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Prove-that-2-n-2-gt-n-2-for-n-N-by-mathematical-induction-

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Question Number 126124 by ZiYangLee last updated on 17/Dec/20
Prove that 2^n +2>n^(2 ) for n∈N by mathematical induction.
$$\mathrm{Prove}\:\mathrm{that}\:\mathrm{2}^{{n}} +\mathrm{2}>{n}^{\mathrm{2}\:} \mathrm{for}\:{n}\in\mathbb{N}\:\mathrm{by} \\ $$$$\mathrm{mathematical}\:\mathrm{induction}. \\ $$
Commented by talminator2856791 last updated on 17/Dec/20
i was having the time of my life till i read “by mathematical induction”! hahahhhhahha
$$\:\mathrm{i}\:\mathrm{was}\:\mathrm{having}\:\mathrm{the}\:\mathrm{time}\:\mathrm{of}\:\mathrm{my}\:\mathrm{life}\:\mathrm{till}\:\mathrm{i}\:\mathrm{read}\:“\mathrm{by}\:\mathrm{mathematical}\:\mathrm{induction}''! \\ $$$$\:\mathrm{hahahhhhahha} \\ $$$$\: \\ $$
Commented by ZiYangLee last updated on 30/Dec/20
Good luck haha
$$\mathrm{Good}\:\mathrm{luck}\:\mathrm{haha} \\ $$
Answered by mahdipoor last updated on 17/Dec/20
we know: n≥4⇒n^2 ≥2n+1 (∗) ∀n≥4⇒if 2^n ≥n^2 ⇒2×2^n ≥2×n^2 ⇒ 2^(n+1) ≥n^2 +n^2 ⇒^∗ 2^(n+1) ≥n^2 +2n+1 ⇒ 2^(n+1) ≥(n+1)^2 (∗∗) for n=4 ⇒2^n ≥n^2 ⇒^(∗∗) for: n=4+1=5⇒2^n ≥n^2 ⇒^(∗∗) for: n=5+1=6⇒2^n ≥n^2 .... ⇒for:∀n≥4⇒2^n ≥n^2 ⇒2^n +2>n^2 for:n=1,2,3⇒2^n +2>n^2 ⇒⇒for:∀n∈N⇒2^n +2>n^2
$${we}\:{know}:\:{n}\geqslant\mathrm{4}\Rightarrow{n}^{\mathrm{2}} \geqslant\mathrm{2}{n}+\mathrm{1}\:\:\:\:\left(\ast\right) \\ $$$$\forall{n}\geqslant\mathrm{4}\Rightarrow{if}\:\mathrm{2}^{{n}} \geqslant{n}^{\mathrm{2}} \Rightarrow\mathrm{2}×\mathrm{2}^{{n}} \geqslant\mathrm{2}×{n}^{\mathrm{2}} \Rightarrow \\ $$$$\mathrm{2}^{{n}+\mathrm{1}} \geqslant{n}^{\mathrm{2}} +{n}^{\mathrm{2}} \:\:\overset{\ast} {\Rightarrow}\:\:\mathrm{2}^{{n}+\mathrm{1}} \geqslant{n}^{\mathrm{2}} +\mathrm{2}{n}+\mathrm{1}\:\Rightarrow \\ $$$$\mathrm{2}^{{n}+\mathrm{1}} \geqslant\left({n}+\mathrm{1}\right)^{\mathrm{2}} \:\:\:\:\left(\ast\ast\right) \\ $$$${for}\:{n}=\mathrm{4}\:\Rightarrow\mathrm{2}^{{n}} \geqslant{n}^{\mathrm{2}} \\ $$$$\:\overset{\ast\ast} {\Rightarrow}{for}:\:{n}=\mathrm{4}+\mathrm{1}=\mathrm{5}\Rightarrow\mathrm{2}^{{n}} \geqslant{n}^{\mathrm{2}} \\ $$$$\:\overset{\ast\ast} {\Rightarrow}{for}:\:{n}=\mathrm{5}+\mathrm{1}=\mathrm{6}\Rightarrow\mathrm{2}^{{n}} \geqslant{n}^{\mathrm{2}} \\ $$$$…. \\ $$$$\Rightarrow{for}:\forall{n}\geqslant\mathrm{4}\Rightarrow\mathrm{2}^{{n}} \geqslant{n}^{\mathrm{2}} \Rightarrow\mathrm{2}^{{n}} +\mathrm{2}>{n}^{\mathrm{2}} \\ $$$${for}:{n}=\mathrm{1},\mathrm{2},\mathrm{3}\Rightarrow\mathrm{2}^{{n}} +\mathrm{2}>{n}^{\mathrm{2}} \\ $$$$\Rightarrow\Rightarrow{for}:\forall{n}\in{N}\Rightarrow\mathrm{2}^{{n}} +\mathrm{2}>{n}^{\mathrm{2}} \\ $$
Answered by physicstutes last updated on 17/Dec/20
prove for n = 1, (2)^1 +2 > 1^2 assume for n = k ⇒ 2^k +2 > k^2 prove for n = k+1. 2^(k+1) +2 = 2^k .2 + 2 = 2(2^k +1) from 2^k +2 > k^2 ⇒ 2(2^(k−1) +1) > k^2 but ∀ n ∈ N, 2^(k−1) < 2^k ⇒ 2(2^k +1) > (k+1)^2 thus true ∀ n ∈ N
$$\mathrm{prove}\:\mathrm{for}\:{n}\:=\:\mathrm{1},\:\:\left(\mathrm{2}\right)^{\mathrm{1}} +\mathrm{2}\:>\:\mathrm{1}^{\mathrm{2}} \\ $$$$\mathrm{assume}\:\mathrm{for}\:{n}\:=\:{k} \\ $$$$\Rightarrow\:\:\:\mathrm{2}^{{k}} +\mathrm{2}\:>\:{k}^{\mathrm{2}} \\ $$$$\mathrm{prove}\:\mathrm{for}\:{n}\:=\:{k}+\mathrm{1}. \\ $$$$\:\mathrm{2}^{{k}+\mathrm{1}} +\mathrm{2}\:=\:\mathrm{2}^{{k}} .\mathrm{2}\:+\:\mathrm{2}\:=\:\mathrm{2}\left(\mathrm{2}^{{k}} +\mathrm{1}\right) \\ $$$$\:\mathrm{from}\:\:\mathrm{2}^{{k}} +\mathrm{2}\:>\:{k}^{\mathrm{2}} \\ $$$$\Rightarrow\:\mathrm{2}\left(\mathrm{2}^{{k}−\mathrm{1}} +\mathrm{1}\right)\:>\:{k}^{\mathrm{2}} \\ $$$$\mathrm{but}\:\forall\:{n}\:\in\:\mathbb{N},\:\mathrm{2}^{{k}−\mathrm{1}} <\:\mathrm{2}^{{k}} \\ $$$$\Rightarrow\:\mathrm{2}\left(\mathrm{2}^{{k}} +\mathrm{1}\right)\:>\:\left({k}+\mathrm{1}\right)^{\mathrm{2}} \\ $$$$\mathrm{thus}\:\mathrm{true}\:\forall\:{n}\:\in\:\mathbb{N} \\ $$

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