To find the sum of the series [tex]\( 1 + 3 + 9 + \ldots \)[/tex] up to the 6[tex]\(^\text{th}\)[/tex] term, we observe that this is a geometric series. A geometric series is characterized by each term being a constant multiple (the common ratio) of the term before it.
Given:
- The first term ([tex]\(a\)[/tex]) is 1.
- The common ratio ([tex]\(r\)[/tex]) is 3.
- The number of terms ([tex]\(n\)[/tex]) is 6.
The formula for the sum of the first [tex]\(n\)[/tex] terms of a geometric series is:
[tex]\[
S_n = a \left(\frac{r^n - 1}{r - 1}\right)
\][/tex]
Substituting the given values into the formula:
[tex]\[
S_6 = 1 \left(\frac{3^6 - 1}{3 - 1}\right)
\][/tex]
Calculate [tex]\(3^6\)[/tex]:
[tex]\[
3^6 = 3 \times 3 \times 3 \times 3 \times 3 \times 3 = 729
\][/tex]
Then,
[tex]\[
S_6 = \frac{729 - 1}{3 - 1} = \frac{728}{2} = 364
\][/tex]
So, the sum of the series up to the 6[tex]\(^\text{th}\)[/tex] term is:
[tex]\[
364
\][/tex]
Thus, the correct answer is:
b. [tex]\(364\)[/tex]
