Question

HELP!! Using the geometric sequence 2,2/3,2/9,. ,2/243, find the approximate sum of the geometric series. Round your answer to the nearest whole number.

Answer choices:

4

3

1

6

Answer 1

we can always get the common ratio of a geometric sequence by simply dividing the current term by the previous term, in this case for the sake of simplicity let's divide 2/3 by 2, so

`\cfrac{2}{3}/ 2\implies \cfrac{2}{3}/ \cfrac{2}{1}\implies \cfrac{2}{3}\cdot \cfrac{1}{2}\implies \cfrac{1}{3}\impliedby \textit{common ratio}`

so the sequence more or less looks like
`2~~,~~\cfrac{2}{3}~~,~~\cfrac{2}{9}~~,~~\cfrac{2}{81}~~,~~\cfrac{2}{243}`

so we'd want the sum of the first 5 terms

`\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} \end{cases} \\\\\\ S_5=2\left( \cfrac{1~~ - ~~((1)/(3))^5}{1~~ - ~~(1)/(3)} \right)\implies S_5=2\left( \cfrac{~~(242)/(243)~~}{(2)/(3)} \right)\implies S_5=\cfrac{242}{81}\implies S_5=2(80)/(81)`