Sequences
1. Reminder about linear sequences
“the interval between values is constant” – this tells us it is a “linear” sequence (a bit like a straight
line on a graph). The nth term formula looks like an+b where a and b are numbers, eg 2n+5.
How to find the formula
Position n
Value t
Increase:


1
6
2
11
+5
3
16
+5
4
21
+5
5
26
n
?
+5
The “+5” increase tells us the formula for the nth term must be t = 5n + something
Now think how to get the first term t= 6 when n = 1.
, we need to have the “something” = 1.

Hence the sequence 6, 11, 16, 21, 26… has the formula 5n + 1
Examples
(a) Sequence 1, 4, 7, 10, 13, the increase is +3 each time so we need 3n + something.
To make the first term = 1, we think
, the formula is 3n – 2
(b) Sequence 5, 3, 1, -1, -3, the increase is -2 each time and the formula is -2n +7
2. What do “linear” and “quadratic” mean?
Look at these examples.

2n  1 is a linear expression since the highest power is just n . The values plot as a strqight
line.

n2  3n  4 is a quadratic expression since the highest power is n 2 . When plotted, all
quadratics follow a curve known as a parabola.
You won’t get questions on them, so just for reference:

n3  n2  3n  4 is a cubic expression (highest power n3 )

n4  7n3  5n2  3n  4 is a quartic expression (highest power n 4 )

n5  2n4  7n3  5n2  3n  4 is a quintic expression (highest power n5 )
These follow more interesting curves!
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3. Quadratic sequences - finding the values from the formula:
A quadratic sequence has n 2 in its formula, e.g. 3n2  5n  2 , in general like
a, b, c are numbers ("coefficients")
For instance, n2 makes a sequence 12 , 22 ,32 , 42 ,... so the values are 1, 4, 9, 16, 25,…
makes a sequence 101, 104, 109, 116, 125, …
makes a sequence 2, 6, 12, 20, 30, …
I know you can work out the values in your head or on a calculator but a good way of doing it (you’ll
see why later) is to make a table showing two separate sequences:


the actual quadratic part ( an 2 values)
a linear sequence (bn  c)
and then add ing them.
Example 1.
Find the first 6 terms in the sequence having n2  3n  1 as the nth term.
Position “n”
2
n
3n  1
1
1
4
2
4
7
3
9
10
4
16
13
5
25
16
6
36
19
1+4 = 5
4+7=11
19
29
41
55
Add to get
n2  3n  1
 The sequence is 5, 11, 19, 29, 41, 55,…
Example 2.
Find the first 6 terms in the sequence having 3n2  5n  2 as the nth term.
Position “n”
2
3n
5n  2
1
2
3 1  3
3  2  12
-3
-8
3
27
-13
12+(-8) = 4
14
2
2
4
48
-18
5
75
-23
6
108
-28
30
52
80
Add to get
3n2  5n  2 3+(-3) = 0
 The sequence is 0, 4, 14, 30, 52, 80,…
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4. How to find the nth term formula from the list of numbers.
Any quadratic sequence e.g. 2n2  3n  1 can be written as a pure n 2 part (  2n2 ) plus a linear
part ( 3n  1 ). We need to find each formula separately.
(i) Find how many n 2 you need
(ii) subtract the n 2 part from the sequence values. You are then left with a linear sequence that is
easy to identify.
(i) finding the n 2 coefficient.
Make a table showing first and second order differences, e.g. for the sequence 8, 13, 20, 29, 40
Position n
Value
1
8
2
13
First order difference: +5
3
20
+7
4
29
+9
5
40
n
?
+11
Not constant, hence not a linear
sequence.
Second order difference
+2
+2
+2
Increasing by a regular amount each
step, hence quadratic
 Halve the second difference (= 2) to find the multiplier for n 2
 Here,
2
 1 so the sequence must be 1n2  bn  c
2
(ii) finding the linear part
We make another table starting with sequence values 8, 13, 20, 29, 40 from above. We now know
the formula must be like 1n2  bn  c .
Position “n”
Sequence
1
8
2
13
3
20
4
29
5
40
6
53
n2  bn  c
n2
1
4
9
16
25
36
8-1=7
13-4=9
20-9=11
29-16 = 13
40-25=15
53-36=17
Subtract to find
what we “need”
for bn+c
The “linear sequence” part 7, 9, 11, 13, 15, 17 goes up in steps of 2 and follows the formula 2n  5 .
The original sequence adds pairs values from the n 2 sequence and the 2n  5 sequence.
2
 the complete formula is n  2n  5
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Another example: find the nth term formula for the sequence 14, 31, 58, 95, 142, 199
 First we find whether we need 1n 2 , 2n 2 , 3n 2 or whatever, by finding the second order
difference and halving it:
Position n
Value
1
14
2
31
First order difference: +17
Second order difference
3
58
4
95
+27
+10
+37
+10
5
142
6
199
n
?
+47
+10
10
 5 so the sequence is 5n 2 + something
2
 Now we take 5n 2 from each value in the sequence. This is just a way of findind what extra
we would add to the square terms to get the complete sequence.
Position “n”
Sequence
1
14
2
31
3
58
4
95
5
142
6
199
5n2  bn  c
5n 2
5×1=5
5×4=20
5×9=45
5×16=80
5×25=125
5×36=180
9
11
13
15
17
19
Subtract to
leave bn+c
The bn+c sequence goes in steps of +2, with first term 9, so must be 2n+7
The original sequence adds pairs values from the 5n 2 sequence and the 2n  7 sequence.
 the complete formula is 5n2  2n  7 .
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