Question

Calcule o somatório

$\mathsf{\displaystyle\sum_{k=0}^n \frac{2^k}{(2^k+1)\cdot (2^{k+1}+1)}}$

e expresse a fórmula fechada em função de n.

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Dica: Decomponha o somando em frações.

Answer 1

Olá Lukyo.


Calcule o somatório

$\mathsf{\displaystyle\sum_{k=0}^n \dfrac{2^k}{(2^k+1)\cdot(2^{k+1}+1)}}$

e expresse a fórmula fechada em função de n.

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Propriedades usadas

$\star~\boxed{\boxed{\mathsf{\Delta a_k = a_{k+1}-a_k}}}\\\\\\\\\star~\boxed{\boxed{\mathsf{\displaystyle\sum_{k=p}^n \Delta a_k=a_{n+1}-a_p}}}$

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Nessa expressão podemos notar a presença de termos consecutivos e semelhantes. Isso me induz a querer tentar desmembrar na diferença de termos consecutivos, para que eu possa aplicar a propriedade telescópica.


Organizando a expressão

$\mathsf{\dfrac{2^k}{(2^k+1)\cdot(2^{k+1}+1)}}$

Some e subtraia $\mathsf{2^k}$

$\mathsf{\dfrac{2^k+2^k-2^k}{(2^k+1)\cdot(2^{k+1}+1)}}\\\\\\\mathsf{\dfrac{2\cdot2^k-2^k}{(2^k+1)\cdot(2^{k+1}+1)}}\\\\\\\mathsf{\dfrac{2^{k+1}-2^k}{(2^k+1)\cdot(2^{k+1}+1)}}$

Some e subtraia $\mathsf{2^k\cdot2^{k+1}}$

$\mathsf{\dfrac{2^{k+1}-2^k+2^k\cdot2^{k+1}-2^k\cdot2^{k+1}}{(2^k+1)\cdot(2^{k+1}+1)}}\\\\\\\mathsf{\dfrac{2^k\cdot2^{k+1}+2^{k+1}-2^k\cdot2^{k+1}-2^k}{(2^k+1)\cdot(2^{k+1}+1)}}$

Coloque $\mathsf{2^{k+1}}$ e $\mathsf{2^k}$ em evidência

$\mathsf{\dfrac{2^{k+1}\cdot(2^k+1)-2^k\cdot(2^{k+1}+1)}{(2^k+1)\cdot(2^{k+1}+1)}}\\\\\\\mathsf{\dfrac{2^{k+1}\cdot(2^k+1)}{(2^k+1)\cdot(2^{k+1}+1)}-\dfrac{2^k\cdot(2^{k+1}+1)}{(2^k+1)\cdot(2^{k+1}+1)}}$

Simplifique os fatores comuns $\mathsf{2^{k}+1}$ e $\mathsf{2^{k+1}+1}$

$\mathsf{\dfrac{2^{k+1}\cdot(2^k+1)}{(2^k+1)\cdot(2^{k+1}+1)}-\dfrac{2^k\cdot(2^{k+1}+1)}{(2^k+1)\cdot(2^{k+1}+1)}}\\\\\\\mathsf{\dfrac{2^{k+1}}{(2^{k+1}+1)}-\dfrac{2^k}{(2^k+1)}}}}$

Obtemos a diferença de dois termos consecutivos.

$\mathsf{Tomando~a_k=\dfrac{2^k}{2^k+1}}$,

Pela propriedade telescópica, temos

$\mathsf{\displaystyle\sum_{k=0}^n a_{k+1}-a_k=\displaystyle\sum_{k=0}^n\dfrac{2^{k+1}}{2^{k+1}+1}-\dfrac{2^k}{2^k+1}=\dfrac{2^{n+1}}{2^{n+1}+1}-\dfrac{2^0}{2^0+1}}\\\\=\\\\\boxed{\mathsf{\displaystyle\sum_{k=0}^n\dfrac{2^{k+1}}{2^{k+1}+1}-\dfrac{2^k}{2^k+1}=\dfrac{2^{n+1}}{2^{n+1}+1}-\dfrac{1}{2}}}$



Dúvidas? comente.