Question

Find a general function for the sequence: −3,6,−9,12,−15,18,…



an=(−1)n−1(3n)
an=(−1)n(3n)
an=−3n
an=(−1)n3(n−1)
an=(−3)n−1n

Answer 1

To find a general function for the given sequence \(-3, 6, -9, 12, -15, 18, \ldots\), let's examine the pattern.

1. **Identify the terms**:
- For \( n = 1\), \( a_1 = -3 \)
- For \( n = 2\), \( a_2 = 6 \)
- For \( n = 3\), \( a_3 = -9 \)
- For \( n = 4\), \( a_4 = 12 \)
- For \( n = 5\), \( a_5 = -15 \)
- For \( n = 6\), \( a_6 = 18 \)

2. **Noticing signs**:
The sequence alternates signs:
- The 1st, 3rd, and 5th terms are negative.
- The 2nd, 4th, and 6th terms are positive.

This pattern suggests a factor of \((-1)^n\) or \((-1)^{n-1}\) for the sign.

3. **Identify the multiples**:
The absolute values of the terms increase in increments of 3:
- The absolute values are \( 3, 6, 9, 12, 15, 18 \), which are \( 3 \times 1, 3 \times 2, 3 \times 3, 3 \times 4, 3 \times 5, 3 \times 6 \).

4. **General term construction**:
Therefore, the general term can be expressed as:
\[
a_n = (-1)^{n-1} \cdot 3n
\]
This accounts for both the alternating signs (by using \((-1)^{n-1}\)) and the sequence of multiples of 3 (by using \(3n\)).

5. **Verifying against options**:
Now let's compare this form with the provided options:
- \( a_n = (-1)^{n-1}(3n) \)
- \( a_n = (-1)^{n}(3n) \)
- \( a_n = -3n \)
- \( a_n = (-1)^{n}3(n-1) \)
- \( a_n = (-3)^{n-1}n \)

The correct answer matches the first option:
\[
a_n = (-1)^{n-1}(3n)
\]

Thus, the general function for the sequence is:

\[
\boxed{ a_n = (-1)^{n-1}(3n) }
\]