To determine which polynomial sequence shows that the terms of the 2nd differences are constant, we need to calculate the first differences and then the second differences for each sequence. The sequence that has constant second differences corresponds to a quadratic polynomial.
Let's calculate them for each choice:
A. {2, 6, 11, 14}
1st differences:
- \(6 - 2 = 4\)
- \(11 - 6 = 5\)
- \(14 - 11 = 3\)
1st differences: {4, 5, 3}
2nd differences:
- \(5 - 4 = 1\)
- \(3 - 5 = -2\)
2nd differences: {1, -2} (not constant)
B. {1, 7, 12, 19}
1st differences:
- \(7 - 1 = 6\)
- \(12 - 7 = 5\)
- \(19 - 12 = 7\)
1st differences: {6, 5, 7}
2nd differences:
- \(5 - 6 = -1\)
- \(7 - 5 = 2\)
2nd differences: {-1, 2} (not constant)
C. {0, 4, 16, 34}
1st differences:
- \(4 - 0 = 4\)
- \(16 - 4 = 12\)
- \(34 - 16 = 18\)
1st differences: {4, 12, 18}
2nd differences:
- \(12 - 4 = 8\)
- \(18 - 12 = 6\)
2nd differences: {8, 6} (not constant)
D. {-6, -1, 14, 39}
1st differences:
- \(-1 - (-6) = 5\)
- \(14 - (-1) = 15\)
- \(39 - 14 = 25\)
1st differences: {5, 15, 25}
2nd differences:
- \(15 - 5 = 10\)
- \(25 - 15 = 10\)
2nd differences: {10, 10} (constant)
Based on the calculations, the sequence that shows that the terms of the 2nd differences are constant is D. {-6, -1, 14, 39}.
1 answer