Quadratic Sequences (Cambridge (CIE) IGCSE International Maths)
Revision Note
Written by: Mark Curtis
Reviewed by: Dan Finlay
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 = × 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
Examiner Tips and Tricks
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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