Answer:
D. [tex]f(n)=2\cdot 2^{n-1}[/tex]
Step-by-step explanation:
From the given table, you have that
[tex]a_1 =2\\ \\a_2=4\\ \\a_3=8\\ \\a_4=16\\ \\a_5=32\\ \\a_6=64[/tex]
If [tex]a_n[/tex] are terms of geometric series, then
[tex]a_n=a_{n-1}\cdot r[/tex]
Since
[tex]4=2\cdot 2\ \ \{a_2=2a_1\}\\ \\8=4\cdot 2\ \ \{a_3=2a_2\}\\ \\16=8\cdot 2\ \ \{a_4=2a_3\}\\ \\...,[/tex]
you can state that [tex]r=2[/tex] and [tex]a_1=2.[/tex]
The explicit formula for the geometric series is
[tex]f(n)=a_1\cdot r^{n-1},[/tex]
so
[tex]f(n)=2\cdot 2^{n-1}[/tex]
Hence, option D is correct
