Answer:
[tex]a_n=7(2^{n-1})[/tex]Explanation:
Given the sequence with the recursive formula:
[tex]\begin{gathered} a_1=7 \\ a_n=2a_{n-1} \end{gathered}[/tex]First, we determine the first three terms in the sequence.
[tex]\begin{gathered} a_2=2a_{2-1}=2a_1=2\times7=14 \\ a_3=2a_{3-1}=2a_2=2\times14=28 \end{gathered}[/tex]Therefore, the first three terms of the sequence are: 7, 14 and 28.
This is a geometric sequence where:
• The first term, a=7
,• The common ratio, r =14/7 = 2
We use the formula for the nth term of a GP.
[tex]\begin{gathered} a_n=ar^{n-1} \\ a_n=7\times2^{n-1} \end{gathered}[/tex]The explicit formula for the sequence is:
[tex]a_n=7(2^{n-1})[/tex]