Question

Writeanequationtodescribethesequencebelow.Usentorepresentthepositionofaterminthesequence,wheren = 1forthefirstterm.
229 , –229/3, 229/9, ...
Write your answer using proper fractions, improper fractions, and integers.

an = _(_)^ n^ -^ 1

Answer 1

To find the equation for the sequence given by the terms \(229, -\frac{229}{3}, \frac{229}{9}, \ldots\), let's examine how the terms progress.

1. The first term is \(229\), which can be expressed as \(229 = \frac{229}{1}\).
2. The second term is \(-\frac{229}{3}\).
3. The third term is \(\frac{229}{9}\).

### Identify the Pattern

We can rewrite the terms to see their fractions more clearly:

- \(a_1 = \frac{229}{1}\)
- \(a_2 = \frac{-229}{3} = \frac{229 \cdot (-1)}{3^1}\)
- \(a_3 = \frac{229}{9} = \frac{229}{3^2}\)

### Generalizing the Sequence

From the sequence:

- The numerators are consistently \(229\).
- The denominators follow a pattern of powers of \(3\) with signs alternating based on the position.

We note that:
- Position \(n=1\) has a sign of \((-)^0\) (positive)
- Position \(n=2\) has a sign of \((-)^1\) (negative)
- Position \(n=3\) has a sign of \((-)^0\) (positive), and so on.

The sign can be expressed as \((-1)^{n-1}\) which alternates based on the term's position.

### Putting it Together

Thus, we can express \(a_n\) as follows:

\[
a_n = 229 \cdot \frac{(-1)^{n-1}}{3^{n-1}}
\]

### Final Equation

So the equation for the nth term of the sequence is:

\[
a_n = 229 \cdot (-1)^{n-1} \cdot \frac{1}{3^{n-1}}
\]

This can be written as:

\[
a_n = \frac{229 \cdot (-1)^{n-1}}{3^{n-1}}
\]

Putting everything in one line according to your requested format:

\[
a_n = 229 \cdot \left(-\frac{1}{3}\right)^{n-1}
\]

This describes the \(n\)th term of the sequence correctly.