Question

A soma dos seis primeiros termos da pg ( 1 \ 3,1/6,1/2,...) é igual a?

Answer 1

$P.G. \left(\dfrac{1}{3};\ \dfrac{1}{6};\ \dfrac{1}{12};\ \cdots\right)\\\\\\q=\dfrac{a_n}{a_{n-1}},\ \text{para }n\in\mathbb{N}\text{ e }n\ge2\\\\\\q=\dfrac{a_2}{a_{2-1}}\longrightarrow{q}=\dfrac{a_2}{a_1}\longrightarrow{q}=\dfrac{\dfrac{1}{6}}{\dfrac{1}{3}}\longrightarrow{q}=\dfrac{1}{6}\cdot\dfrac{3}{1}\longrightarrow{q}=\dfrac{1}{2}\\\\\\S_n=\dfrac{a_1(q^n-1)}{q-1},\text{ para }n\in\mathbb{N}^*\\\\\\S_6=\dfrac{\dfrac{1}{3}\left(\left(\dfrac{1}{2}\right)^2-1\right)}{\dfrac{1}{2}-1}\longrightarrow$

$S_6=\dfrac{\dfrac{1}{3}\left(\left(\dfrac{1}{64}\right)-1\right)}{-\dfrac{1}{2}}\longrightarrow{S}_6=\dfrac{\dfrac{1}{3}\left(-\dfrac{63}{64}\right)}{-\dfrac{1}{2}}\longrightarrow{S}_6=\dfrac{-\dfrac{21}{64}}{-\dfrac{1}{2}}\longrightarrow\\\\\\\boxed{S_6=-\dfrac{21}{32}}$