Question

Determine a soma dos termos da
P.G. ( √2 , 2 , √8 , 4 , √32 , 8 , √128 , 16 ).

Answer 1

$a_{2}=2\\a_{3}=\sqrt{8}=2\sqrt{2}\\\\q=\dfrac{a_{3}}{a_{2}}=\dfrac{2\sqrt{2}}{2}=\sqrt{2}$
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A soma dos 'n' primeiros termos de uma PG é dada por:

$S_{n}=\dfrac{a_{1}(q^{n}-1)}{q-1}$

Logo, a soma dos 8 primeiros termos será:

$S_{8}=\dfrac{a_{1}(q^{8}-1)}{q-1}\\\\\\S_{8}=\dfrac{\sqrt{2}([\sqrt{2}]^{8}-1)}{\sqrt{2}-1}\\\\\\S_{8}=\dfrac{\sqrt{2}(2^{\frac{8}{2}}-1)}{\sqrt{2}-1}\\\\\\S_{8}=\dfrac{\sqrt{2}(2^{4}-1)}{\sqrt{2}-1}\\\\\\S_{8}=\dfrac{\sqrt{2}(16-1)}{\sqrt{2}-1}\\\\\\S_{8}=\dfrac{15\sqrt{2}}{\sqrt{2}-1}$

Racionalizando:

$S_{8}=\dfrac{15\sqrt{2}(\sqrt{2}+1)}{(\sqrt{2}-1)(\sqrt{2}+1)}\\\\\\S_{8}=\dfrac{15\sqrt{2}\cdot\sqrt{2}+15\sqrt{2}}{(\sqrt{2})^{2}-1^{2}}\\\\\\S_{8}=\dfrac{15\cdot2+15\sqrt{2}}{2-1}\\\\\\\boxed{\boxed{S_{8}=15(2+\sqrt{2})}}$