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Quadratic Patterns: Detecting Constant Second Differences

Not every sequence grows by a constant amount. Some sequences accelerate. Their first differences change, but the second differences stay constant. That pattern is a strong signal of a quadratic relationship.

This guide develops second differences as a diagnostic tool, connects them to quadratic nth terms, shows how to build a rule from data, and explains why a constant second difference is the sequence version of constant curvature in a quadratic graph.

Start with the differences

Consider:

2, 6, 12, 20, 30, ...
First differences: 4, 6, 8, 10
Second differences: 2, 2, 2

The first differences are not constant, so the sequence is not arithmetic. The second differences are constant, suggesting a quadratic nth term.

Why quadratic sequences have constant second differences

Suppose Tn=an²+bn+c. The first difference is linear in n. Taking another difference removes that linear change and leaves a constant value of 2a.

Therefore:

constant second difference = 2a

This gives the coefficient of n² immediately.

Worked example 1: identify the quadratic coefficient

For the sequence 2, 6, 12, 20, 30, the constant second difference is 2.

2a=2
a=1

So begin with n². The sequence n² gives 1, 4, 9, 16, 25. Compare with the target sequence: the differences are 1, 2, 3, 4, 5, which is n.

Tn=n²+n

Check n=5: 25+5=30.

Worked example 2: a more general quadratic

Find the nth term of 5, 12, 23, 38, 57, …

First differences: 7, 11, 15, 19
Second differences: 4, 4, 4

Since 2a=4, a=2. Start with 2n²: 2, 8, 18, 32, 50. Subtract these from the sequence to get 3, 4, 5, 6, 7, which is n+2.

Tn=2n²+n+2

A systematic coefficient method

For Tn=an²+bn+c, use the constant second difference to find a. Then subtract an² from the original sequence. The remainder should be linear, so find its nth term as bn+c.

This decomposition turns a difficult-looking rule into one quadratic layer plus one linear layer.

Worked example 3: reverse use

For Tn=n²+3n, find the 12th term.

T12=12²+3(12)=144+36=180

Now test whether 70 belongs to the sequence:

n²+3n=70
n²+3n-70=0
(n+10)(n-7)=0

The positive integer solution is n=7, so 70 is the 7th term.

Visual quadratic growth

Quadratic sequences often appear when a growing figure gains an increasing number of tiles each stage. For example, square numbers 1, 4, 9, 16, … have first differences 3, 5, 7, … and constant second difference 2.

The increasing first difference is visible in the border added to each larger square.

Connection to graphs

If term number n is plotted against Tn, a quadratic sequence lies on a parabola rather than a straight line. Constant first difference corresponds to constant gradient; constant second difference corresponds to a gradient that changes at a constant rate.

Common errors

Stopping after seeing changing first differences. Compute second differences before deciding the structure.

Using second difference as a directly. The second difference is 2a, so halve it.

Assuming every non-linear-looking sequence is quadratic. Constant second differences are the key clue.

Forgetting to test the final rule. Verify several terms, not only one.

Diagnostic table

Observed mistakeLikely issueRepair
Calls sequence arithmetic because differences “look patterned”Difference order confusionCheck whether first differences are actually constant
Uses a=second differenceQuadratic coefficient ruleUse 2a=second difference
Finds n² part but cannot finishResidual linear structureSubtract an² and solve the remainder

Practice

  1. Find the nth term of 3, 8, 15, 24, 35, …
  2. Find the nth term of 6, 15, 28, 45, 66, …
  3. For Tn=2n²-n, find T10.
  4. Determine whether 90 is in the sequence Tn=n²+2n.

Answers

1. n²+2n. 2. 2n²+3n+1. 3. 190. 4. n²+2n=90 gives n=9, so yes.

Connected routes

Use From Pattern Spotting to a General Term and Arithmetic Sequences as prerequisites, then continue to Recursive Rules and Explicit Rules. Return to the Mathematics Learning Hub.