Maths Gcse Edexcel Higher
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Quadratic-Sequences Edexcel Higher
Exam code:1MA1
Quadratic sequences
What is a quadratic sequence?
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A quadratic sequence has an n th term formula that involves n2
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The second differences are constant (the same)
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These are the differences between the first differences
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For example, 3, 9, 19, 33, 51, …
1st Differences: 6, 10, 14, 18, …2nd Differences: 4, 4, 4, …
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The sequence with the n th term formula n2 are the square numbers
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1, 4, 9, 16, 25, 36, 49, …
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From 12, 22, 32, 42, …
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How do I find the nth term formula for any quadratic sequence?
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STEP 1
Work out the sequences of first and second differences-
e.g. for the sequence 1, 10, 23, 40, 61
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|
sequence |
1 |
10 |
23 |
40 |
61 |
|
first difference |
+9 |
+13 |
+17 |
+21 |
|
|
second difference |
+4 |
+4 |
+4 |
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STEP 2
Divide the second difference by 2 to find the coefficient of n2-
e.g. a = 4 ÷ 2 = 2
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STEP 3
Write out the first three or four terms of an2 and subtract the terms from the corresponding terms of the given sequence-
e.g. for the sequence 1, 10, 23, 40, 61
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|
sequence |
1 |
10 |
23 |
40 |
|
2n2 |
2 |
8 |
18 |
32 |
|
difference |
-1 |
2 |
5 |
8 |
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STEP 4
Work out the nth term of these differences to find the bn + c part of the formula-
e.g. the nth term of -1, 2, 5, 8, … is bn + c = 3n − 4
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STEP 5
Find an2 + bn + c by adding together this linear nth term to an2 term-
e.g. an2 + bn + c = 2n2 + 3n − 4
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Examiner Tips and Tricks
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You must learn the square numbers from 12 to 152
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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