Answer :

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]