Respuesta :
[tex]\bf \qquad \qquad \textit{sum of a finite geometric sequence} \\\\ \displaystyle S_n=\sum\limits_{i=1}^{n}\ a_1\cdot r^{i-1}\implies S_n=a_1\left( \cfrac{1-r^n}{1-r} \right)\quad \begin{cases} n=\textit{last term's}\\ \qquad position\\ a_1=\textit{first term}\\ r=\textit{common ratio}\\ \cline{1-1} n=4\\ a_1=2\\ r=\frac{1}{3} \end{cases}[/tex]
[tex]\bf S_4=2\left( \cfrac{1-\left( \frac{1}{3} \right)^4}{1-\frac{1}{3}} \right)\implies S_4 = 2\left( \cfrac{1-\frac{1}{81}}{\frac{2}{3}} \right)\implies S_4 = 2\left( \cfrac{\frac{80}{81}}{~~\frac{2}{3}~~} \right) \\\\\\ S_4=2\left( \cfrac{40}{27} \right)\implies S_4=\cfrac{80}{27}\implies S_4=2\frac{26}{27}[/tex]
Answer:
80/27.
Step-by-step explanation:
Sum of n terms = a1 * (1 - r^n) / (1 - r)
Sum of 4 terms = 2 * (1 -(1/3)^4) / ( 1 - 1/3)
= 2 * 80/81 / 2/3
= 160 / 81 * 3/2
= 480/ 162
= 80/27 (answer
