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Question Number 126124 by ZiYangLee last updated on 17/Dec/20

$$\mathrm{Prove}\mathrm{that}{2}^n+2>n^2\mathrm{for}n\in \mathbb{N}\mathrm{by}$$ $$\mathrm{mathematical}\mathrm{induction}.$$

Commented bytalminator2856791 last updated on 17/Dec/20

$$\mathrm{i}\mathrm{was}\mathrm{having}\mathrm{the}\mathrm{time}\mathrm{of}\mathrm{my}\mathrm{life}\mathrm{till}\mathrm{i}\mathrm{read}\mathrm{♮}\mathrm{by}\mathrm{mathematical}\mathrm{induction}\epsilon !$$ $$\mathrm{hahahhhhahha}$$ $$$$

Commented byZiYangLee last updated on 30/Dec/20

$$\mathrm{Good}\mathrm{luck}\mathrm{haha}$$

Answered by mahdipoor last updated on 17/Dec/20

$$weknow:n⩾4\Rightarrow n^2⩾2n+1(\ast )$$ $$\mathrm{\forall }n⩾4\Rightarrow if{2}^n⩾n^2\Rightarrow 2\times 2^n⩾2\times n^2\Rightarrow$$ $$2^{n+1}⩾n^2+n^2\stackrel{\ast }{\Rightarrow }{2}^{n+1}⩾n^2+2n+1\Rightarrow$$ $$2^{n+1}⩾{(n+1)}^2(\ast \ast )$$ $$forn=4\Rightarrow 2^n⩾n^2$$ $$\stackrel{\ast \ast }{\Rightarrow }for:n=4+1=5\Rightarrow 2^n⩾n^2$$ $$\stackrel{\ast \ast }{\Rightarrow }for:n=5+1=6\Rightarrow 2^n⩾n^2$$ $$....$$ $$\Rightarrow for:\mathrm{\forall }n⩾4\Rightarrow 2^n⩾n^2\Rightarrow 2^n+2>n^2$$ $$for:n=1,2,3\Rightarrow 2^n+2>n^2$$ $$\Rightarrow \Rightarrow for:\mathrm{\forall }n\in N\Rightarrow 2^n+2>n^2$$

Answered by physicstutes last updated on 17/Dec/20

$$\mathrm{prove}\mathrm{for}n=1,{\left(2\right)}^1+2>1^2$$ $$\mathrm{assume}\mathrm{for}n=k$$ $$\Rightarrow 2^k+2>k^2$$ $$\mathrm{prove}\mathrm{for}n=k+1.$$ $$2^{k+1}+2=2^k.2+2=2(2^k+1)$$ $$\mathrm{from}{2}^k+2>k^2$$ $$\Rightarrow 2(2^{k-1}+1)>k^2$$ $$\mathrm{but}\mathrm{\forall }n\in \mathbb{N},2^{k-1}<2^k$$ $$\Rightarrow 2(2^k+1)>{(k+1)}^2$$ $$\mathrm{thus}\mathrm{true}\mathrm{\forall }n\in \mathbb{N}$$

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