Question

Encontre a fórmula fechada para a relação de recorrência a seguir.


$\mathsf{a_1=\dfrac{1}{3}}\\\\\\\\\mathsf{a_{n+1}=\dfrac{1}{3}\cdot(2-a_n)~~~~~~~~n\geq1}$


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Por favor responder de forma detalhada. Respostas com brincadeiras serão eliminadas.

Answer 1

Encontrar a lei de formação na forma fechada da sequência definida recursivamente. Reescrevendo de uma forma diferente, mas equivalente à que fora dada no enunciado:

        $\mathsf{a_n}=\left\{\!\begin{array}{ll} \mathsf{\dfrac{1}{3}\,,}&\quad\mathsf{se~~n=1}\\\\ \mathsf{\dfrac{1}{3}\cdot (2-a_{n-1})=\dfrac{2}{3}-\dfrac{1}{3}\,a_{n-1},}&\quad\mathsf{se~~n>1} \end{array}\right.\qquad\qquad\mathsf{onde~~n\in\mathbb{N}^*}$


Vamos tentar encontrar um padrão de formação para a sequência:

•  Para  n = 2:

      $\mathsf{a_2=\dfrac{2}{3}-\dfrac{1}{3}\,a_1}\\\\\\ \mathsf{a_2=\dfrac{2}{3}-\dfrac{1}{3}\cdot \dfrac{1}{3}}\\\\\\ \mathsf{a_2=\dfrac{2}{3}-\left(\dfrac{1}{3}\right)^{\!\!2}}$


•   Para  n = 3:

     $\mathsf{a_3=\dfrac{2}{3}-\dfrac{1}{3}\,a_2}\\\\\\ \mathsf{a_3=\dfrac{2}{3}-\dfrac{1}{3}\cdot \left[\dfrac{2}{3}-\left(\dfrac{1}{3}\right)^{\!\!2}\right]}\\\\\\ \mathsf{a_3=\dfrac{2}{3}-\dfrac{1}{3}\cdot \dfrac{2}{3}+\left(\dfrac{1}{3}\right)^{\!\!3}}\\\\\\ \mathsf{a_3=\dfrac{2}{3}\cdot \left(1-\dfrac{1}{3}\right)+\left(\dfrac{1}{3}\right)^{\!\!3}}$


•   Para  n = 4:

     $\mathsf{a_4=\dfrac{2}{3}-\dfrac{1}{3}\,a_3}\\\\\\ \mathsf{a_4=\dfrac{2}{3}-\dfrac{1}{3}\cdot \left[\dfrac{2}{3}\cdot \left(1-\dfrac{1}{3}\right)+\left(\dfrac{1}{3}\right)^{\!\!3}\right]}\\\\\\ \mathsf{a_4=\dfrac{2}{3}\cdot 1-\dfrac{1}{3}\cdot \dfrac{2}{3}\cdot \left(1-\dfrac{1}{3}\right)-\left(\dfrac{1}{3}\right)^{\!\!4}}\\\\\\ \mathsf{a_4=\dfrac{2}{3}\cdot \left[1-\dfrac{1}{3}\cdot \left(1-\dfrac{1}{3}\right)\right]-\left(\dfrac{1}{3}\right)^{\!\!4}}\\\\\\ \mathsf{a_4=\dfrac{2}{3}\cdot \left(1-\dfrac{1}{3}+\dfrac{1}{3^2}\right)-\left(\dfrac{1}{3}\right)^{\!\!4}}$


Ao que tudo indica, o n-ésimo termo seria

     (2/3) multiplicado por uma soma de uma P.G. alternante de razão  (– 1/3),  subtraído de uma potência de  (– 1/3).

A base negativa garante a alternância dos sinais dos termos do somatório e da exponencial.


Hipótese:

      $\mathsf{a_n=\displaystyle\dfrac{2}{3}\cdot \sum_{k=0}^{n-2}\left(\!-\,\dfrac{1}{3}\right)^{\!\!k}-\left(\!-\,\dfrac{1}{3}\right)^{\!\!n}}$


Usando a fórmula da soma de uma P.G., temos

     $\mathsf{a_n=\dfrac{2}{3}\cdot \dfrac{(-\frac{1}{3})^k}{(-\frac{1}{3})-1}\bigg|_{k=0}^{k=(n-2)+1}-\left(\!-\,\dfrac{1}{3}\right)^{\!\!n}}\\\\\\ \mathsf{a_n=\dfrac{2}{3}\cdot \dfrac{(-\frac{1}{3})^k}{-\frac{4}{3}}\bigg|_{k=0}^{k=n-1}-\left(\!-\,\dfrac{1}{3}\right)^{\!\!n}}\\\\\\ \mathsf{a_n=\dfrac{2}{3}\cdot \left(\!-\,\dfrac{3}{4} \right )\cdot \left(\!-\,\dfrac{1}{3}\right)^{\!\!k}\bigg|_{k=0}^{k=n-1}-\left(\!-\,\dfrac{1}{3}\right)^{\!\!n}}\\\\\\ \mathsf{a_n=-\,\dfrac{1}{2}\cdot \left[\left(\!-\,\dfrac{1}{3}\right)^{\!\!{n-1}}-\left(\!-\,\dfrac{1}{3}\right)^{\!\!0}\right]-\left(\!-\,\dfrac{1}{3}\right)^{\!\!n}}$

     $\mathsf{a_n=-\,\dfrac{1}{2}\cdot \left[\left(\!-\,\dfrac{1}{3}\right)^{\!\!{n}}\cdot \left(\!-\,\dfrac{1}{3}\right)^{\!\!{-1}}-1\right]-\left(\!-\,\dfrac{1}{3}\right)^{\!\!n}}\\\\\\ \mathsf{a_n=-\,\dfrac{1}{2}\cdot \left[\left(\!-\,\dfrac{1}{3}\right)^{\!\!{n}}\cdot (-3)-1\right]-\left(\!-\,\dfrac{1}{3}\right)^{\!\!n}}\\\\\\ \mathsf{a_n=\dfrac{3}{2}\cdot \left(\!-\,\dfrac{1}{3}\right)^{\!\!{n}}+\dfrac{1}{2}-\left(\!-\,\dfrac{1}{3}\right)^{\!\!n}}\\\\\\ \mathsf{a_n=\left(\dfrac{3}{2}-1\right )\cdot \left(\!-\,\dfrac{1}{3}\right)^{\!\!{n}}+\dfrac{1}{2}}\\\\\\ \mathsf{a_n=\dfrac{1}{2}\cdot \left(\!-\,\dfrac{1}{3}\right)^{\!\!{n}}+\dfrac{1}{2}}$

     $\mathsf{a_n=\dfrac{1}{2}\cdot \left[\left(\!-\,\dfrac{1}{3}\right)^{\!\!n}+1\right]\qquad\quad n=1,\,2,\,3,\,\ldots}$

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Verificando se a fórmula encontrada satisfaz a relação de recorrência dada:

•   Para  n = 1:

$\mathsf{a_1=\dfrac{1}{2}\cdot \left[\left(\!-\,\dfrac{1}{3}\right)^{\!\!1}+1\right]}\\\\\\ \mathsf{a_1=\dfrac{1}{2}\cdot \left[-\,\dfrac{1}{3}+1\right]}\\\\\\ \mathsf{a_1=\dfrac{1}{\diagup\!\!\!\! 2}\cdot \dfrac{\diagup\!\!\!\! 2}{3}}\\\\\\ \mathsf{a_1=\dfrac{1}{3}}\qquad\quad\checkmark$


•   Para  n > 1:

     $\mathsf{\dfrac{1}{3}\cdot (2-a_{n-1})}\\\\\\ =\mathsf{\dfrac{1}{3}\cdot \left(2-\dfrac{1}{2}\cdot \left[\left(\!-\,\dfrac{1}{3}\right)^{\!\!{n-1}}+1\right]\right)}\\\\\\ =\mathsf{\dfrac{2}{3}-\dfrac{1}{3}\cdot \dfrac{1}{2}\cdot \left[\left(\!-\,\dfrac{1}{3}\right)^{\!\!n-1}+1\right]}\\\\\\ =\mathsf{\dfrac{2}{3}+\dfrac{1}{2}\cdot \left(\!-\,\dfrac{1}{3}\right)\cdot \left(\!-\,\dfrac{1}{3}\right)^{\!\!n-1}-\dfrac{1}{6}}\\\\\\ =\mathsf{\dfrac{1}{2}\cdot \left(\!-\,\dfrac{1}{3}\right)^{\!\!{n}}+\dfrac{2}{3}-\dfrac{1}{6}}$

     $=\mathsf{\dfrac{1}{2}\cdot \left(\!-\,\dfrac{1}{3}\right)^{\!\!{n}}+\dfrac{1}{2}}\\\\\\ =\mathsf{\dfrac{1}{2}\cdot \left[\left(\!-\,\dfrac{1}{3}\right)^{\!\!{n}}+1\right]}$

     $=\mathsf{a_n\qquad\quad\checkmark}$

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Resposta:

     $\boxed{\begin{array}{c}\mathsf{a_n=\dfrac{1}{2}\cdot \left[\left(\!-\,\dfrac{1}{3}\right)^{\!\!{n}}+1\right]}\end{array}}\qquad\quad \mathsf{n=1,\,2,\,3,\,\ldots}$


Bons estudos! :-)