Question

(50 PONTOS) Sobre uma sequência numérica $a_{n},$ o operador primeira diferença posterior de $a_{n}$ é definido como

$\Delta(a_{n})=a_{n+1}-a_{n},\;\;\;\;n \in \mathbb{N}$

Sabendo que

$\bullet\;\;\Delta(n)=1\;\;\;\text{ e }\;\;\;\displaystyle\sum\limits_{n=0}^{k}{\beta^{n}}=\dfrac{\beta^{k+1}-1}{\beta-1}$

com $\beta \not \in \{0,\,1\};$


e que pela propriedade telescópica, temos

$\displaystyle\sum\limits_{n=0}^{k}\Delta(a_{n}\cdot b_{n})=a_{k+1}\cdot b_{k+1}-a_{0}\cdot b_{0}$


use a propriedade da diferença do produto de duas sequências

$\bullet\;\;\Delta(a_{n}\cdot b_{n})=\Delta(a_{n})\cdot b_{n}+a_{n+1}\cdot \Delta(b_{n})$

e obtenha uma fórmula fechada para a soma

$\displaystyle\sum\limits_{n=0}^{k}{n\,\beta^{n}}$
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Resposta: $\beta^{k+1}\cdot \left[\dfrac{k+1}{\beta-1}-\dfrac{\beta}{(\beta-1)^{2}} \right ]+\dfrac{\beta}{(\beta-1)^{2}}$

Answer 1

Definindo as sequências

$a_{n}=n\\b_{n}=\beta^n$

Temos que

$\Delta(a_{`n}\cdot b_{n})=\Delta(a_{n})\cdot b_{n}+a_{n+1}\cdot\Delta(b_{n})\\\\\Delta(a_{`n}\cdot b_{n})=\Delta(n)\cdot\beta^{n}+(n+1)\cdot(\beta^{n+1}-\beta^{n})\\\\\Delta(a_{`n}\cdot b_{n})=1\cdot\beta^{n}+n\cdot\beta^{n+1}-n\cdot\beta^{n}+\beta^{n+1}-\beta^{n}\\\\\Delta(a_{`n}\cdot b_{n})=(\beta^{n}-\beta^{n})+n\cdot\beta^{n+1}-n\cdot\beta^{n}+\beta^{n+1}\\\\\Delta(a_{`n}\cdot b_{n})=n\cdot\beta^{n}\cdot(\beta-1)+\beta^{n+1}$

Logo:

$\displaystyle\sum\limits_{n=0}^{k}\Delta(a_{`n}\cdot b_{n})=\sum\limits_{n=0}^{k}n\cdot\beta^{n}\cdot(\beta-1)+\beta^{n+1}\\\\\\\sum\limits_{n=0}^{k}\Delta(a_{n}\cdot b_{n})=\sum\limits_{n=0}^{k}n\cdot\beta^{n}\cdot(\beta-1)+\sum\limits_{n=0}^{k}\beta^{n+1}\\\\\\\sum\limits_{n=0}^{k}\Delta(a_{n}\cdot b_{n})=(\beta-1)\sum\limits_{n=0}^{k}n\cdot\beta^{n}+\beta\sum\limits_{n=0}^{k}\beta^{n}$

Como

$\displaystyle\sum\limits_{n=0}^{k}\Delta(a_{n}\cdot b_{n})=a_{k+1}\cdot b_{k+1}-a_{0}\cdot b_{0}\\\\\\\sum\limits_{n=0}^{k}\Delta(a_{n}\cdot b_{n})=(k+1)\cdot(\beta^{k+1})-0\cdot\beta^{0}\\\\\\\boxed{\boxed{\sum\limits_{n=0}^{k}\Delta(a_{n}\cdot b_{n})=(k+1)\cdot\beta^{k+1}}}$

e

$\displaystyle\sum\limits_{n=0}^{k}\beta^{n}=\dfrac{\beta^{k+1}-1}{\beta-1}$

Ao substituir, temos:

$\displaystyle\sum\limits_{n=0}^{k}\Delta(a_{n}\cdot b_{n})=(\beta-1)\sum\limits_{n=0}^{k}n\cdot\beta^{n}+\beta\sum\limits_{n=0}^{k}\beta^{n}\\\\\\(k+1)\cdot\beta^{k+1}=(\beta-1)\sum\limits_{n=0}^{k}n\cdot\beta^{n}+\beta\dfrac{\beta^{k+1}-1}{\beta-1}\\\\\\(\beta-1)\sum\limits_{n=0}^{k}n\cdot\beta^{n}=(k+1)\cdot\beta^{k+1}-\beta\dfrac{\beta^{k+1}-1}{\beta-1}\\\\\\(\beta-1)\sum\limits_{n=0}^{k}n\cdot\beta^{n}=\dfrac{(\beta-1)(k+1)\beta^{k+1}-\beta^{k+1}\cdot\beta+\beta}{\beta-1}$

$\displaystyle(\beta-1)\cdot\sum\limits_{n=0}^{k}n\cdot\beta^{n}=\dfrac{(\beta-1)(k+1)\beta^{k+1}}{\beta-1}-\dfrac{\beta}{\beta-1}\beta^{k+1}+\dfrac{\beta}{\beta-1}\\\\\\(\beta-1)\sum\limits_{n=0}^{k}n\cdot\beta^{n}=(k+1)\beta^{k+1}-\dfrac{\beta}{\beta-1}\beta^{k+1}+\dfrac{\beta}{\beta-1}\\\\\\\sum\limits_{n=0}^{k}n\cdot\beta^{n}=\dfrac{k+1}{\beta-1}\beta^{k+1}-\dfrac{\beta}{(\beta-1)^{2}}\beta^{k+1}+\dfrac{\beta}{(\beta-1)^{2}}$

Colocando $\beta^{`k+1}$ em evidência:

$\boxed{\boxed{\sum\limits_{n=0}^{k}n\cdot\beta^{n}=\beta^{k+1}\cdot\left[\dfrac{k+1}{\beta-1}-\dfrac{\beta}{(\beta-1)^{2}}\right]+\dfrac{\beta}{(\beta-1)^{2}}}}$