Question

1. write the explicit rule for each geometric sequence.

\

$$\begin{array}{|c|c|c|c|c|c|} \\hline n & 1 & 2 & 3 & 4 & 5 \\\\ \\hline f(n) & 7 & 14 & 28 & 56 & 112 \\\\ \\hline \\end{array}$$

first, identify the first term of the sequence. \\(f(1) = \square\\).

then, find the common ratio \\(r\\). \\(r = \frac{\square}{\square} = \square\\).

remember, to find the common ratio, divide any term by the term before it.

use the first term \\(f(1)\\) and the common ratio \\(r\\) to write the explicit rule.
\\(f(n) = f(1) \cdot (r)^{n-1}\\)

\\(f(n) = \underline{\quad\quad\quad\quad}\\)

the explicit rule for the geometric sequence is: \underline{\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad}

Explanation:

Identify the first term

From the table, when \(n = 1\), \(f(1) = 7\).

Find the common ratio

Divide the second term by the first term:

$$ r = \frac{f(2)}{f(1)} = \frac{14}{7} = 2 $$

Write the explicit rule

Substitute \(f(1) = 7\) and \(r = 2\) into the formula \(f(n) = f(1) \cdot (r)^{n-1}\):

$$ f(n) = 7 \cdot (2)^{n-1} $$

Answer:

1. Write the explicit rule for each geometric sequence.

First, identify the first term of the sequence.
\(f(1) =\) <blank>7</blank>.

Then, find the common ratio \(r\).
\(r =\) <blank>\(\frac{14}{7}\)</blank> \(=\) <blank>2</blank>.

Use the first term \(f(1)\) and the common ratio \(r\) to write the explicit rule.
\(f(n) = f(1) \cdot (r)^{n-1}\)

\(f(n) =\) <blank>\(7 \cdot (2)^{n-1}\)</blank>

The explicit rule for the geometric sequence is: <blank>\(f(n) = 7 \cdot (2)^{n-1}\)</blank>