Question

Calcule o somatório

$\mathsf{\displaystyle\sum_{k=0}^n}$ 3^(2^k) · [3^(2^k) – 1]

e expresse a fórmula fechada em função de n.

Answer 1

Olá Lukyo.



Propriedade telescópica

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

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Organizando e desenvolvendo a expressão.

$\mathsf{\displaystyle\sum_{k=0}^n~3^{2^k}\cdot(3^{2^k}-1)=\displaystyle\sum_{k=0}^n~(3^{2^k})^2-3^{2^k}=\displaystyle\sum_{k=0}^n~3^{2^k\cdot2^1}-3^{2^k}}\\\\=\\\\\mathsf{\displaystyle\sum_{k=0}^n 3^{2^{k+1}}-3^{2^k}}$

Tomando $\mathsf{a_k=3^{2^k}}$ , temos

$\mathsf{\Delta a_k=a_{k+1}-a_k}\\\\\mathsf{\Delta a_k=3^{2^{k+1}}-3^{2^k}}$

$\mathsf{\displaystyle\sum_{k=0}^n 3^{2^{k+1}}-3^{2^k}=\displaystyle\sum_{k=0}^n \Delta a_k=a_{n+1}-a_0}\\\\=\\\\\mathsf{\displaystyle\sum_{k=0}^n~3^{2^{k+1}}-3^{2^k}=3^{2^{n+1}}-3^{2^0}}\\\\\\\mathsf{\displaystyle\sum_{k=0}^n~3^{2^{k+1}}-3^{2^k}=3^{2^{n+1}}-3^{1}}\\\\\\\boxed{\mathsf{\displaystyle\sum_{k=0}^n~3^{2^{k+1}}-3^{2^k}=3^{2^{n+1}}-3^{}}}$


Dúvidas? comente.