^(k+1) >2k^2 2^(k+1) >k^2 +k^2 −2k+1+2k−1 2^(k+1) >k^2 −2k−1+(k+1)^2 2^(k+1) >(k−1)^2 −2+(k+1)^2 ∵ k≥5⇒k−1≥4⇒(k−1)^2 ≥16 (k−1)^2 −2≥14>0 ∴(k−1)^2 −2+(k+1)^2 >(k+1)^2 Hence, 2^(k+1) >(k+1)^2 . ∴ P(k)⇒P(k+1) ∵ P(5) is true, by P.M.I, 2^n >n^2 ∀n≥5,n∈Z. (ii) Let P(n): n!>3^(n−1) , n≥5,n∈Z. P(5): 5!=120>3^4 =81 ∴ P(n) true for n=5. Suppose P(n) true for n=k (k≥5): k!>3^(k−1) P(k+1): k!>3^(k−1) ×(k+1>0): (k+1)!>(k+1)3^(k−1) ∵ k≥5⇒k+1≥6⇒(k+1)3^(k−1) ≥6×3^(k−1) (k+1)3^(k−1) ≥2×3^k >3^k Hence, (k+1)!>3^([k+1]−1) . ∴ P(k)⇒P(k+1) Since P(5) is true, by P.M.I, k!>3^(k−1) ∀n≥5,n∈Z.](https://www.tinkutara.com/question/Q2756.png)
$$\left({Skeletons}\:{of}\:{proofs}\right) \\ $$$$\left({i}\right)\:{Let}\:{P}\left({n}\right):\:\mathrm{2}^{{n}} >{n}^{\mathrm{2}} \:,\:\forall{n}\geqslant\mathrm{5},{n}\in\mathbb{Z}. \\ $$$${P}\left(\mathrm{5}\right):\:\mathrm{2}^{\mathrm{5}} =\mathrm{32}>\mathrm{25}=\mathrm{5}^{\mathrm{2}} \\ $$$$\Rightarrow{P}\left({n}\right)\:{true}\:{for}\:{n}=\mathrm{5}. \\ $$$$ \\ $$$${Supppose}\:{P}\left({n}\right)\:{true}\:{for}\:{n}={k}\:\left({k}\geqslant\mathrm{5}\right): \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{2}^{{k}} >{k}^{\mathrm{2}} \\ $$$${P}\left({k}+\mathrm{1}\right):\:\mathrm{2}^{{k}} >{k}^{\mathrm{2}} \\ $$$$×\mathrm{2}:\:\:\:\:\:\:\:\:\:\:\mathrm{2}^{{k}+\mathrm{1}} >\mathrm{2}{k}^{\mathrm{2}} \\ $$$$\:\:\:\:\mathrm{2}^{{k}+\mathrm{1}} >{k}^{\mathrm{2}} +{k}^{\mathrm{2}} −\mathrm{2}{k}+\mathrm{1}+\mathrm{2}{k}−\mathrm{1} \\ $$$$\:\:\:\:\:\mathrm{2}^{{k}+\mathrm{1}} >{k}^{\mathrm{2}} −\mathrm{2}{k}−\mathrm{1}+\left({k}+\mathrm{1}\right)^{\mathrm{2}} \\ $$$$\mathrm{2}^{{k}+\mathrm{1}} >\left({k}−\mathrm{1}\right)^{\mathrm{2}} −\mathrm{2}+\left({k}+\mathrm{1}\right)^{\mathrm{2}} \\ $$$$\because\:{k}\geqslant\mathrm{5}\Rightarrow{k}−\mathrm{1}\geqslant\mathrm{4}\Rightarrow\left({k}−\mathrm{1}\right)^{\mathrm{2}} \geqslant\mathrm{16} \\ $$$$\left({k}−\mathrm{1}\right)^{\mathrm{2}} −\mathrm{2}\geqslant\mathrm{14}>\mathrm{0} \\ $$$$\therefore\left({k}−\mathrm{1}\right)^{\mathrm{2}} −\mathrm{2}+\left({k}+\mathrm{1}\right)^{\mathrm{2}} >\left({k}+\mathrm{1}\right)^{\mathrm{2}} \\ $$$${Hence},\:\mathrm{2}^{{k}+\mathrm{1}} >\left({k}+\mathrm{1}\right)^{\mathrm{2}} . \\ $$$$\therefore\:{P}\left({k}\right)\Rightarrow{P}\left({k}+\mathrm{1}\right) \\ $$$$\because\:{P}\left(\mathrm{5}\right)\:{is}\:{true},\:{by}\:{P}.{M}.{I},\:\mathrm{2}^{{n}} >{n}^{\mathrm{2}} \: \\ $$$$\forall{n}\geqslant\mathrm{5},{n}\in\mathbb{Z}. \\ $$$$ \\ $$$$\left({ii}\right)\:{Let}\:{P}\left({n}\right):\:{n}!>\mathrm{3}^{{n}−\mathrm{1}} ,\:{n}\geqslant\mathrm{5},{n}\in\mathbb{Z}. \\ $$$${P}\left(\mathrm{5}\right):\:\mathrm{5}!=\mathrm{120}>\mathrm{3}^{\mathrm{4}} =\mathrm{81} \\ $$$$\therefore\:{P}\left({n}\right)\:{true}\:{for}\:{n}=\mathrm{5}. \\ $$$$ \\ $$$${Suppose}\:{P}\left({n}\right)\:{true}\:{for}\:{n}={k}\:\left({k}\geqslant\mathrm{5}\right): \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{k}!>\mathrm{3}^{{k}−\mathrm{1}} \\ $$$${P}\left({k}+\mathrm{1}\right):\:\:{k}!>\mathrm{3}^{{k}−\mathrm{1}} \\ $$$$×\left({k}+\mathrm{1}>\mathrm{0}\right):\:\:\left({k}+\mathrm{1}\right)!>\left({k}+\mathrm{1}\right)\mathrm{3}^{{k}−\mathrm{1}} \\ $$$$\because\:{k}\geqslant\mathrm{5}\Rightarrow{k}+\mathrm{1}\geqslant\mathrm{6}\Rightarrow\left({k}+\mathrm{1}\right)\mathrm{3}^{{k}−\mathrm{1}} \geqslant\mathrm{6}×\mathrm{3}^{{k}−\mathrm{1}} \\ $$$$\left({k}+\mathrm{1}\right)\mathrm{3}^{{k}−\mathrm{1}} \geqslant\mathrm{2}×\mathrm{3}^{{k}} >\mathrm{3}^{{k}} \\ $$$${Hence},\:\left({k}+\mathrm{1}\right)!>\mathrm{3}^{\left[{k}+\mathrm{1}\right]−\mathrm{1}} . \\ $$$$\therefore\:{P}\left({k}\right)\Rightarrow{P}\left({k}+\mathrm{1}\right) \\ $$$${Since}\:{P}\left(\mathrm{5}\right)\:{is}\:{true},\:{by}\:{P}.{M}.{I}, \\ $$$${k}!>\mathrm{3}^{{k}−\mathrm{1}} \:\forall{n}\geqslant\mathrm{5},{n}\in\mathbb{Z}. \\ $$
Commented by RasheedAhmad last updated on 26/Nov/15
$$\mathcal{N}{ice}! \\ $$