Quadratic Sequences and Second Differences
Recognise a quadratic sequence from its constant second difference, and find the nth term rule in the form an² + bn + c.
What a learner can do afterwards
- Take first and second differences to decide a sequence is quadratic
- Use half the second difference as the coefficient of n²
- Find the rest of the rule by comparing the sequence with the n² terms
1 · Read
You have seen the nth term of linear sequences: a constant first difference, and a rule that is something times n plus something. And you can now read off the regions of a quadratic inequality, which is exactly the tool you will need when you ask when a sequence passes a given value. Today the sequence itself bends. A quadratic sequence follows a rule of the form an squared + bn + c, and its fingerprint is a constant second difference: the difference between the first differences. Read that one row and the whole rule is within reach.
In a linear sequence the first difference is constant. In a quadratic sequence the first difference is not constant, but it changes by the same amount each time, so the second difference is constant. There is a shortcut that ties the two together: the second difference is always twice the coefficient of n squared. A rule with n squared has a second difference of 2, with 2n squared it is 4, with 5n squared it is 10. So the moment you can read off a constant second difference, halve it, and you have a.
To find b and c, peel off the n squared part. Write the sequence of n squared terms, or 2n squared terms if a is 2, under the given terms and subtract row by row. What remains is a linear sequence, because the curved part is gone. Find its rule the way you do for any linear sequence: its first difference is b, and checking the first term gives c. If a is negative, the n squared part you peel off is negative, so the subtraction becomes an addition. Add the two rules together for the full nth term, and test it at n = 1.
Find the nth term of 4, 9, 16, 25, 36. The first differences are 5, 7, 9, 11, and the second difference is 2, so a is 1. Peel off the n squared terms 1, 4, 9, 16, 25 and subtract: 3, 5, 7, 9, 11. That remainder is linear with first difference 2, so its rule is 2n + 1. The full rule is n squared + 2n + 1. Check: at n = 1, 1 + 2 + 1 = 4, the first term. Bonus: this rule is (n + 1) squared, so the sequence is just the perfect squares shifted by one position.
The coefficient need not be 1. Find the nth term of 4, 9, 18, 31, 48. The first differences are 5, 9, 13, 17, and the second difference is 4, so a is 2. Peel off the 2n squared terms 2, 8, 18, 32, 50 and subtract: 2, 1, 0, -1, -2. The remainder decreases by 1 each time, so its rule is -n + 3. The full rule is 2n squared - n + 3. Check at n = 1: 2 - 1 + 3 = 4. The subtraction row is where signs go wrong most often, so keep the two sequences lined up term by term.
The second difference can be negative. Find the nth term of 12, 8, 2, -6, -16. The first differences are -4, -6, -8, -10, and the second difference is -2, so a is -1. Peeling off the negative n squared part means adding the n squared terms 1, 4, 9, 16, 25: 13, 12, 11, 10, 9. The remainder decreases by 1, so its rule is -n + 14. The full rule is -n squared - n + 14. Check at n = 1: -1 - 1 + 14 = 12. A negative a just means the sequence eventually turns and heads down.
Take the second difference to test whether a sequence is quadratic, halve it to get the n squared coefficient, peel that part off, and find the linear rule of what is left to complete the nth term.
2 · Watch
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Where it sits
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Where this leads
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24 questions wait behind this lesson, each with its answer explained. Every answer feeds the sky: stars light as they are learned, and dim when it is time to come back.