Quadratic Sequences (Cambridge (CIE) IGCSE International Maths)

Revision Note

Quadratic Sequences

What is a quadratic sequence?

  • A quadratic sequence has an n th term formula that involves n2

  • The second differences are constant (the same)

    • These are the differences between the first differences

    • For example,   3, 9, 19, 33, 51, …
      1st Differences: 6, 10, 14, 18, ...

      2nd Differences:  4,   4,   4, ...

How do I find the nth term formula for a simple quadratic sequence?

  • The sequence with the n th term formula n2 are the square numbers 

    • 1, 4, 9, 16, 25, 36, 49, ...

      • From 12, 22, 32, 42, ...

  • Finding the n th term formula comes from comparing sequences to the square numbers

    • 2, 5, 10, 17, 26, 37, 50, ... has the formula n2 + 1

      • Each term is 1 more than the square numbers

    • 2, 8, 18, 32, 50, 72, 98, ... has the formula 2n2

      • Each term is double a square number

  • It may be a simple combination of both

    • For example, doubling then adding 1

      • 3, 9, 19, 33, 51, 73, 99, ... has the formula 2n2 + 1

  • Some sequences are just the square numbers but starting later

    • 16, 25, 36, 49, ... has the formula (n + 3)2

      • Substitute in n = 1, n = 2, n = 3 to see why

  • You can also use second differences to help find the n th term

    • For the simple quadratic sequence an2 + b

      • a  is half of the second difference

    • E.g. for the sequence 3, 9, 19, 33, 51, ...

      • The second difference is 4, so a = 1 half × 4, a = 2

      • Compare the sequence 2n2, (2, 8, 18, 32, 50, ...) to the original sequence

      • The original sequence has the n th term rule 2n2 + 1

Exam Tip

  • You must learn the square numbers from 12 to 152

Worked Example

For the sequence 6, 9, 14, 21, 30, ....

(a) Find a formula for the nth term.

Method 1

Find the first and second differences

Sequence:   6,   9,   14,   21,   30
     First differences:   3,    5,     7,     9, ...           
    Second differences:   2,    2,     2, ...               

The second differences are constant so this must be a quadratic sequence
Compare each term to terms in the sequence n2 (the square numbers)

           n2 :  1,   4,   9,   16,   25,  ... 
 Sequence:  6,   9,   14,   21,   30, ...      

Each term is 5 more than the terms in n2, so add 5 to n2

nth term = n2 + 5

Method 2

Find the first and second differences

Sequence:   6,   9,   14,   21,   30
     First differences:   3,    5,     7,     9, ...           
    Second differences:   2,    2,     2, ...               

Halve the second difference, this is the value of a
Write down the sequence an2
Compare this to the original sequence

           1n2 :  1,   4,   9,   16,   25,  ... 
 Sequence:  6,   9,   14,   21,   30, ...      

Each term is 5 more than the terms in n2, so add 5 to n2

nth term = n2 + 5

(b) Hence, find the 20th term of the sequence.

Substitute n = 20 into n2 + 5

(20)2 + 5 = 400 + 5

The 20th term is 405

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Mark Curtis

Author: Mark Curtis

Mark graduated twice from the University of Oxford: once in 2009 with a First in Mathematics, then again in 2013 with a PhD (DPhil) in Mathematics. He has had nine successful years as a secondary school teacher, specialising in A-Level Further Maths and running extension classes for Oxbridge Maths applicants. Alongside his teaching, he has written five internal textbooks, introduced new spiralling school curriculums and trained other Maths teachers through outreach programmes.