Question

Calcule a soma dos oito primeiros termos (27,9,3,...)

Answer 1

Olá

Nessa P.G., deveremos encontrar a razão

$r=\dfrac{a_2}{a_1}\\\\\\ r =\dfrac{9}{27}\\\\\\ r =\dfrac{1}{3}$

Descoberta a razão, usemos a fórmula da soma dos termos da P.G.

$S_n=\dfrac{a_1\cdot( q^{n}-1)}{q-1}$

Substituamos os valores

$S_8=\dfrac{27\cdot\left(\dfrac{1}{3}^{8}-1\right)}{\dfrac{1}{3}-1}$

Simplifiquemos a fração

$S_8=\dfrac{27\cdot\left(\dfrac{1}{6561}-1\right)}{\left(\dfrac{-2}{3}\right)}\\\\\\ S_8=\dfrac{27\cdot\left(\dfrac{-6560}{6561}\right)}{\left(\dfrac{-2}{3}\right)}\\\\\\ S_8=\dfrac{\left(\dfrac{-6560}{243}\right)}{\left(\dfrac{-2}{3}\right)}$

Use a fórmula
$\boxed{\dfrac{\left(\dfrac{a}{b}\right)}{\left(\dfrac{c}{d}\right)}=\dfrac{a\cdot d}{b\cdot c}}$ para simplificar os valores

$S_8=\dfrac{-6560\cdot3}{243\cdot(-2)}\\\\\\ S_8=\dfrac{-19680}{-486}$

Simplifique a fração por um fator (-6)

$\boxed{S_8=\dfrac{3280}{81}}~~\checkmark$