To find the first five terms of the quadratic polynomial sequence given by the formula \( a_n = 3n^2 - 2 \), we will substitute the values of \( n \) from 1 to 5 into the formula.
Step 1: Calculate the First Five Terms
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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 \( n = 5 \): \[ a_5 = 3(5^2) - 2 = 3(25) - 2 = 75 - 2 = 73 \]
The first five terms of the sequence are:
\[ 1, 10, 25, 46, 73 \]
Step 2: Calculate the First Differences
The first differences are calculated by subtracting each term from the subsequent term:
- \( a_2 - a_1 = 10 - 1 = 9 \)
- \( a_3 - a_2 = 25 - 10 = 15 \)
- \( a_4 - a_3 = 46 - 25 = 21 \)
- \( a_5 - a_4 = 73 - 46 = 27 \)
The first differences are:
\[ 9, 15, 21, 27 \]
Step 3: Calculate the Second Differences
The second differences are calculated by subtracting each first difference from the subsequent first difference:
- \( 15 - 9 = 6 \)
- \( 21 - 15 = 6 \)
- \( 27 - 21 = 6 \)
The second differences are:
\[ 6, 6, 6 \]
Summary
- The first five terms of the quadratic polynomial sequence are: \( 1, 10, 25, 46, 73 \)
- The first differences are: \( 9, 15, 21, 27 \)
- The second differences are: \( 6, 6, 6 \)
Thus, we can conclude that in a quadratic sequence, the second differences are constant, confirming that our sequence is indeed quadratic.