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Original Question
Writeanequationtodescribethesequencebelow.Usentorepresentthepositionofaterminthesequence,wheren = 1forthefirstterm –3, 12,–48,....Asked by hello
Writeanequationtodescribethesequencebelow.Usentorepresentthepositionofaterminthesequence,wheren = 1forthefirstterm.
1 , –3 , 9 , ...
Write your answer using decimals and integers.
an =
(
)n - 1
1 , –3 , 9 , ...
Write your answer using decimals and integers.
an =
(
)n - 1
Answers
Answered by
GPT-4o mini
To find an equation that describes the given sequence \( 1, -3, 9, \ldots \), we first need to determine the pattern of the sequence.
Let's analyze the sequence:
- The first term (\( n = 1 \)) is \( 1 \).
- The second term (\( n = 2 \)) is \( -3 \).
- The third term (\( n = 3 \)) is \( 9 \).
We can look for a pattern by noting that:
1. **Differences between terms**:
- \( -3 - 1 = -4 \)
- \( 9 - (-3) = 12 \)
The first difference is \( -4 \) and the second difference is \( 12 \), suggesting that the sequence may not be linear and possibly quadratic.
2. **Quadratic Assumption**:
Let's assume the sequence can be modeled by a quadratic function of the form:
\[
a_n = An^2 + Bn + C
\]
3. **Using the terms**:
We can set up equations by plugging in the first three values of \( n \):
- For \( n = 1 \):
\[
A(1)^2 + B(1) + C = 1 \quad \text{(1)}
\]
- For \( n = 2 \):
\[
A(2)^2 + B(2) + C = -3 \quad \text{(2)}
\]
- For \( n = 3 \):
\[
A(3)^2 + B(3) + C = 9 \quad \text{(3)}
\]
4. **Simplifying the equations**:
From equation (1):
\[
A + B + C = 1
\]
From equation (2):
\[
4A + 2B + C = -3
\]
From equation (3):
\[
9A + 3B + C = 9
\]
5. **Solving the system of equations**:
We can eliminate \( C \) by subtracting the first equation from the others:
Subtract (1) from (2):
\[
(4A + 2B + C) - (A + B + C) = -3 - 1
\]
Gives:
\[
3A + B = -4 \quad \text{(4)}
\]
Subtract (1) from (3):
\[
(9A + 3B + C) - (A + B + C) = 9 - 1
\]
Gives:
\[
8A + 2B = 8
\]
Dividing by 2:
\[
4A + B = 4 \quad \text{(5)}
\]
6. **Subtract (4) from (5)**:
\[
(4A + B) - (3A + B) = 4 - (-4)
\]
Simplifies to:
\[
A = 8
\]
7. **Substituting back to find \( B \)**:
Substitute \( A = 8 \) in (4):
\[
3(8) + B = -4
\]
Gives:
\[
B = -4 - 24 = -28
\]
8. **Finding \( C \)**:
Substitute \( A = 8 \) and \( B = -28 \) back in (1):
\[
8 - 28 + C = 1
\]
Gives:
\[
C = 1 + 28 - 8 = 21
\]
9. **Final equation**:
Thus, the formula for the sequence is:
\[
a_n = 8n^2 - 28n + 21
\]
So, the equation that describes the sequence is:
\[
a_n = 8n^2 - 28n + 21
\]
Let's analyze the sequence:
- The first term (\( n = 1 \)) is \( 1 \).
- The second term (\( n = 2 \)) is \( -3 \).
- The third term (\( n = 3 \)) is \( 9 \).
We can look for a pattern by noting that:
1. **Differences between terms**:
- \( -3 - 1 = -4 \)
- \( 9 - (-3) = 12 \)
The first difference is \( -4 \) and the second difference is \( 12 \), suggesting that the sequence may not be linear and possibly quadratic.
2. **Quadratic Assumption**:
Let's assume the sequence can be modeled by a quadratic function of the form:
\[
a_n = An^2 + Bn + C
\]
3. **Using the terms**:
We can set up equations by plugging in the first three values of \( n \):
- For \( n = 1 \):
\[
A(1)^2 + B(1) + C = 1 \quad \text{(1)}
\]
- For \( n = 2 \):
\[
A(2)^2 + B(2) + C = -3 \quad \text{(2)}
\]
- For \( n = 3 \):
\[
A(3)^2 + B(3) + C = 9 \quad \text{(3)}
\]
4. **Simplifying the equations**:
From equation (1):
\[
A + B + C = 1
\]
From equation (2):
\[
4A + 2B + C = -3
\]
From equation (3):
\[
9A + 3B + C = 9
\]
5. **Solving the system of equations**:
We can eliminate \( C \) by subtracting the first equation from the others:
Subtract (1) from (2):
\[
(4A + 2B + C) - (A + B + C) = -3 - 1
\]
Gives:
\[
3A + B = -4 \quad \text{(4)}
\]
Subtract (1) from (3):
\[
(9A + 3B + C) - (A + B + C) = 9 - 1
\]
Gives:
\[
8A + 2B = 8
\]
Dividing by 2:
\[
4A + B = 4 \quad \text{(5)}
\]
6. **Subtract (4) from (5)**:
\[
(4A + B) - (3A + B) = 4 - (-4)
\]
Simplifies to:
\[
A = 8
\]
7. **Substituting back to find \( B \)**:
Substitute \( A = 8 \) in (4):
\[
3(8) + B = -4
\]
Gives:
\[
B = -4 - 24 = -28
\]
8. **Finding \( C \)**:
Substitute \( A = 8 \) and \( B = -28 \) back in (1):
\[
8 - 28 + C = 1
\]
Gives:
\[
C = 1 + 28 - 8 = 21
\]
9. **Final equation**:
Thus, the formula for the sequence is:
\[
a_n = 8n^2 - 28n + 21
\]
So, the equation that describes the sequence is:
\[
a_n = 8n^2 - 28n + 21
\]
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