To find the first five terms of the quadratic polynomial sequence given by the formula \( a_n = 3n^2 - 2 \), we will evaluate it for \( n = 1, 2, 3, 4, \) and \( 5 \).
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For \( n = 1 \): \[ a_1 = 3(1^2) - 2 = 3(1) - 2 = 3 - 2 = 1 \]
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For \( n = 2 \): \[ a_2 = 3(2^2) - 2 = 3(4) - 2 = 12 - 2 = 10 \]
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For \( n = 3 \): \[ a_3 = 3(3^2) - 2 = 3(9) - 2 = 27 - 2 = 25 \]
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For \( n = 4 \): \[ a_4 = 3(4^2) - 2 = 3(16) - 2 = 48 - 2 = 46 \]
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For \( n = 5 \): \[ a_5 = 3(5^2) - 2 = 3(25) - 2 = 75 - 2 = 73 \]
Now we have the first five terms of the sequence: \[ 1, 10, 25, 46, 73 \]
Next, we calculate the first differences. The first difference \( \Delta_1 \) is calculated as \( a_{n+1} - a_n \):
- \( \Delta_1(1) = a_2 - a_1 = 10 - 1 = 9 \)
- \( \Delta_1(2) = a_3 - a_2 = 25 - 10 = 15 \)
- \( \Delta_1(3) = a_4 - a_3 = 46 - 25 = 21 \)
- \( \Delta_1(4) = a_5 - a_4 = 73 - 46 = 27 \)
So, the first differences are: \[ 9, 15, 21, 27 \]
Next, we will calculate the second differences \( \Delta_2 \) which are the differences of the first differences:
- \( \Delta_2(1) = \Delta_1(2) - \Delta_1(1) = 15 - 9 = 6 \)
- \( \Delta_2(2) = \Delta_1(3) - \Delta_1(2) = 21 - 15 = 6 \)
- \( \Delta_2(3) = \Delta_1(4) - \Delta_1(3) = 27 - 21 = 6 \)
Thus, the second differences are: \[ 6, 6, 6 \]
To summarize:
- The first five terms of the sequence are \( 1, 10, 25, 46, 73 \).
- The first differences are \( 9, 15, 21, 27 \).
- The second differences are \( 6, 6, 6 \).