The simpliest way to answer the question is by iteration, substituting n with the numbers 1, 2, 3, 4 to find the first 4 numbers. So we have the formula;
[tex]A_n=3(2)^{n-1}[/tex]When n = 1,
[tex]\begin{gathered} A_n=3(2)^{n-1} \\ A_1=3(2)^{1-1} \\ A_1=3(2)^0 \\ A_1=3 \end{gathered}[/tex]When n = 2,
[tex]\begin{gathered} A_n=3(2)^{n-1} \\ A_2=3(2)^{2-1} \\ A_2=3(2)^1 \\ A_2=6^{} \end{gathered}[/tex]When n = 3,
[tex]\begin{gathered} A_n=3(2)^{n-1} \\ A_3=3(2)^{3-1} \\ A_3=3(2)^2 \\ A_3=12 \end{gathered}[/tex]When n = 4,
[tex]\begin{gathered} A_n=3(2)^{n-1} \\ A_4=3(2)^{4-1} \\ A_4=3(2)^3 \\ A_4=24 \end{gathered}[/tex]Therefore when n = {1, 2, 3, 4}, An = {3, 6, 12, 24}, making 3, 6, 12, 24 the first 4 numbers of our formula.
Therefore the answer is LETTER C.
