After you have learned to solve problems with arithmetic and quadratic sequences, you may be asked to solve problems with cubic sequences. As the name implies, cubic sequences rely on powers no higher than 3 to find the next term in the sequence. Depending on the complexity of the sequence, quadratic, linear and constant terms may also be included. The general form for finding the nth term in a cubic sequence is an^3 + bn^2 + cn + d.
Check that the sequence you have is a cubic sequence by taking the difference between each consecutive pair of numbers (called the "method of common differences"). Continue to take the differences of the differences three times total, at which point all the differences should be equal.
Sequence: 11, 27, 59, 113, 195, 311 Differences: 16 32 54 82 116 16 22 28 34 6 6 6
Set up a system of four equations with four variables to find the coefficients a, b, c and d. Use the values given in the sequence as if they were points on a graph in the form (n, nth term in sequence). It is easiest to start with the first 4 terms, as they are usually smaller or simpler numbers to work with.
Example: (1, 11), (2, 27), (3, 59), (4, 113) Plug in to: an^3 + bn^2 + cn + d = nth term in sequence a + b + c + d = 11 8a + 4b + 2c + d = 27 27a + 9b + 3c + d = 59 64a + 16b + 4c + d = 113
Solve the system of 4 equations using your favorite method.
In this example, the results are: a = 1, b = 2, c = 3, d = 5.
Write the equation for the nth term in a sequence using your newly found coefficients.
Example: nth term in the sequence = n^3 + 2n^2 + 3n + 5
Plug in your desired value of n into the equation and calculate the nth term in the sequence.
Example: n = 10 10^3 + 2_10^2 + 3_10 + 5 = 1235
About the Author
Based in Tucson, Ariz., Elio Lewis has served as a writing tutor since 2005. He received a B.A. in linguistics from the University of Arizona.