Quadratic Sequences (Cambridge (CIE) IGCSE Maths): Revision Note
Exam code: 0580 & 0980
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 nth term formula n2 are the square numbers
1, 4, 9, 16, 25, 36, 49, ...
From 12, 22, 32, 42, ...
Some sequences are just the square numbers but with a different starting number
16, 25, 36, 49, ... has the formula (n + 3)2
Substitute in n = 1, n = 2, n = 3 to see why
0, 1, 4, 9, ... has the formula (n - 1)2
Finding the nth 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 the above
For example, doubling then adding 1
3, 9, 19, 33, 51, 73, 99, ... has the formula 2n2 + 1
For example, subtracting 2 from square numbers starting with 4
2, 7, 14, 23, ... has the formula (n + 1)2 - 2
How do I find the nth term formula for any quadratic sequence?
STEP 1
Work out the sequences of first and second differencese.g. for the sequence 1, 10, 23, 40, 61
sequence | 1 | 10 | 23 | 40 | 61 |
first difference | +9 | +13 | +17 | +21 | |
second difference | +4 | +4 | +4 |
STEP 2
Divide the second difference by 2 to find the coefficient of n2e.g. a = 4 ÷ 2 = 2
STEP 3
Write out the first three or four terms of an2 and subtract the terms from the corresponding terms of the given sequencee.g. for the sequence 1, 10, 23, 40, 61
sequence | 1 | 10 | 23 | 40 |
2n2 | 2 | 8 | 18 | 32 |
difference | -1 | 2 | 5 | 8 |
STEP 4
Work out the nth term of these differences to find the bn + c part of the formulae.g. the nth term of -1, 2, 5, 8, ... is bn + c = 3n − 4
STEP 5
Find an2 + bn + c by adding together this linear nth term to an2 terme.g. an2 + bn + c = 2n2 + 3n − 4
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.
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
Sequence: 6, 9, 14, 21, 30, ...
1n2 : 1, 4, 9, 16, 25, ...
Differences: 5, 5, 5, 5, 5, ...
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
Worked Example
For the sequence 5, 7, 11, 17, 25, ....
(a) Find a formula for the nth term.
Start by finding the first and second differences
Sequence: 5, 7, 11, 17, 25
First differences: 2, 4, 6, 8, ...
Second difference: 2, 2, 2, ...
Hence
a = 2 ÷ 2 = 1
Now write down an2 (just n2 in this case as a = 1) and subtract the terms from the original sequence
sequence: 5, 7, 11, 17, ...
an2. : 1, 4, 9, 16, ...
difference: 4, 3, 2, 1, ...
Work out the nth term of these differences to give you bn + c
bn + c = −n + 5
Add an2 and bn + c together to give you the nth term of the sequence
nth term = n2 − n + 5
(b) Hence find the 20th term of the sequence.
Substitute n = 20 into n2 − n + 5
(20)2 − 20 + 5 = 400 − 15
20th term = 385
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