To determine the recursive formula for the given geometric sequence [tex]\( \frac{1}{2}, \frac{3}{4}, \frac{9}{8}, \frac{27}{16}, \ldots \)[/tex], we need to identify the pattern and structure of the sequence.
Observing the sequence, we see the following terms:
- 1st term: [tex]\( a_1 = \frac{1}{2} \)[/tex]
- 2nd term: [tex]\( a_2 = \frac{3}{4} \)[/tex]
- 3rd term: [tex]\( a_3 = \frac{9}{8} \)[/tex]
- 4th term: [tex]\( a_4 = \frac{27}{16} \)[/tex]
The ratio [tex]\( \frac{a_{n+1}}{a_n} \)[/tex] in a geometric sequence is constant. Let's calculate this common ratio [tex]\( r \)[/tex].
[tex]\[
r = \frac{a_2}{a_1} = \frac{\frac{3}{4}}{\frac{1}{2}} = \frac{3}{4} \times \frac{2}{1} = \frac{3 \times 2}{4} = \frac{3}{2}
\][/tex]
[tex]\[
r = \frac{a_3}{a_2} = \frac{\frac{9}{8}}{\frac{3}{4}} = \frac{9}{8} \times \frac{4}{3} = \frac{9 \times 4}{8 \times 3} = \frac{36}{24} = \frac{3}{2}
\][/tex]
[tex]\[
r = \frac{a_4}{a_3} = \frac{\frac{27}{16}}{\frac{9}{8}} = \frac{27}{16} \times \frac{8}{9} = \frac{27 \times 8}{16 \times 9} = \frac{216}{144} = \frac{3}{2}
\][/tex]
Thus, the common ratio [tex]\( r \)[/tex] is [tex]\(\frac{3}{2}\)[/tex].
This means each term in the sequence is [tex]\(\frac{3}{2}\)[/tex] times the previous term.
To write the recursive formula:
- The first term is [tex]\( a_1 = \frac{1}{2} \)[/tex].
- Each subsequent term is produced by multiplying the previous term by [tex]\(\frac{3}{2}\)[/tex].
Thus, the recursive formula for the sequence is:
[tex]\[
a_1 = \frac{1}{2}
\][/tex]
[tex]\[
a_n = a_{n-1} \cdot \frac{3}{2} \quad \text{for} \quad n \geq 2
\][/tex]
Among the given options, option B describes this recursive formula correctly:
[tex]\[
\text{B) } a_1 = \frac{1}{2} \quad ; \quad a_n = a_{n-1} \cdot \frac{3}{2} \quad \text{for} \quad n \geq 2
\][/tex]
So, the correct answer is [tex]\( \boxed{2} \)[/tex].
