Question

Propriedades dos somatórios:

Sabendo que

$\mathsf{\displaystyle\sum_{k=0}^n}$ k · 10^k = (1/81) · [(9n – 1) · 10^(n + 1) + 10],

obtenha uma fórmula fechada para

$\mathsf{\displaystyle\sum_{k=0}^n}$ k · 10^(– k).

Answer 1

Olá Lukyo


No final da resposta terá um anexo com todas as propriedades utilizadas para elucidar melhor a resposta

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Pela propriedade do somatório, através de uma troca de ordem de variável, podemos somar de frente para trás e de trás para frente, que o resultado se mantém, ou seja

$\mathsf{\displaystyle\sum_{k=0}^{n}a_n=\displaystyle\sum_{k=0}^n a_{n-k}}$

Portanto, temos a seguinte igualdade

$\mathsf{\displaystyle\sum_{k=0}^n k\cdot10^{k}=\displaystyle\sum_{k=0}^n (n-k)\cdot10^{n-k}}$

Vamos mexer nessa igualdade, de forma que apareça o somatório que queremos achar, ou seja

$\mathsf{\displaystyle\sum_{k=0}^n k\cdot10^{-k}}$

Organizando e desenvolvendo a igualdade

$\mathsf{\displaystyle\sum_{k=0}^n k\cdot10^{k}=\displaystyle\sum_{k=0}^n (n-k)\cdot10^{n}\cdot10^{-k}}\\\\\\\mathsf{\displaystyle\sum_{k=0}^n k\cdot10^{k}=10^n\cdot \displaystyle\sum_{k=0}^n (n-k)\cdot10^{-k}}\\\\\\\mathsf{\displaystyle\sum_{k=0}^n k\cdot10^{k}=10^n\cdot \displaystyle\sum_{k=0}^n (n\cdot 10^{-k}-k\cdot10^{-k})}$

$\mathsf{\displaystyle\sum_{k=0}^nk\cdot10^k=10^n\cdot\displaystyle\sum_{k=0}^n n\cdot10^{-k}-10^n\cdot \displaystyle\sum_{k=0}^nk\cdot10^{-k}}$


Isole $\mathsf{\displaystyle\sum_{k=0}^n k\cdot10^{-k}}$


$\mathsf{\displaystyle\sum_{k=0}^nk\cdot10^k=10^n\cdot\displaystyle\sum_{k=0}^n n\cdot10^{-k}-10^n\cdot \displaystyle\sum_{k=0}^nk\cdot10^{-k}}\\\\\\\\\mathsf{\displaystyle\sum_{k=0}^nk\cdot10^k+10^n\cdot \displaystyle\sum_{k=0}^nk\cdot10^{-k}=10^n\cdot \displaystyle\sum_{k=0}^nn\cdot10^{-k}}\\\\\\\\\mathsf{10^n\cdot\displaystyle\sum_{k=0}^nk\cdot10^{-k}=10^n\displaystyle\sum_{k=0}^nn\cdot10^{-k}-\displaystyle\sum_{k=0}^nk\cdot10^k}$

Multiplique ambos os lados por $\mathsf{\dfrac{1}{10^n}}$

$\mathsf{\Big(\dfrac{1}{\diagup\!\!\!\!\!\!10^n}\Big)\cdot\diagup\!\!\!\!\!\!10^n\cdot\displaystyle\sum_{k=0}^nk\cdot10^{-k}=\Big(\dfrac{1}{\diagup\!\!\!\!\!10^n}\Big)\cdot\diagup\!\!\!\!\!10^n\displaystyle\sum_{k=0}^nn\cdot10^{-k}-\Big(\dfrac{1}{10^n}\Big)\cdot\displaystyle\sum_{k=0}^nk\cdot10^k}$

$\mathsf{\displaystyle\sum_{k=0}^nk\cdot10^{-k}=\displaystyle\sum_{k=0}^nn\cdot10^{-k}-\Big(\dfrac{1}{10^n}\Big)\cdot\displaystyle\sum_{k=0}^nk\cdot10^k}$

Achando a fórmula fechada do somatório $\mathsf{\displaystyle\sum_{k=0}^nn\cdot10^{-k}}}$

Pela propriedade telescópica, temos que 

$\mathsf{\displaystyle\sum_{k=0}^n\Delta a_{k}=a_{n+1}-a_0}}}$

Tomando $\mathsf{a_k=10^{-k}}$

$\mathsf{\displaystyle\sum_{k=0}^n\Delta a_k=\displaystyle\sum_{k=0}^n 10^{-(k+1)}-10^{-k}=10^{-(n+1)}-10^{0}}\\\\\\\\\mathsf{\displaystyle\sum_{k=0}^n 10^{-k-1}-10^{-k}=10^{-n-1}-1}$

$\mathsf{\displaystyle\sum_{k=0}^n10^{-k-1}-10^{-k}=10^{-n-1}-1}\\\\\\\mathsf{\displaystyle\sum_{k=0}^n 10^{-k}\cdot(10^{-1}-1)=10^{-n-1}-1}$

$\mathsf{\displaystyle\sum_{k=0}^n 10^{-k}\cdot(10^{-1}-1)=10^{-n-1}-1}\\\\\\\\\mathsf{\displaystyle\sum_{k=0}^n \Big(\dfrac{n}{10^{-1}-1}\Big)\cdot 10^{-k}\cdot(10^{-1}-1)=\Big(\dfrac{n}{10^{-1}-1}\Big)\cdot10^{-n-1}-1}\\\\\\\\\mathsf{\displaystyle\sum_{k=0}^n n\cdot10^{-k}=n\cdot\big(\dfrac{10^{-n-1}-1}{10^{-1}-1}\Big)}$

$\mathsf{\displaystyle\sum_{k=0}^n n\cdot10^{-k}=n\cdot\Big(\dfrac{10^{-n-1}-1}{\dfrac{1}{10}-\dfrac{10}{10}}\Big)}\\\\\\\mathsf{\displaystyle\sum_{k=0}^n n\cdot10^{-k}=n\cdot\Big(\dfrac{10^{-n-1}-1}{\dfrac{-9}{10}}\Big)}\\\\\\\mathsf{\displaystyle\sum_{k=0}^n n\cdot10^{-k}=n\cdot\Big(10^{-n-1}-1\cdot\Big(\dfrac{10}{-9}\Big)}\\\\\\\mathsf{\displaystyle\sum_{k=0}^n n\cdot10^{-k}=n\cdot\Big(\dfrac{10^{-n-1+1}-10}{-9}\Big)}$

$\mathsf{\displaystyle\sum_{k=0}^n n\cdot10^{-k}=n\cdot\Big(\dfrac{10-10^{-n}}{9}\Big)}$

Como já temos a fórmula fechada dos dois somatórios do lado direito, basta substitui-los na equação

$\mathsf{\displaystyle\sum_{k=0}^nk\cdot10^{-k}=n\cdot\Big(\dfrac{10-10^{-n}}{9}\Big)-\Big(\dfrac{1}{10^n}\Big)\cdot\Big[\dfrac{(9n-1)\cdot10^{n+1}+10}{81}\Big]}\\\\\\\\\mathsf{\displaystyle\sum_{k=0}^nk\cdot10^{-k}=n\cdot\Big(\dfrac{10-10^{-n}}{9}\Big)-\Big(\dfrac{1}{10^n}\Big)\cdot\Big[\dfrac{(9n-1)\cdot10^{n+1}+10}{81}\Big]}$

$\mathsf{\displaystyle\sum_{k=0}^nk\cdot10^{-k}=n\cdot\Big(\dfrac{10-10^{-n}}{9}\Big)-\Big[\dfrac{9n\cdo10^{n+1}-10^{n+1}+10}{81\cdot10^n}\Big]}\\\\\\\\\mathsf{\displaystyle\sum_{k=0}^nk\cdot10^{-k}=\dfrac{9\cdot10^n}{9\cdot10^n}\cdot\Big(\dfrac{n\cdot10-n\cdot10^{-n}}{9}\Big)-\Big[\dfrac{9n\cdot10^{n+1}-10^{n+1}+10}{81\cdot10^n}\Big]}$

$\mathsf{\displaystyle\sum_{k=0}^nk\cdot10^{-k}=\Big(\dfrac{9n\cdot10^{n+1}-9n\cdot10^{-n+n}}{81\cdot10^n}\Big)-\Big[\dfrac{9n\cdot10^{n+1}-10^{n+1}+10}{81\cdot10^n}\Big]}\\\\\\\\\mathsf{\displaystyle\sum_{k=0}^nk\cdot10^{-k}=\dfrac{\diagup\!\!\!\!\!9n\cdot10^{n+1}-9n-\diagup\!\!\!\!\!\!9n\cdot10^{n+1}+10^{n+1}-10}{81\cdot10^n}}\\\\\\\\\boxed{\mathsf{\displaystyle\sum_{k=0}^nk\cdot10^{-k}=\dfrac{-9n+10^{n+1}-10}{81\cdot10^n}}}$

Encontrada a formula fechada que pede o enunciado


Dúvidas? comente.