Which equation could be used to calculate the sum of the geometric series?

[tex]\[
\frac{1}{3}+\frac{2}{9}+\frac{4}{27}+\frac{8}{81}+\frac{16}{243}
\][/tex]

A. [tex]\( S_5 = \frac{\frac{1}{3}\left(1-\left(\frac{2}{3}\right)^5\right)}{\left(1-\frac{2}{3}\right)} \)[/tex]

B. [tex]\( S_5 = \frac{\frac{2}{3}\left(1-\left(\frac{1}{3}\right)^5\right)}{\left(1-\frac{1}{3}\right)} \)[/tex]



Answer :

To solve for the sum of the given geometric series:
[tex]\[ \frac{1}{3}+\frac{2}{9}+\frac{4}{27}+\frac{8}{81}+\frac{16}{243} \][/tex]
we need to identify the first term [tex]\(a\)[/tex], the common ratio [tex]\(r\)[/tex], and the number of terms [tex]\(n\)[/tex].

1. The first term [tex]\(a\)[/tex] is [tex]\(\frac{1}{3}\)[/tex].
2. The common ratio [tex]\(r\)[/tex] can be determined by dividing the second term by the first term: [tex]\( \frac{\frac{2}{9}}{\frac{1}{3}} = \frac{2}{3} \)[/tex].
3. The number of terms [tex]\(n\)[/tex] is [tex]\(5\)[/tex].

Next, we use the formula for the sum of the first [tex]\(n\)[/tex] terms of a geometric series:
[tex]\[ S_n = a \frac{1-r^n}{1-r} \][/tex]

Plugging in the values we identified:
- [tex]\(a = \frac{1}{3}\)[/tex]
- [tex]\(r = \frac{2}{3}\)[/tex]
- [tex]\(n = 5\)[/tex]

The formula becomes:
[tex]\[ S_5 = \frac{\frac{1}{3}\left(1-\left(\frac{2}{3}\right)^5\right)}{1-\frac{2}{3}} \][/tex]

Thus, the equation that could be used to calculate the sum of the geometric series is:
[tex]\[ S_5=\frac{\frac{1}{3}\left(1-\left(\frac{2}{3}\right)^5\right)}{\left(1-\\frac{2}{3}\right)} \][/tex]
Evaluating this expression will give the sum of the series.