Quadratic Sequences and the Difference Method: IGCSE Maths
A quadratic sequence is a number pattern where the differences between consecutive terms are not constant, but the differences between THOSE differences are. The classic example is the square numbers: 1, 4, 9, 16, 25. This page covers how to recognise a quadratic sequence using the difference method, how to continue one by extending the differences, and how to substitute into a given quadratic formula to find specific terms. It is part of the Sequences chapter of the Edexcel IGCSE Maths course.
What is a quadratic sequence?
A linear sequence has a constant first difference. The sequence \(3, 7, 11, 15, \ldots\) goes up by 4 each time, so its first differences are \(4, 4, 4, \ldots\).
A quadratic sequence does not. The first differences keep changing, but if you take the differences of the differences, those become constant. The most familiar example is the square numbers:
The gaps between square numbers are \(3, 5, 7, 9, 11, \ldots\). These first differences keep growing, but the gaps between THOSE (the second differences) are all \(2\). That constant second difference is the fingerprint of a quadratic sequence.
Continuing a quadratic sequence
The difference method builds the next term by working from the bottom row up:
- Write the first differences underneath the sequence.
- Write the second differences underneath those.
- If the second differences are constant, the sequence is quadratic. Extend the second-difference row by adding one more value.
- Use the new second difference to find the next first difference.
- Add the new first difference to the last term of the sequence to get the next term.
Repeat steps 4 and 5 for each additional term you need. The method generalises: if the second differences are not constant either, you can take third differences, and so on.
Substituting into a quadratic formula
Sometimes a sequence is given to you directly as a formula, such as:
To find the value of any term, replace \(n\) with the position number and evaluate. For example:
- 1st term: \(1^2 + 3 = 4\)
- 2nd term: \(2^2 + 3 = 7\)
- 5th term: \(5^2 + 3 = 28\)
- 10th term: \(10^2 + 3 = 103\)
Substitution works for any quadratic formula, including ones with two terms like \(n^2 + n\) or three terms like \(2n^2 - n + 5\). Apply BIDMAS carefully: square first, then multiply by any coefficient, then add or subtract.
Worked examples
💡 Example 1: continue a quadratic sequence
Find the next two terms of the sequence \(2, 6, 12, 20, 30, \ldots\)
Sequence: \(2, \; 6, \; 12, \; 20, \; 30\)
1st differences: \(4, \; 6, \; 8, \; 10\)
2nd differences: \(2, \; 2, \; 2\) which are constant.
Extend 2nd differences: next is \(2\).
Next 1st difference: \(10 + 2 = 12\).
Next term: \(30 + 12 = 42\).
Then again: 2nd diff \(2\), 1st diff \(12 + 2 = 14\), term \(42 + 14 = 56\).
What's happening?
Each new term is built by adding the next first difference. The first difference itself grows by the constant second difference. Work upwards from the bottom row.
💡 Example 2: substitute into a formula
The nth term of a sequence is given by \(n^2 + 5\). Find the 1st, 4th, and 10th terms.
1st term: \(1^2 + 5 = 1 + 5 = 6\)
4th term: \(4^2 + 5 = 16 + 5 = 21\)
10th term: \(10^2 + 5 = 100 + 5 = 105\)
The first ten terms of the sequence are: \(6, 9, 14, 21, 30, 41, 54, 69, 86, 105\).
What's happening?
Replace \(n\) with the position number. Square it first, then add 5. The position number sits inside the square; do not square the whole expression.
💡 Example 3: is a value in the sequence?
Is \(101\) a term of the sequence with nth term \(n^2 + 5\)? If so, which term?
Set the formula equal to the target value:
\(n^2 + 5 = 101\)
\(n^2 = 96\)
\(n = \sqrt{96} \approx 9.798\)
\(n\) is not a positive integer, so \(101\) is NOT a term of this sequence.
What's happening?
Substitute and rearrange to solve for \(n\). For a number to be in the sequence, \(n\) must be a positive whole number. A non-integer value of \(n\) means the target is not in the sequence.
💡 Example 4: spotting a quadratic pattern
A pattern of dots is built up in stages. Stage 1 has 3 dots, stage 2 has 8 dots, stage 3 has 15 dots, stage 4 has 24 dots, stage 5 has 35 dots. Show that this is a quadratic sequence and find the number of dots at stage 6.
Sequence: \(3, \; 8, \; 15, \; 24, \; 35\)
1st differences: \(5, \; 7, \; 9, \; 11\)
2nd differences: \(2, \; 2, \; 2\)
Second differences are constant, so the sequence is quadratic.
Extend: next 2nd diff is \(2\), next 1st diff is \(11 + 2 = 13\).
Stage 6 dots: \(35 + 13 = 48\).
What's happening?
Real-world quadratic patterns crop up in geometry (areas, dot patterns, handshake problems). Use the difference method to confirm the sequence is quadratic and to find the next stage.
🔑 Key points
- A quadratic sequence has a constant SECOND difference, not a constant first difference.
- To continue a quadratic sequence, extend the constant second-difference row, then build upwards.
- To find a specific term from a given quadratic formula, substitute the position number for \(n\).
- To test whether a value is in a sequence, set the formula equal to the value and solve for \(n\). The value is in the sequence only if \(n\) is a positive whole number.
- Square numbers \(1, 4, 9, 16, 25, \ldots\) are the simplest quadratic sequence (nth term \(= n^2\)).
⚠️ Common pitfalls
- Calling a sequence quadratic just because the first differences are not constant. The second differences must also be constant.
- Forgetting BIDMAS when substituting. In \(n^2 + 5\) you square \(n\) first, then add 5, giving \(n^2 + 5\), NOT \((n+5)^2\).
- Squaring a coefficient by mistake. In \(2n^2\), only \(n\) is squared. So \(2 \times 4^2 = 2 \times 16 = 32\), not \(64\).
- Accepting a non-integer or negative \(n\) when checking if a value is in the sequence. \(n\) must be a positive whole number.
- Stopping after one row of differences. Always check the second differences before deciding the sequence type.