Question

calcule a soma dos sete primeiros termos da P.G. (3,12,48...).

Answer 1

$(3,12,48,\dots)$

$a_1=3 \ \ \ \ \ q=\dfrac{12}{3}=4$

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

$S_7=\dfrac{3\cdot(4^{7}-1)}{4-1}=\dfrac{3\cdot(16.384-1)}{3}=\dfrac{3\cdot16.383}{3}=\boxed{16.383}$