Question

Sebastian writes the recursive formula f(x+1)=4f(x) to represent a geometric sequence whose second term is 12.

Answer 1

Answer:

B. f(x)=3(4)^x-1

Step-by-step explanation:

In order to know which explicit formula can be used to model the same sequence you have to calculate f(x) in:

f(x+1) and 4f(x)

and prove if they have the same results in both sides (f(x+1) = 4f(x))

Then, if you take $f(x)=3(4)^(x-1)$

1. f(x+1)

$f(x+1)=3(4)^((x+1)-1)$=$f(x+1)=3(4)^(x)$

2. 4f(x)

$4f(x)$=$4*3(4)^(x-1)$=$12(4)^(x-1)$

if you evaluate both sides (f(x+1) and 4f(x)) for x=0, x=1, x=2, you have:

x=0

$f(x+1)=3(4)^(x+1)-1$=$3(4)^0$=3

$4f(x)$=$4*3(4)^(x-1)$=$12(4)^-1$=3

x=1

$f(x+1)=3(4)^((x+1)-1)$=$3(4)^1$=12

$4f(x)$=$4*3(4)^(x-1)$=$12(4)^0$=12

x=2

$f(x+1)=3(4)^((x+1)-1)$=$3(4)^2$=48

$4f(x)$=$4*3(4)^(x-1)$=$12(4)^1$=48

Then you can say for: f(x)=3(4)^x-1 is fulfilled that f(x+1) = 4f(x)

Answer 2

B., each time you multiply by 4 so you are increasing by 4, exponentially, you can also literally eliminate all the other choices