Explanation:
If we have a recursive expression with the form
[tex]a_n=a_{n-1}\cdot c[/tex]
Then, the explicit formula is
[tex]a_n=a_1\cdot c^{n-1}[/tex]
Therefore, for each option, we get:
[tex]\begin{gathered} a_n=a_{n-1}\cdot2\text{ with a}_1=1 \\ \text{ Then} \\ a_n=1\cdot2^{n-1}=2^{n-1} \end{gathered}[/tex][tex]\begin{gathered} a_n=a_{n-1}\cdot-2\text{ with a}_1=2 \\ \text{ Then} \\ a_n=2\cdot(-3)^{n-1} \end{gathered}[/tex][tex]\begin{gathered} a_n=a_{n-1}\cdot4\text{ with a}_1=-1 \\ \text{ Then} \\ a_n=-1\cdot4^{n-1}=-4^{n-1} \end{gathered}[/tex][tex]\begin{gathered} a_n=a_{n-1}\cdot2\text{ with a}_1=-3 \\ \text{ Then} \\ a_n=-3\cdot2^{n-1} \end{gathered}[/tex]
Answer:
Therefore, the answer is:
