Question

Use the table to write an explicit rule and a recursive rule for the sequence.

Answer 1

Given the table:

n 0 1 2 3

f(n) 0.12 0.36 1.08 3.24

Let's write the explicit and recursive rule for the sequence.

Let's determine if the sequence is a geometric sequence.

$\begin{gathered} r=\frac{3.24}{1.08}=3 \\ \\ r=\frac{1.08}{0.36}=3 \\ \\ r=\frac{0.36}{0.12}=3 \end{gathered}$

The sequence has a common ratio of 3.

Therefore, it is a geometric sequence.

For the explicit formula of a geometric sequence, apply the formula:

$a_n=a_1r^{n-1}$

Where:

a1 is the first term = 0.36

r is the common ratio = 3

Hence, we have the explicit rule:

$f(n)=0.36(3)^{n-1}$

• Recursive rule:

Form the recursive rule of a sequence, we have:

$\begin{gathered} \begin{cases}{f(1)=0.36} \\ {} \\ {f(n)=3(f(n)-1);\text{ n>0}}\end{cases} \\ \\ \end{gathered}$

ANSWER:

• Explicit formula:

$f(n)=0.36(3)^{n-1}$

• Recursive formula:

$\begin{gathered} f(1)=0.36 \\ \\ f(n)=3(f(n)-1);\text{ n>0} \end{gathered}$