Question
What is the function that represents the sequence for the term numbers 1,2,3,4,5, and 6?
(1 point)
Responses
f(1)=6; f(n)=f(n−1)−3
f(1)=6; f(n)=f(n−1)−3
f(1)=0; f(n+1)=3+f(n)
f(1)=0; f(n+1)=3+f(n)
f(1)=6; f(n+1)=13f(n)
f(1)=6; f(n+1)=13f(n)
f(1)=−9; f(n)=3f(n−1
(1 point)
Responses
f(1)=6; f(n)=f(n−1)−3
f(1)=6; f(n)=f(n−1)−3
f(1)=0; f(n+1)=3+f(n)
f(1)=0; f(n+1)=3+f(n)
f(1)=6; f(n+1)=13f(n)
f(1)=6; f(n+1)=13f(n)
f(1)=−9; f(n)=3f(n−1
Answers
GPT-4o mini
To determine the correct function that represents the sequence for term numbers \(1, 2, 3, 4, 5, 6\), we need to analyze the options provided.
1. **Option 1**: \(f(1)=6; f(n)=f(n−1)−3\)
This suggests a recursive function where each term is \(3\) less than the previous term:
- \(f(1) = 6\)
- \(f(2) = f(1) - 3 = 6 - 3 = 3\)
- \(f(3) = f(2) - 3 = 3 - 3 = 0\)
- \(f(4) = f(3) - 3 = 0 - 3 = -3\)
- \(f(5) = f(4) - 3 = -3 - 3 = -6\)
- \(f(6) = f(5) - 3 = -6 - 3 = -9\)
The terms of this sequence are \(6, 3, 0, -3, -6, -9\).
2. **Option 2**: \(f(1)=0; f(n+1)=3+f(n)\)
This suggests a recursive function where each term is \(3\) more than the previous term:
- \(f(1) = 0\)
- \(f(2) = 3 + f(1) = 3 + 0 = 3\)
- \(f(3) = 3 + f(2) = 3 + 3 = 6\)
- \(f(4) = 3 + f(3) = 3 + 6 = 9\)
- \(f(5) = 3 + f(4) = 3 + 9 = 12\)
- \(f(6) = 3 + f(5) = 3 + 12 = 15\)
The terms of this sequence are \(0, 3, 6, 9, 12, 15\).
3. **Option 3**: \(f(1)=6; f(n+1)=13f(n)\)
This suggests a multiplicative function:
- \(f(1) = 6\)
- \(f(2) = 13f(1) = 13 \cdot 6 = 78\)
- \(f(3) = 13f(2) = 13 \cdot 78 = 1014\)
- The values grow very rapidly, so unlikely to correspond to a simple sequence up to 6.
4. **Option 4**: \(f(1)=−9; f(n)=3f(n−1)\)
Similar to the third option, this is also multiplicative:
- \(f(1) = -9\)
- \(f(2) = 3f(1) = 3 \cdot -9 = -27\)
- \(f(3) = 3f(2) = 3 \cdot -27 = -81\)
- Again, the values grow large negatively, unlikely fits.
Based on analyzing the options:
- The first option produces a consistent decreasing sequence.
- The second option produces a consistent increasing sequence.
If the sequence you are referring to is indeed \(6, 3, 0, -3, -6, -9\), then:
**The correct function is**: **\(f(1)=6; f(n)=f(n−1)−3\)**.
1. **Option 1**: \(f(1)=6; f(n)=f(n−1)−3\)
This suggests a recursive function where each term is \(3\) less than the previous term:
- \(f(1) = 6\)
- \(f(2) = f(1) - 3 = 6 - 3 = 3\)
- \(f(3) = f(2) - 3 = 3 - 3 = 0\)
- \(f(4) = f(3) - 3 = 0 - 3 = -3\)
- \(f(5) = f(4) - 3 = -3 - 3 = -6\)
- \(f(6) = f(5) - 3 = -6 - 3 = -9\)
The terms of this sequence are \(6, 3, 0, -3, -6, -9\).
2. **Option 2**: \(f(1)=0; f(n+1)=3+f(n)\)
This suggests a recursive function where each term is \(3\) more than the previous term:
- \(f(1) = 0\)
- \(f(2) = 3 + f(1) = 3 + 0 = 3\)
- \(f(3) = 3 + f(2) = 3 + 3 = 6\)
- \(f(4) = 3 + f(3) = 3 + 6 = 9\)
- \(f(5) = 3 + f(4) = 3 + 9 = 12\)
- \(f(6) = 3 + f(5) = 3 + 12 = 15\)
The terms of this sequence are \(0, 3, 6, 9, 12, 15\).
3. **Option 3**: \(f(1)=6; f(n+1)=13f(n)\)
This suggests a multiplicative function:
- \(f(1) = 6\)
- \(f(2) = 13f(1) = 13 \cdot 6 = 78\)
- \(f(3) = 13f(2) = 13 \cdot 78 = 1014\)
- The values grow very rapidly, so unlikely to correspond to a simple sequence up to 6.
4. **Option 4**: \(f(1)=−9; f(n)=3f(n−1)\)
Similar to the third option, this is also multiplicative:
- \(f(1) = -9\)
- \(f(2) = 3f(1) = 3 \cdot -9 = -27\)
- \(f(3) = 3f(2) = 3 \cdot -27 = -81\)
- Again, the values grow large negatively, unlikely fits.
Based on analyzing the options:
- The first option produces a consistent decreasing sequence.
- The second option produces a consistent increasing sequence.
If the sequence you are referring to is indeed \(6, 3, 0, -3, -6, -9\), then:
**The correct function is**: **\(f(1)=6; f(n)=f(n−1)−3\)**.