Question

Usando indução matemática, prove que 2/3+2/3^2 +⋯+2/3^n =1-1/3^n

Answer 1

Olá Cristiele.


- Usando indução, prove que 

$\mathsf{\dfrac{2}{3^1}+\dfrac{2}{3^2}+...+\dfrac{2}{3^n}=1-\dfrac{1}{3^n}}$

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Verificando se é válido para n = 1

$\mathsf{\dfrac{2}{3}=1-\dfrac{1}{3^1}}\\\\\\\mathsf{\dfrac{2}{3}=\dfrac{3}{3}-\dfrac{1}{3}}\\\\\\\mathsf{\dfrac{2}{3}=\dfrac{2}{3}~~\checkmark}$

Vamos assumir por H.I, que é valido para n = k

$\mathsf{\dfrac{2}{3^1}+\dfrac{2}{3^2}+...+\dfrac{2}{3^k}=1-\dfrac{1}{3^k}}$

Com isso, queremos verificar se será válido para n = k + 1, onde deverá resultar em


$\mathsf{\dfrac{2}{3^1}+\dfrac{2}{3^2}+...+\dfrac{2}{3^k}+\dfrac{2}{3^{k+1}}=1-\dfrac{1}{3^{k+1}}}$

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Aplicaremos agora de fato a indução.

Por H.I, temos que

$\mathsf{\dfrac{2}{3^1}+\dfrac{2}{3^2}+...+\dfrac{2}{3^k}=1-\dfrac{1}{3^{k}}}$

Some $\mathsf{\dfrac{2}{3^{k+1}}}$ em ambos os lados


$\mathsf{\dfrac{2}{3^1}+\dfrac{2}{3^2}+...+\dfrac{2}{3^k}+\dfrac{2}{3^{k+1}}=1-\dfrac{1}{3^{k}}+\dfrac{2}{3^{k+1}}}\\\\\\\mathsf{\dfrac{2}{3^1}+\dfrac{2}{3^2}+...+\dfrac{2}{3^k}+\dfrac{2}{3^{k+1}}=1-\dfrac{1}{3^{k}}\cdot\dfrac{3}{3}+\dfrac{2}{3^{k+1}}}\\\\\\\mathsf{\dfrac{2}{3^1}+\dfrac{2}{3^2}+...+\dfrac{2}{3^k}+\dfrac{2}{3^{k+1}}=1-\dfrac{3}{3^{k+1}}+\dfrac{2}{3^{k+1}}}$

$\mathsf{\dfrac{2}{3^1}+\dfrac{2}{3^2}+...+\dfrac{2}{3^k}+\dfrac{2}{3^{k+1}}=1-\dfrac{3}{3^{k+1}}+\dfrac{2}{3^{k+1}}}\\\\\\\mathsf{\dfrac{2}{3^1}+\dfrac{2}{3^2}+...+\dfrac{2}{3^k}+\dfrac{2}{3^{k+1}}=1+\dfrac{-3+2}{3^{k+1}}}}\\\\\\\mathsf{\dfrac{2}{3^1}+\dfrac{2}{3^2}+...+\dfrac{2}{3^k}+\dfrac{2}{3^{k+1}}=1-\dfrac{1}{3^{k+1}}~\checkmark}$


Concluímos o que queríamos demonstrar.


Dúvidas? comente.