Cubic Sequence

A number sequence where the n^th term follows a rule of the form an³ + bn² + cn + d, with a ≠ 0 and a, b, c, and d as constants.

It builds on quadratic sequences; just as quadratics have constant second differences, cubic sequences have constant third differences.

• Third differences are obtained by subtracting each second difference from the one that follows it.
• The defining property of a cubic sequence is that its third differences are constant and non-zero.

Example: Consider the sequence 2, 10, 30, 68, 130, …

Solution:

Terms: 2 10 30 68 130

First differences: 8 20 38 62

Second differences: 12 18 24

Third differences: 6 6

The third differences are all 6, confirming that this is a cubic sequence.

Some Real-Life examples are:

1. 3D Construction and Packaging

When objects grow in three dimensions, quantities follow a cubic pattern.
Example: 1,8,27,64...1, 8, 27, 64...1,8,27,64... (blocks in a cube)

2. Fluid Dynamics: Wind Power

Wind energy depends on the cube of wind speed (P∝v³)
Use: Engineers use this to choose efficient locations for wind turbines.

3. Cost Functions

Production costs can follow a cubic pattern.
Use: Helps find the most profitable production level before costs rise sharply.

General Formula

The n^th term of any cubic sequence is expressed as:

T(n) = an³ + bn² + cn + d

• a : rate of change of second differences; a ≠ 0.
• b : rate of change of first differences.
• c : rate of change of terms.
• d : constant term.
• n : position of the term (starting from n = 1).

Note : Third difference = 6a, so a = (third difference) / 6.

Common Applications

• Physics: Describing volume and displacement in three-dimensional motion.
• Engineering: Modeling deflection curves of beams under distributed loads.
• Computer Science: Analyzing algorithms with O(n³) time complexity.
• Economics: Total cost functions where marginal cost is quadratic take a cubic form.

Generating Terms of a Cubic Sequence

Terms are generated by substituting n = 1, 2, 3, … into T(n) = an³ + bn² + cn + d. Each substitution provides the value of the term at that position.

Example: Find the first 5 terms for T(n) = 2n³ − n² + 3n − 1.

Solution:

n	Calculation	T(n)
1	2(1) − 1(1) + 3(1) − 1	3
2	2(8) − 1(4) + 3(2) − 1	17
3	2(27) − 1(9) + 3(3) − 1	53
4	2(64) − 1(16) + 3(4) − 1	123
5	2(125) − 1(25) + 3(5) − 1	239

So, the terms are 3, 17, 53, 123, 239.

Finding the n^th Term of a Quadratic Sequence

Given only the terms of a cubic sequence, the goal is to reverse-engineer the formula T(n) = an³ + bn² + cn + d by determining a, b, c, and d.

• a = (third difference) / 6
• b, c, d are solved using simultaneous equations formed from T(1), T(2), and T(3)

Steps to determine the n^th term:

Step 1: Compute the first differences by subtracting consecutive terms.

Step 2: Compute the second differences by subtracting consecutive first differences.

Step 3: Compute the third differences by subtracting consecutive second differences.

Step 4: Determine a use:

a = (constant third difference) / 6

Step 5: Substitute a into T(1), T(2), and T(3) to get three equations in b, c, and d.

Step 6: Solve the simultaneous equations to find b, c, and d.

Step 7: Write the complete formula and verify it using a term not used in the working.

Example: Find the n^th term for the sequence: 3, 15, 43, 93, 171, …

Solution:

Step 1: First differences.

Terms: 3 15 43 93 171

First differences: 12 28 50 78

Step 2: Find the second differences.

First differences: 12 28 50 78

Second differences: 16 22 28

Step 2: Find the third differences.

Second differences: 16 22 28

Third differences: 6 6

Step 4: Find a.

a = 6 / 6 = 1

So, T(n) = n³ + bn² + cn + d.

Step 5: Set up equations using T(1), T(2), and T(3).

T(1) = 1 + b + c + d = 3 → b + c + d = 2 … (i)

T(2) = 8 + 4b + 2c + d = 15 → 4b + 2c + d = 7 … (ii)

T(3) = 27 + 9b + 3c + d = 43 → 9b + 3c + d = 16 … (iii)

Step 6: Solve simultaneously.

Subtracting (i) from (ii): 3b + c = 5 … (iv)

Subtracting (ii) from (iii): 5b + c = 9 … (v)

Subtracting (iv) from (v): 2b = 4 → b = 2

Substituting into (iv): 6 + c = 5 → c = −1

Substituting into (i): 2 − 1 + d = 2 → d = 1

So, T(n) = n³ + 2n² − n + 1.

Quadratic vs Cubic Sequences

Solved Examples

Example 1: In the cubic sequence 3, __, 33, 72, 135, …, find the missing second term and the n^th term formula.

Solution:

Let T(2) = k, So:
First Differences are :(k − 3), (33 − k), 39, 63.
Second differences: (33 − k) − (k − 3) = 24.
Third differences: (6 + k) − (36 − 2k) = 18 − k.

Since third differences must be constant:
3k − 30 = 18 − k
o, k = 12 => Third Difference = 6.

Now, a = 6 / 6 = 1.

T(n) = n³ + bn² + cn + d.

Setting up equations from the first three terms:

T(1): 1 + b + c + d = 3 → b + c + d = 2 … (i)
T(2): 8 + 4b + 2c + d = 12 → 4b + 2c + d = 4 … (ii)
T(3): 27 + 9b + 3c + d = 33 → 9b + 3c + d = 6 … (iii)

(ii) − (i): 3b + c = 2 … (iv)
(iii) − (ii): 5b + c = 2 … (v)
(v) − (iv): 2b = 0 → b = 0

Substituting into (iv): c = 2

Substituting into (i): d = 0

So, T(n) = n³ + 2n.

Example 2: The n^th term of a cubic sequence is T(n) = n³ + n².

• (a) Find the 6th term.
• (b) Which term of the sequence has the value 80?

Solution:

(a) Substituting n = 6 directly into the formula:

T(6) = 6³ + 6² = 216 + 36 = 252.

(b) Setting T(n) = 80:

n³ + n² = 80

n²(n + 1) = 80

Testing positive integers: n = 4 => 16 × 5 = 80

So, the 4th term equals 80.

Practice Problems

Problem 1: Determine whether the sequence 1, 9, 35, 91, 189, … is a cubic sequence. Justify your answer using differences.

Problem 2: The n^th term of a sequence is given by T(n) = 2n³ − n² + 4. Write down the first 5 terms of the sequence.

Problem 3: Find the n^th term formula for the cubic sequence: 5, 14, 35, 74, 137, …

Problem 4: The n^th term of a cubic sequence is T(n) = n³ − 2n. Find all positive integer values of n for which T(n) < 100.

Problem 5: A cubic sequence has its first four terms as 4, 10, 28, and 64. The n^th term is of the form an³ + bn² + cn + d. Find the complete formula and hence determine the smallest value of n for which T(n) exceeds 500.