Question and Answers Forum

All Questions      Topic List

None Questions

Previous in All Question      Next in All Question      

Previous in None      Next in None      

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 bytalminator2856791 last updated on 17/Dec/20

i was having the time of my life till i read ♮by mathematical inductionε! hahahhhhahha

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

Commented byZiYangLee 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

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

Terms of Service

Privacy Policy

Contact: info@tinkutara.com