Olimpíadas ⇒ (Olimpíada do Pará-12)- Indução matemática Tópico resolvido

[K1llua]

Bom dia, retirei essa questão do livro "Elementos da matemática- Vol. 03"

Observe que:

[tex3]12=3\times 4 [/tex3]
[tex3]1122=33\times 34[/tex3]
[tex3]111222=333\times 334[/tex3]
[tex3]11112222=3333\times 3334[/tex3]

Demonstre que [tex3]\underset{n}{\underbrace{111...11}}\underset{n}{\underbrace{222...22}}=\underset{n}{\underbrace{333...33}}\times \underset{n-1}{\underbrace{333...33}}4[/tex3]

Alguém poderia me ajudar? Por favor.

[jedi]

[tex3]F(n)=\underset{n}{\underbrace{111...11}}\underset{n}{\underbrace{222...22}}[/tex3]

[tex3]100.F(n)=\underset{n}{\underbrace{111...11}}\underset{n}{\underbrace{222...22}00}[/tex3]

[tex3]100.F(n)-10^{n+1}=\underset{n+1}{\underbrace{111...11}}\underset{n-1}{\underbrace{222...22}00}[/tex3]

[tex3]100.F(n)-10^{n+1}+22=\underset{n+1}{\underbrace{111...11}}\underset{n+1}{\underbrace{222...22}}[/tex3]

[tex3]100.F(n)-10^{n+1}+22=F(n+1)[/tex3]

[tex3]G(n)=\underset{n}{\underbrace{333...33}}\times \underset{n-1}{\underbrace{333...33}}4[/tex3]

[tex3]100.G(n)=\underset{n}{\underbrace{333...33}}\times \underset{n-1}{\underbrace{333...33}}4\times100[/tex3]

[tex3]100.G(n)=\underset{n}{\underbrace{333...33}0}\times \underset{n-1}{\underbrace{333...33}}40[/tex3]

[tex3]100.G(n)=(\underset{n+1}{\underbrace{333...33}}-3)\times (\underset{n}{\underbrace{333...33}}4+6)[/tex3]

[tex3]100.G(n)=\underset{n+1}{\underbrace{333...33}}\times (\underset{n}{\underbrace{333...33}}4)-3.(\underset{n}{\underbrace{333...33}}4)+6.(\underset{n+1}{\underbrace{333...33}})-18[/tex3]

[tex3]100.G(n)=\underset{n+1}{\underbrace{333...33}}\times (\underset{n}{\underbrace{333...33}}4)-3.(\underset{n+1}{\underbrace{333...33}}+1)+6.(\underset{n+1}{\underbrace{333...33}})-18[/tex3]

[tex3]100.G(n)=\underset{n+1}{\underbrace{333...33}}\times (\underset{n}{\underbrace{333...33}}4)-3.(\underset{n+1}{\underbrace{333...33}})+6.(\underset{n+1}{\underbrace{333...33}})-21[/tex3]

[tex3]100.G(n)=\underset{n+1}{\underbrace{333...33}}\times (\underset{n}{\underbrace{333...33}}4)-3.(\underset{n+1}{\underbrace{333...33}})-21[/tex3]

[tex3]100.G(n)=\underset{n+1}{\underbrace{333...33}}\times (\underset{n}{\underbrace{333...33}}4)-(\underset{n+1}{\underbrace{999...99}})-21[/tex3]

[tex3]100.G(n)=\underset{n+1}{\underbrace{333...33}}\times (\underset{n}{\underbrace{333...33}}4)+(10^{n+1}-1)-21[/tex3]

[tex3]100.G(n)=\underset{n+1}{\underbrace{333...33}}\times (\underset{n}{\underbrace{333...33}}4)+10^{n+1}-22[/tex3]

[tex3]100.G(n)-10^{n+1}+22=G(n+1)[/tex3]

portanto se [tex3]F(n)=G(n)[/tex3] então [tex3]F(n+1)=G(n+1)[/tex3]; [tex3]F(n+2)=G(n+2)[/tex3]....

como [tex3]F(1)=G(1)[/tex3]

então conluímos que a relação é validade para todo n.

[K1llua]

[tex3]F(n)=\underset{n}{\underbrace{111...11}}\underset{n}{\underbrace{222...22}}[/tex3]

[tex3]100.F(n)=\underset{n}{\underbrace{111...11}}\underset{n}{\underbrace{222...22}00}[/tex3]

[tex3]100.F(n)-10^{n+1}=\underset{n+1}{\underbrace{111...11}}\underset{n-1}{\underbrace{222...22}00}[/tex3]

[tex3]100.F(n)-10^{n+1}+22=\underset{n+1}{\underbrace{111...11}}\underset{n+1}{\underbrace{222...22}}[/tex3]

[tex3]100.F(n)-10^{n+1}+22=F(n+1)[/tex3]

[tex3]G(n)=\underset{n}{\underbrace{333...33}}\times \underset{n-1}{\underbrace{333...33}}4[/tex3]

[tex3]100.G(n)=\underset{n}{\underbrace{333...33}}\times \underset{n-1}{\underbrace{333...33}}4\times100[/tex3]

[tex3]100.G(n)=\underset{n}{\underbrace{333...33}0}\times \underset{n-1}{\underbrace{333...33}}40[/tex3]

[tex3]100.G(n)=(\underset{n+1}{\underbrace{333...33}}-3)\times (\underset{n}{\underbrace{333...33}}4+6)[/tex3]

[tex3]100.G(n)=\underset{n+1}{\underbrace{333...33}}\times (\underset{n}{\underbrace{333...33}}4)-3.(\underset{n}{\underbrace{333...33}}4)+6.(\underset{n+1}{\underbrace{333...33}})-18[/tex3]

[tex3]100.G(n)=\underset{n+1}{\underbrace{333...33}}\times (\underset{n}{\underbrace{333...33}}4)-3.(\underset{n+1}{\underbrace{333...33}}+1)+6.(\underset{n+1}{\underbrace{333...33}})-18[/tex3]

[tex3]100.G(n)=\underset{n+1}{\underbrace{333...33}}\times (\underset{n}{\underbrace{333...33}}4)-3.(\underset{n+1}{\underbrace{333...33}})+6.(\underset{n+1}{\underbrace{333...33}})-21[/tex3]

[tex3]100.G(n)=\underset{n+1}{\underbrace{333...33}}\times (\underset{n}{\underbrace{333...33}}4)-3.(\underset{n+1}{\underbrace{333...33}})-21[/tex3]

[tex3]100.G(n)=\underset{n+1}{\underbrace{333...33}}\times (\underset{n}{\underbrace{333...33}}4)-(\underset{n+1}{\underbrace{999...99}})-21[/tex3]

[tex3]100.G(n)=\underset{n+1}{\underbrace{333...33}}\times (\underset{n}{\underbrace{333...33}}4)+(10^{n+1}-1)-21[/tex3]

[tex3]100.G(n)=\underset{n+1}{\underbrace{333...33}}\times (\underset{n}{\underbrace{333...33}}4)+10^{n+1}-22[/tex3]

[tex3]100.G(n)-10^{n+1}+22=G(n+1)[/tex3]

portanto se [tex3]F(n)=G(n)[/tex3] então [tex3]F(n+1)=G(n+1)[/tex3]; [tex3]F(n+2)=G(n+2)[/tex3]....

como [tex3]F(1)=G(1)[/tex3]

então conluímos que a relação é validade para todo n.

Obrigada!