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:
- 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.
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Quadratic Assumption: Let's assume the sequence can be modeled by a quadratic function of the form: \[ a_n = An^2 + Bn + C \]
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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 \( n = 3 \): \[ A(3)^2 + B(3) + C = 9 \quad \text{(3)} \]
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Simplifying the equations: From equation (1): \[ A + B + C = 1 \]
From equation (2): \[ 4A + 2B + C = -3 \]
From equation (3): \[ 9A + 3B + C = 9 \]
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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: \[ 4A + B = 4 \quad \text{(5)} \]
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Subtract (4) from (5): \[ (4A + B) - (3A + B) = 4 - (-4) \] Simplifies to: \[ A = 8 \]
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Substituting back to find \( B \): Substitute \( A = 8 \) in (4): \[ 3(8) + B = -4 \] Gives: \[ B = -4 - 24 = -28 \]
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Finding \( C \): Substitute \( A = 8 \) and \( B = -28 \) back in (1): \[ 8 - 28 + C = 1 \] Gives: \[ C = 1 + 28 - 8 = 21 \]
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Final equation: Thus, the formula for the sequence is: \[ a_n = 8n^2 - 28n + 21 \]
So, the equation that describes the sequence is: \[ a_n = 8n^2 - 28n + 21 \]