FS
O
O
PR
PA
G
E
5
O
R
R
EC
TE
D
Recurrence
relations
5.1 Kick off with CAS
U
N
C
5.2 Generating the terms of a first-order recurrence relation
c05RecurrenceRelations.indd 220
5.3 First-order linear recurrence relations
5.4 Graphs of first-order recurrence relations
5.5 Review
12/09/15 4:08 AM
5.1 Kick off with CAS
Generating terms of sequences with CAS
Value
1
7
O
Term
FS
First-order recurrence relations relate a term in a sequence to the previous term in
the same sequence. You can use recurrence relations to generate all of the terms in a
sequence, given a starting (or initial) value.
1 Use CAS to generate a table of the first 10 terms of a sequence if the starting
value is 7 and each term is formed by adding 3 to the previous term.
O
2
3
PR
4
5
E
6
G
7
PA
8
9
D
10
2 Use CAS to generate the first 10 terms of a sequence if the starting value is 14
Term
Value
14
EC
1
TE
and each term is formed by subtracting 5 from the previous term.
2
U
N
C
O
R
R
3
Please refer to
the Resources
tab in the Prelims
section of your
eBookPLUS for
a comprehensive
step-by-step
guide on how to
use your CAS
technology.
c05RecurrenceRelations.indd 221
4
5
6
7
8
9
10
3 Use CAS to generate the first 10 terms of a sequence where the starting value is
1.5 and each term is formed by multiplying the previous term by 2.
4 Use CAS to generate the first 10 terms of a sequence where the starting value is
4374 and each term is formed by dividing the previous term by 3.
12/09/15 4:08 AM
5.2
Generating the terms of a first-order
recurrence relation
A first-order recurrence relation relates a term in a sequence to the previous term
in the same sequence, which means that we only need an initial value to be able to
generate all remaining terms of a sequence.
In a recurrence relation the nth term is represented by un, with the next term after
un being represented by un + 1, and the term directly before un being represented
by un − 1.
The initial value of the sequence is represented by either u 0 or u1.
We can define the sequence 1, 5, 9, 13, 17, … with the following equation:
un + 1 = un + 4
u1 = 1.
This expression is read as ‘the next term is the previous term plus 4, starting at 1’.
Or, transposing the above equation, we get:
un + 1 − un = 4
u1 = 1.
Unit 3
AOS R&FM
Topic 1
Concept 1
O
O
FS
Generating a
sequence
Concept summary
Practice questions
PR
Interactivity
Initial values and
first-order recurrence
relations
int-6262
PA
G
E
A first-order recurrence relation defines a relationship between
two successive terms of a sequence, for example, between:
un + 1, the next term.
un, the previous term
Another notation that can be used is:
un, the next term.
un − 1, the previous term
The first term can be represented by either u0 or u1.
1
The following equations each define a sequence. Which of them are first-order
recurrence relations (defining a relationship between two consecutive terms)?
a un = un − 1 + 2
u1 = 3
2n
b un = 4 + 2
n = 1, 2, 3, …
3fn
c fn + 1 = 3
n = 1, 2, 3, …
C
O
R
R
WORKED
EXAMPLE
EC
TE
D
Throughout this topic we will use either notation format as short hand for next term,
previous term and first term.
term.
THINK
N
a The equation contains the consecutive
U
terms un and un − 1 to describe a pattern
with a known term.
WRITE
a This is a first-order recurrence relation. It has
the pattern
un = un − 1 + 2
and a starting or first term of u1 = 3.
b The equation contains only the u n term.
b This is not a first-order recurrence relation because
c The equation contains the consecutive
c This is an incomplete first-order recurrence relation.
There is no un + 1 or un − 1 term.
terms f n and f n + 1 to describe a pattern but
has no known first or starting term.
222
n = 1, 2, 3, …
it does not describe the relationship between two
consecutive terms.
It has no first or starting term, so a sequence cannot
be commenced.
MATHS QUEST 12 FURTHER MATHEMATICS VCE Units 3 and 4
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Given a fully defined first-order recurrence relation (pattern and a known term)
we can generate the other terms of the sequence.
Starting term
2
Write the first five terms of the sequence defined by the first-order
recurrence relation:
generate the next term, u1, using the pattern:
The next term is 3 × the previous term + 5.
2 Now we can continue generating the next term,
u0 = 2
u2 = 33uu1 + 5
= 3 × 11 + 5
= 38
u3 = 3u2 + 5
= 3 × 38 + 5
= 119
u4 = 3u3 + 5
= 3 × 119 + 5
= 362
The sequence is 2, 11, 38, 119, 362.
3
A sequence is defined by the first-order recurrence relation:
un +
1
= 2un − 3
n = 1, 2, 3, …
If the fourth term of the sequence is −29, that is, u4 = −29, then what is the
second term?
U
N
C
WORKED
EXAMPLE
O
R
R
3 Write your answer.
EC
TE
D
PA
u2, and so on.
un = 3un − 1 + 5
u1 = 3u0 + 5
=3×2+5
= 11
E
1 Since we know the u 0 or starting term, we can
PR
WRITE
G
THINK
u0 = 2.
O
un = 3un − 1 + 5
O
WORKED
EXAMPLE
FS
Earlier, it was stated that a starting term was required to fully define a sequence.
As can be seen below, the same pattern with a different starting point gives a different
set of numbers.
un + 1 = un + 2
u1 = 3
gives 3, 5, 7, 9, 11, …
un + 1 = un + 2
u1 = 2
gives 2, 4, 6, 8, 10, …
THINK
1 Transpose the equation to make the previous term,
un, the subject.
2 Use u 4 to find u3 by substituting into the
transposed equation.
WRITE
un =
un + 1 + 3
2
u4 + 3
2
−29 + 3
=
2
= −13
u3 =
Topic 5 RECURRENCE RELATIONS
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223
12/09/15 4:08 AM
u3 + 3
2
−13 + 3
=
2
= −5
3 Use u3 to find u2.
u2 =
4 Write your answer.
The second term, u2, is −5.
EXERCISE 5.2 Generating the terms of a first-order recurrence relation
The following equations each define a sequence. Which of them
are first-order recurrence relations (defining a relationship between two
consecutive terms)?
a un = un − 1 + 6 u1 = 7 n = 1, 2, 3, ...
b un = 5n + 1
n = 1, 2, 3, ...
2 The following equations each define a sequence. Which of them are first-order
recurrence relations?
n = 1, 2, 3, ...
a un = 4un − 1 − 3
b fn + 1 = 5fn − 8 f1 = 0 n = 1, 2, 3, ...
3 WE2 Write the first five terms of the sequence defined by the first-order
recurrence relation:
un = 4un − 1 + 3 u0 = 5.
WE1
FS
1
PA
G
E
PR
O
O
PRACTISE
4 Write the first five terms of the sequence defined by the first-order
D
recurrence relation:
A sequence is defined by the first-order recurrence relation:
un + 1 = 2un − 1 n = 1, 2, 3, ...
If the fourth term of the sequence is 5, that is, u 4 = 5, then what is the
second term?
WE3
EC
5
f0 = −2.
TE
fn + 1 = 5fn − 6
R
6 A sequence is defined by the first-order recurrence relation:
C
O
R
un + 1 = 3un + 7 n = 1, 2, 3, ...
If the seventh term of the sequence is 34, that is, u7 = 34, then what is the
fifth term?
U
N
CONSOLIDATE
NSOLIDATE
NSOLIDAT
E
224
7 Which of the following equations are complete first-order recurrence relation?
a
b
c
d
e
f
g
h
i
j
un = 2 + n
un = un − 1 − 1
un = 1 − 3un − 1
un − 4un − 1 = 5
un = −un − 1
un = n + 1
un = 1 − un − 1
un = an − 1
f n + 1 = 3f n − 1
pn = pn − 1 + 7
u0 = 2
u0 = 2
u1 = 2
u 0 = 21
u2 = 2
u0 = 7
MATHS QUEST 12 FURTHER MATHEMATICS VCE Units 3 and 4
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8 Write the first five terms of each of the following sequences.
a un = un − 1 + 2
u0 = 6
b un = un − 1 − 3
c un = 1 + un − 1
u 0 = 23
d u n + 1 = u n − 10
u0 = 5
u1 = 7
9 Write the first five terms of each of the following sequences.
a u n = 3u n − 1
u0 = 1
b u n = 5u n − 1
c u n = −4u n − 1
u0 = 1
d u n + 1 = 2u n
u 0 = −2
u1 = −1
10 Write the first five terms of each of the following sequences.
11
D
TE
14
PA
G
E
13
PR
O
O
12
u0 = 1
b u n = 3u n − 1 − 2
u1 = 5
Write the first seven terms of each of the following sequences.
a u n = −u n − 1 + 1
u0 = 6
b u n + 1 = 5u n − 1
u1 = 1
Which of the sequences is generated by the following first-order recurrence
relation?
un = 3un − 1 + 4
u0 = 2
A 2, 3, 4, 5, 6, …
B 2, 6, 10, 14, 18, …
C 2, 10, 34, 106, 322, …
D 2, 11, 47, 191, 767, …
E 6, 10, 14, 18, 22, …
Which of the sequences is generated by the following first-order recurrence
relation?
un + 1 = 22uun − 1
u1 = −3
A −3, 5, 9, 17, 33, …
B −3, −5, −9, −17, −33, …
C −3, 5, −3, 5, −3, …
D −3, −8, −14, −26, −54, …
E −3, −7, −15, −31, −63,
63, …
A sequence is defined by the first-order recurrence relation:
un + 1 = 3un + 1
n = 1, 2, 3, …
If the fourth term is 67 (that is, u 4 = 67), what is the second term?
FS
a u n = 2u n − 1 + 1
EC
15 A sequence is defined by the first-order recurrence relation:
un + 1 = 4un − 5
n = 1, 2, 3, …
R
If the third term is −41 (that is, u3 = −41), what is the first term?
16 For the sequence defined in question 15, if the seventh term is −27, what is the
O
R
fifth
fi
fth term?
U
N
C
MASTER
MASTE
MAST
ER
R
17 A sequence is defined by the first-order recurrence relation:
un + 1 = 5un − 10
n = 1, 2, 3, …
If the third term is −10, the first term is:
5
−14
A
B
C0
D2
E4
6
6
18 Write the first-order recurrence relations for the following descriptions of
a sequence and generate the first five terms of the sequence.
1
a The next term is 3 times the previous term, starting at .
4
b Next year’s attendance at a motor show is 2000 more than the previous year’s
attendance, with a first year attendance of 200 000.
c The next term is the previous term less 7, starting at 100.
d The next day’s total sum is double the previous day’s sum less 50, with a first
day sum of $200.
Topic 5 RECURRENCE RELATIONS
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225
12/09/15 4:08 AM
5.3
First-order linear recurrence relations
First-order linear recurrence relations with a common
difference
Consider the sequence 3, 7, 11, 15, 19, …
The common difference, d, is the value between consecutive terms in the
sequence:
d = u2 − u1 = u3 − u2 = u 4 − u3 = …
d = 7 − 3 = 11 − 7 = 15 − 11 = +4
The common difference is +4.
This sequence may be defined by the first-order linear recurrence relation:
un + 1 − un = 4
u1 = 3.
Rewriting this, we obtain:
un + 1 = un + 4
u1 = 3.
Unit 3
AOS R&FM
Topic 1
FS
Concept 2
PR
O
O
First-order linear
recurrence
relations
Concept summary
Practice questions
A sequence with a common difference of d
d may be defined by
a first-order linear recurrence relation of the form:
Interactivity
First-order
recurrence relations
with a common
difference
int-6264
(or un + 1 − un = d)
E
un + 1 = un + d
G
where d is the common difference and for
PA
d > 0 it is an increasing sequence
4
Express each of the following sequences as first-order recurrence relations.
TE
WORKED
EXAMPLE
D
d < 0 it is a decreasing sequence.
a 7, 12, 17, 22, 27, …
THINK
O
R
a 1 Write the sequence.
R
EC
9, −
−15,
15, …
b 9, 3, −3, −9,
C
2 Check for a common difference.
N
3 There is a common difference of 5
U
and the first term is 7.
b 1 Write the sequence.
a 7, 12, 17, 22, 27, …
d = u 4 − u3
= 22 − 17
=5
d = u 3 − u2
= 17 − 12
=5
d = u 2 − u1
= 12 − 7
=5
The first-order recurrence relation is given by:
un+1 = un + d
un+1 = un + 5
u1 = 7
b 9, 3, −3, −9, −15, …
d = u 3 − u2
= −3 − 3
= −6
2 Check for a common difference.
d = u 4 − u3
= −9 − −3
= −6
3 There is a common difference of −6
The first-order recurrence relation is given by:
un + 1 = un − 6
u1 = 9
and the first term is 9.
226
WRITE
d = u 2 − u1
=3−9
= −6
MATHS QUEST 12 FURTHER MATHEMATICS VCE Units 3 and 4
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12/09/15 4:08 AM
WORKED
EXAMPLE
5
Express the following sequence as a first-order recurrence relation.
un = −3n − 2
n = 1, 2, 3, 4, 5, …
THINK
WRITE
1 Generate the sequence using the given rule.
n = 1, 2, 3, 4, 5, …
n=1
2 There is a common difference of −3 and the
The first order difference equation is:
un + 1 = un − 3
u1 = −5
G
PA
first term is −5.
Write the first-order recurrence relation.
E
PR
O
O
FS
un = −3n − 2
u1 = −3 × 1 − 2
= −3 − 2
= −5
n=2
u2 = −3 × 2 − 2
= −66 − 2
= −8
−8
u3 = −3
−3 × 3 − 2
n=3
= −9
−9 − 2
=−
−11
n=4
u4 = −3 × 4 − 2
= −12 − 2
= −14
The sequence is −5,
5, −8, −11, −14, …
D
First-order linear recurrence relations with a common ratio
U
N
C
TE
EC
O
R
R
Interactivity
First-order
recurrence relations
with a common ratio
int-6263
Not all sequences have a common difference (i.e. a sequence increasing or decreasing
by adding or subtracting the same number to find your next term). You might also find
that a sequence increases or decreases by multiplying the terms by a common ratio.
Consider the geometric 1, 3, 9, 27, 81, … The common ratio is found by:
u2 u3 u4
=
=
=...
u1 u2 u3
9 27
3
=...
R= = =
1
3
9
R=
=3
The common ratio is 3.
This sequence may be defined by the first-order linear recurrence relation:
un + 1 = 3un
u1 = 1
A sequence with a common ratio of R may be defined by a first-order
linear recurrence relation of the form:
un + 1 = Run
where R is the common ratio
R > 1 is an increasing sequence
0 < R < 1 is a decreasing sequence
R < 0 is a sequence alternating between positive and
negative values.
Topic 5 RECURRENCE RELATIONS
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12/09/15 4:08 AM
WORKED
EXAMPLE
6
Express each of the following sequences as first-order recurrence relations.
b 3, −6, 12, −24, 48 …
a 1, 5, 25, 125, 625, …
THINK
WRITE
a R=
a 1 There is a common ratio of 5 and the
first term is 1.
5 25 125
=
=
=…
1
5
25
=5
u1 = 1
The first-order recurrence relation is given by:
un + 1 = 5un
u1 = 1
−6 12 −24
=
=
=…
b R =
3
−6
12
b 1 There is a common ratio of −2 and the
first term is 3.
O
FS
2 Write the first-order recurrence relation.
O
= −2
The first-order recurrence relation is given by:
un + 1 = −2un u1 = 3
7
Express each of the following sequences as first-order recurrence relations.
a un = 2(7)n − 1
n = 1, 2, 3, 4, …
b un = −3(2)n − 1
n = 1, 2, 3, 4, …
PA
WORKED
EXAMPLE
G
E
2 Write the first-order recurrence relation.
PR
u1 = 3
WRITE
D
THINK
a n = 1, 2, 3, 4, …
n=1
R
O
R
C
=2×1
n=2
=2
u1 = 2(7) 2 − 1
= 2 × 71
=2×7
n=3
= 14
u1 = 2(7) 3 − 1
= 2 × 72
U
N
un = 2(7)n − 1
u1 = 2(7) 1 − 1
= 2 × 70
EC
TE
a 1 Generate the sequence using the given rule.
= 2 × 49
n=4
= 98
u1 = 2(7) 4 − 1
= 2 × 73
= 2 × 343
2 There is a common ratio of 7 and the first
term is 2.
228
R = 7, u1 = 2
= 686
MATHS QUEST 12 FURTHER MATHEMATICS VCE Units 3 and 4
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12/09/15 4:08 AM
The first-order recurrence relation is given by:
un + 1 = 7un u1 = 2
3 Write the first-order recurrence relation.
b 1 Generate the sequence using the given rule.
b n = 1, 2, 3, 4, …
n=1
un = −3(2)n − 1
u1 = −3(2) 1 − 1
= −3 × 20
= −3 × 1
= −3
u1 = −3(2) 2 − 1
n=2
FS
= −3 × 21
= −33 × 2
O
= −6
−6
u1 = −3(2)
−3(2) 3 − 1
O
n=3
PA
G
E
n=4
PR
= −3
−3 × 22
2 There is a common ratio of 2 and the first
D
term is −3.
= −12
u1 = −3(2) 4 − 1
= −3 × 23
= −3 × 8
= −24
R = 2, u1 = −3
The first-order recurrence relation is given by:
un + 1 = 2un u1 = −3
EC
TE
3 Write the first-order recurrence relation.
= −3 × 4
EXERCISE 5.3 First-order linear recurrence relations
Express each of the following sequences as first-order recurrence relations.
a 1, 4, 7, 10, 13, …
b 12, 8, 4, 0, −4, …
2 Express each of the following sequences as first-order recurrence relations.
a −24, −19, −14, −9, −4, …
b 55, 66, 77, 88, 99, …
3 WE5 Express the following sequence as a first-order recurrence relation.
un = −5n − 3 n = 1, 2, 3, 4, 5, …
WE4
R
1
U
N
C
O
R
PRACTISE
4 Express the sequence as a first-order recurrence relation.
1
n + 4 n = 1, 2, 3, 4, 5, …
2
5 WE6 Express each of the following sequences as first-order recurrence relations.
a 2, 10, 20, 40, 80, …
b 4, −8, 16, −32, 64, …
6 Express each of the following sequences as first-order recurrence relations.
a 3, 15, 45, 225, 1125, …
b 200, −100, 50, −25, 12.5, …
7 WE7 Express each of the following sequences as first-order recurrence relations.
a un = 2(5) n − 1 n = 1, 2, 3, 4, …
b un = −3(4) n − 1
n = 1, 2, 3, 4, …
un =
Topic 5 RECURRENCE RELATIONS
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12/09/15 4:08 AM
8 Express each of the following sequences as first-order recurrence relations.
11
EC
O
R
R
15
TE
D
14
PA
13
G
E
12
O
10
O
9
PR
CONSOLIDATE
n = 1, 2, 3, 4, …
b un = −2(3) n − 1
n = 1, 2, 3, 4, …
Express each of the following sequences as first-order recurrence relations.
a 1, 3, 5, 7, 9, …
b 3, 10, 17, 24, 31, …
c 12, 5, −2, −9, −16, …
d 1, 0.5, 0, −0.5, −1, … e 2, 6, 10, 14, 18, …
f −2, 2, 6, 10, 14, …
g 6, 1, −4, −9, −14, …
h 4, 10.5, 17, 23.5, 30, …
The sequence −6, −3, 0, 3, 6, … can be defined by the first-order
recurrence relation:
A un + 1 = un − 3
u 0 = −6
B un + 1 = un + 3
u1 = −
−6
C u n + 1 = 3u n
u1 = −3
D u n + 1 = 3u n − 1
u0 = −
−33
E u n + 1 = 3u n
u1 = 3
Express each of the following sequences as first-order recurrence relations.
a u n = −n − 3
n = 1, 2, 3, …
b u n = 2n + 1
n = 1, 2, 3, …
c u n = 3n − 4
n = 1, 2, 3, …
d u n = −2n + 6
n = 1, 2, 3, …
The sequence defined by un = −2n + 3, n = 1, 2, 3, …, can be defined by the
first-order recurrence relation:
A un + 1 = un − 2
u1 = 1
B un + 1 = −
−2u
−2
2 n
u1 = 3
C u n + 1 = −2u n
un = 1
D un + 1 = un + 3
u1 = 1
E u n + 1 = −2u n + 3
u1 = 1
Express each of the following sequences as a first-order recurrence relation.
a 12, 10, 8, 6, 4, …
b 1, 2, 3, 4, 5, 6, …
Express each of the following sequences as first-order recurrence relations.
a 5, 10, 20, 40, 80, …
b 1, 6, 36, 216, 1296, …
c −3, 3, −3, 3, −3, …
d −3, −12, −48, −192, −768, …
e 2, 6, 18, 54, 162, …
f 5, −5, 5, −5, 5, …
g −2, −8, −32, −
−128,
128, −
−512,
512, …
h 5, −15, 45, −135, 405, …
The sequence −2,
−2, 6, −
−18,
18, 54, −162, … can be defined by the first-order
recurrence relation:
A un + 1 = −
−2u
−2
2u
un
u1 = −2
B u n + 1 = −2u n
u1 = 3
C un + 1 = −
−3u
−3
3u
un
u1 = 3
D u n + 1 = −3u n
u1 = −2
E u n + 1 = 3u
3 n
u1 = 2
Express each of the following sequences as first-order recurrence relations.
a u n = 2(3)n − 1
n = 1, 2, 3, …
b u n = −3(4)n − 1
n = 1, 2, 3, …
n
−
1
n
−
1
c u n = 0.5(−1)
n = 1, 2, 3, …
d u n = 3(5)
n = 1, 2, 3, …
n
−
1
n
−
1
e u n = −5(2)
n = 1, 2, 3, …
f u n = 0.1(−3)
n = 1, 2, 3, …
n
−
1
The sequence un = −4(1)
n = 1, 2, 3, … can be defined by the first-order
recurrence relation:
A un + 1 = un
u1 = −4
B un + 1 = un − 1
u1 = −4
C u n + 1 = 4u n
u1 = 1
D u n + 1 = −4u n + 3 u1 = −4
E u n + 1 = −4u n
u1 = −1
Express each of the following sequences as a first-order recurrence relation.
a −4, 4, −4, 4, −4, …
b 2, −14, 98, −686, 4802, …
c 2, 6, 18, 54, 162, …
d −1, 1, −1, 1, −1, …
e 5, 10, 20, 40, 80, …
FS
a un = 3(2) n − 1
U
N
C
16
17
18
230
MATHS QUEST 12 FURTHER MATHEMATICS VCE Units 3 and 4
c05RecurrenceRelations.indd 230
12/09/15 4:08 AM
MASTER
19 Express each of the following sequences as recurrence relations.
b 15, 10, 5, 0, −5, −10, …
a 4, 10.5, 17, 23.5, 30, …
c 1, 5, 9, 13, 17, 21, …
20 Express each of the geometric sequences as recurrence relations.
n = 1, 2, 3, …
n = 1, 2, 3, …
n = 1, 2, 3, …
b u n = −3(4)n – 1,
d u n = −0.1(8)n – 1,
n = 1, 2, 3, …
n = 1, 2, 3, …
Graphs of first-order recurrence relations
FS
Certain quantities in nature and business may change
in a uniform way (forming a pattern).
This change may be an increase, as in the case of:
un + 1 = un + 2
u1 = 3,
or it may be a decrease, as in the case of:
un + 1 = un − 2
u1 = 3.
These patterns can be modelled by graphs that, in turn,
can be used to recognise patterns in the real world.
A graph of the equation could be drawn to represent a situation, and by using the
graph the situation can be analysed to find, for example, the next term in the pattern.
G
E
PR
O
O
5.4
a u n = 4(2)n – 1,
c u n = 2(−6)n – 1,
e u n = 3.5(−10)n – 1,
PA
First-order recurrence relations: un + 1 = un + b
(arithmetic patterns)
R
0
1 2 3 4 5
Term number
n
O
R
C
N
U
un
5
4
3
2
1
0
An increasing pattern or a positive
common difference gives an upward
straight line.
WORKED
EXAMPLE
Value of term
D
TE
un
5
4
3
2
1
EC
Value of term
The sequences of a first-order recurrence relation un + 1 = un + b are distinguished by
a straight line or a constant increase or decrease.
1 2 3 4 5
Term number
n
A decreasing pattern or a negative
common difference gives a
downward straight line.
On a graph, show the first five terms of the sequence described by the
first-order recurrence relation:
un + 1 = un − 3
u1 = −5.
THINK
1 Generate the values of each of the five
terms of the sequence.
WRITE/DRAW
un + 1 = un − 3
u 2 = u1 − 3
= −5 − 3
= −8
u 4 = u3 − 3
= −11 − 3
= −14
u1 = −5
u 3 = u2 − 3
= −8 − 3
= −11
u 5 = u4 − 3
= −14 − 3
= −17
Topic 5 RECURRENCE RELATIONS
c05RecurrenceRelations.indd 231
231
12/09/15 4:08 AM
2 Graph these first five terms. The value of
Value of term
the term is plotted on the y-axis, and the
term number is plotted on the x-axis.
un
4
0
−4
−8
−12
−16
−20
2 3 4 5
n
Term number
FS
First-order recurrence relations: un + 1 = Run
n
C
N
U
WORKED
EXAMPLE
232
9
D
3 4 5 6
n
Term number
An increasing saw pattern occurs
when the common ratio is a negative
value less than −1 (R < −1).
Value of term
1
O
O
0
1 2 3 4 5 6
Term number
n
A decreasing pattern or a positive
fractional common ratio (0 < R < 1)
gives a downward curved line.
TE
EC
R
0
−2
−4
−6
−8
O
R
Value of term
An increasing pattern or a positive
common ratio greater than 1 (R > 1)
gives an upward curved line.
un
10
8
6
4
2
un
6
5
4
3
2
1
PR
E
1 2 3 4 5 6
Term number
PA
0
Value of term
un
6
5
4
3
2
1
G
Value of term
The sequences of a first-order recurrence relation un + 1 = Run are distinguished by
a curved line or a saw form.
un
10
8
6
4
2
0
−2
−4
−6
−8
1 2 3 4 5 6
n
Term number
A decreasing saw pattern occurs
when the common ratio is a negative
fraction (−1 < R < 0).
On a graph, show the first six terms of the sequence described by the
first-order recurrence relation:
un + 1 = 4un
u1 = 0.5.
MATHS QUEST 12 FURTHER MATHEMATICS VCE Units 3 and 4
c05RecurrenceRelations.indd 232
12/09/15 4:08 AM
THINK
WRITE/DRAW
O
O
Note: The sixth term is not included in
this graph to more clearly illustrate the
relationship between the terms.
un
160
120
80
40
PR
Value of term
2 Graph these terms.
FS
un + 1 = 4un
u1 = 0.5
u2 = 4u1
u3 = 4u2
= 4 × 0.5
=4×2
=2
=8
u4 = 4u3
u5 = 4u4
= 4 × 32
=4×8
= 128
= 32
u6 = 4u5
= 4 × 128
= 512
1 Generate the six terms of the sequence.
0
n
G
E
1 2 3 4 5
Term number
Straight or linear
EC
TE
D
A straight line or linear pattern is
given by first-order recurrence relations of the form
un + 1 = un + d
Value of term
PA
Interpretation of the graph of first-order recurrence relations
y
70
60
50
40
30
20
10
0
O
R
R
1 2 3 4 5 6 7 8 9 10 11
Term number
x
A non-linear pattern is generated
by first-order recurrence relations
of the form un + 1 = Run
Value of term
U
N
C
Non-linear
Non
N
on-linear
-linear (exponential)
y
140
120
100
80
60
40
20
0
1 2 3 4 5 6 7 8 9 10 11
Term number
Topic 5 RECURRENCE RELATIONS
c05RecurrenceRelations.indd 233
x
233
12/09/15 4:08 AM
Starting term
Earlier, the need for a starting term to be given to fully define a sequence was stated.
As can be seen below, the same pattern but a different starting point gives a different
set of numbers.
un + 1 = un + 2
u1 = 3
gives 3, 5, 7, 9, 11, …
un + 1 = un + 2
u1 = 2
gives 2, 4, 6, 8, 10, …
The first five terms of a sequence are plotted on the graph. Write the
first-order recurrence relation that defines this sequence.
0
1 2 3 4 5
Term number
WRITE
PA
THINK
n
E
PR
O
O
FS
un
18
16
14
12
10
8
6
4
2
G
10
Value of term
WORKED
EXAMPLE
1 Read from the graph the first five terms of
The sequence from the graph is:
18, 15, 12, 9, 6, …
un + 1 = un + d
Common difference, d = −3
un + 1 = un − 3
(or un + 1 − un = −3)
D
the sequence.
2 Notice that the graph is linear and there is a
TE
common difference of −33 between each term.
EC
3 Write your answer including the value
u1 = 18
C
The first four terms of a sequence are plotted on the graph. Write the firstorder recurrence relation that defines this sequence.
Value of term
U
N
WORKED
EXAMPLE
O
R
R
of one of the terms (usually the first),
as well as the rule defining the first order
difference equation.
un + 1 = un − 3
un
18
16
14
12
10
8
6
4
2
0
234
1 2 3 4 5
Term number
n
MATHS QUEST 12 FURTHER MATHEMATICS VCE Units 3 and 4
c05RecurrenceRelations.indd 234
12/09/15 4:08 AM
THINK
WRITE
1 Read the terms of the sequence from the graph.
The sequence is 2, 4, 8, 16, …
2 The graph is non-linear and there is a common
un + 1 = R × un
Common ratio, R = 2
un + 1 = 2un
ratio of 2, that is, for the next term, multiply
the previous term by 2.
u1 = 2
4 Write your answer.
un + 1 = 2un
with u1 = 1
C un + 1 = 2un
with u1 = 1
D un + 1 = un + 1
with u1 = 2
E un + 1 = un + 2
with u1 = 2
O
B un + 1 = un + 2
un
10
9
8
7
6
5
4
3
2
1
PR
with u1 = 1
0
1 2 3 4 5
Term number
n
D
PA
Value of term
A un + 1 = un + 1
O
The first five terms of a sequence are plotted on the graph shown. Which of
the following first-order recurrence relations could describe the sequence?
E
12
G
WORKED
EXAMPLE
u1 = 2
FS
3 Define the first term.
WRITE
TE
THINK
1 Eliminate the options systematically.
EC
Examine the first term given by
the graph to decide if it is u1 = 1
or u1 = 2.
R
2 Observe any pattern between each
There is a common difference of 2 or un + 1 = un + 2.
O
R
successive point on the graph.
The coordinates of the first point on the graph are (1, 1).
The first term is u1 = 1.
Eliminate options D and E.
3 Option B gives both the correct pattern The answer is B.
N
C
and first term.
U
EXERCISE 5.4 Graphs of first-order recurrence relations
PRACTISE
1
On a graph, show the first five terms of a sequence described by the
first-order recurrence relation:
un + 1 = un − 1 u1 = −2.
WE8
2 On a graph, show the first five terms of a sequence described by the first-order
recurrence relation:
3
u n + 1 = un + 3
u1 = 1.
On a graph, show the first six terms of a sequence described by the
first-order recurrence relation:
un + 1 = 3un
u1 = 2.
WE9
Topic 5 RECURRENCE RELATIONS
c05RecurrenceRelations.indd 235
235
12/09/15 4:08 AM
4 On a graph, show the first six terms of a sequence described by the first-order
un
16
14
12
10
8
6
4
2
FS
1
un+ 1 = un
u1 = 36.
2
5 WE10 The first five terms of a sequence are plotted on
the graph shown. Write the first-order recurrence relation
that defines this sequence.
Value of term
recurrence relation:
D
EC
O
R
Value of term
O
R
N
C
graph shown. Write the first-order recurrence relation
that defines this sequence.
U
236
The first five terms of a sequence are plotted
on the graph shown. Which of the following first-order
recurrence relations could describe the sequence?
A un + 1 = un + 1 u1 = 2
B un + 1 = un + 2 u1 = 2
C un + 1 = un + 3 u1 = 1
D un + 1 = un + 1 u1 = 1
E un + 1 = un + 2 u1 = 1
Value of term
WE12
1 2 3 4 5
Term number
n
1 2 3 4 5
Term number
n
un
36
30
24
18
12
6
0
9
n
un
12
10
8
6
4
2
0
8 The first four terms of a sequence are plotted on the
1 2 3 4 5
Term number
un
22
20
18
16
14
12
10
8
6
0
Value of term
The first four terms of a sequence are plotted
on the graph shown. Write the first-order recurrence
relation that defines this sequence.
WE11
TE
7
PA
G
E
PR
graph shown. Write the first-order recurrence relation
that defines this sequence.
Value of term
6 The first five terms of a sequence are plotted on the
O
0
1 2 3 4 5
Term number
n
un
12
10
8
6
4
2
0
1 2 3 4 5
Term number
n
MATHS QUEST 12 FURTHER MATHEMATICS VCE Units 3 and 4
c05RecurrenceRelations.indd 236
12/09/15 4:08 AM
10 The first four terms of a sequence are plotted on
Value of term
the graph shown. Which of the following first-order
recurrence relations could describe the sequence?
A un + 1 = 2un u1 = 2
B un + 1 = 2un u1 = 1
C un + 1 = 2un u1 = 0
D un + 1 = 3un u1 = 1
E un + 1 = 3un u1 = 2
0 1 2 3 4 5 n
For each of the following, plot the first five terms of the
Term number
sequence defined by the first-order recurrence relation.
a un + 1 = un + 3
u1 = 1
b un + 1 = un + 7
u1 = 5
c un + 1 = un − 3
u1 = 17
For each of the following, plot the first five terms of the sequence defined by the
first-order recurrence relation.
a u n + 1 = −2u n
u1 = −0.5
b u n = 0.5u n − 1
u1 = 16
c u n + 1 = 2.5u n
u1 = 2
For each of the following, plot the first four terms of the sequence defined by the
first-order recurrence relation.
a u n + 1 = 3u n − 4
u1 = −3
b un = 2
2un − 1 + 0.5 u1 = 2
c u n = 2 + 5u n − 1
u1 = 0.2
For each of the following, plot the first four terms of the sequence defined by the
first-order recurrence relation.
a u n + 1 = 100 − 3u n u1 = 20
b u n + 1 = u n − 50
u1 = 100
c u n = 0.1u n − 1
u1 = 10
On graphs, show the first 5 terms of the sequences defined by the following
recurrence relations.
a u n + 1 = u n + 3,
u1 = 1,
n = 1, 2, 3, …
b u n + 1 = u n − 3,
u1 = 17, n = 1, 2, 3, …
c u n + 1 = u n − 15,
u1 = 75, n = 1, 2, 3, …
d u n + 1 = u n + 10,
u1 = 80, n = 1, 2, 3, …
On graphs, show the first 4 terms of the sequences defined by the following
recurrence relations.
a u n + 1 = 6u n,
u1 = 1,
n = 1, 2, 3, …
b u n + 1 = 3u n,
u1 = −1, n = 1, 2, 3, …
c u n + 1 = 1.5u n,
u 0 = 1,
n = 0, 1, 2, 3, …
d u n + 1 = 0.5u n,
u1 = 10, n = 1, 2, 3, …
For each of the following graphs, write a first-order recurrence relation that
defines the sequence plotted on the graph.
11
b
un
20
16
12
8
4
0
1 2 3 4 5
Term number
n
c
un
100
80
60
40
20
0
1 2 3 4 5
Term number
n
Value of term
a
Value of term
17
Value of term
U
N
C
O
R
R
16
EC
TE
D
15
PA
G
14
E
PR
13
O
O
12
FS
CONSOLIDATE
un
12
10
8
6
4
2
un
8
6
4
2
0
−2
−4
−6
1 2 3 4 5
n
Term number
Topic 5 RECURRENCE RELATIONS
c05RecurrenceRelations.indd 237
237
12/09/15 4:08 AM
18 For each of the following graphs, write a first-order recurrence relation that
defines the sequence plotted on the graph.
1 2 3 4 5
n
n
0
−2
−4
−6
Term number
1 2 3 4 5
n
19 The first five terms of a sequence are plotted on the graph.
FS
Term number
un
50
40
30
20
10
O
1 2 3 4 5
Term number
0
−2
−4
−6
un
8
6
4
2
Value of term
0
c
un
8
6
4
2
Value of term
b
un
40
35
30
25
20
15
10
5
Value of term
Value of term
a
PA
G
E
PR
O
Which of the following first-order recurrence relations
could describe the sequence?
A un + 1 = un − 8
u1 = 8
B un + 1 = un + 8
u1 = 8
0 1 2 3 4 5 n
C un + 1 = un − 8
u1 = 45
Term number
D un + 1 = un + 8
u1 = 45
E u n + 1 = 8u n
u1 = 45
20 Write the first-order recurrence relation that defines the sequence plotted on the
graph shown.
un
5
0
−5
−10
−
−15
−20
−25
3 4 5
n
21 Graphs of the first five terms of first-order recurrence relations are shown below
together with the first-order recurrence relations. Match the graph with the firstorder recurrence relation by writing the letter corresponding to the graph together
with the number corresponding to the first-order recurrence relation.
1 2 3 4 5
n
0
0
e
1 2 3 4 5
Term number
n
Value of term
Value of term
un
20
16
12
8
4
c
un
40
32
24
16
8
Value of term
0
−2
−4
−6
Term number
d
238
b
un
2
Value of term
U
N
C
Value of term
a
1 2 3 4 5
Term number
n
un
5
4
3
2
1
0
f
1 2 3 4 5
Term number
n
un
20
16
12
8
4
0
Value of term
O
R
R
MASTER
EC
TE
D
1
1 2 3 4 5
Term number
n
un
10
8
6
4
2
0
1 2 3 4 5
Term number
n
MATHS QUEST 12 FURTHER MATHEMATICS VCE Units 3 and 4
c05RecurrenceRelations.indd 238
12/09/15 4:08 AM
i un + 1 = un +
1
2
u1 = 8
ii u n + 1 = u n − 2
u1 = 1
1
u1 = 16
2
1
v u n + 1 = 2u n
u1 = 2
vi u n + 1 = u n −
u1 = 4
2
22 The first few terms of four sequences are plotted on the graphs below. Match the
following first-order recurrence relations with the graphs.
i u n + 1 = u n + 2, u1 = 1,
n = 1, 2, 3, …
ii u n + 1 = 3u n,
u1 = 1,
n = 1, 2, 3, …
iii u n + 1 = −4u n,
u1 = −5,
n = 1, 2, 3, …
iv u n + 1 = u n − 4, u1 = 1,
n = 1, 2, 3, …
iv u n + 1 = u n
E
0
1 2 3 4 5
n
G
n
TE
3 4 5
d un
10
8
6
4
2
0
1 2 3 4 5
n
U
N
C
O
R
R
EC
1
D
un
10
5
0
−5
−10
−15
−20
n
PA
c
2 3 4 5
PR
0
−10
−20
−30
−40
−50
−60
−70
−80
30
25
20
15
10
5
O
b un
un
20
10
O
a
u1 = 1
FS
iii u n + 1 = 2u n
Topic 5 RECURRENCE RELATIONS
c05RecurrenceRelations.indd 239
239
12/09/15 4:08 AM
ONLINE ONLY
5.5 Review
www.jacplus.com.au
The Maths Quest Review is available in a customisable
format for you to demonstrate your knowledge of this
topic.
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A summary of the key points covered in this topic is
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opportunity to demonstrate the skills you have
developed to efficiently answer questions using the
most appropriate methods
FS
REVIEW QUESTIONS
PR
O
O
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the links found in the Resources section of your
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D
www.jacplus.com.au
TE
Interactivities
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PA
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U
N
C
O
R
R
EC
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240
MATHS QUEST 12 FURTHER MATHEMATICS VCE Units 3 and 4
c05RecurrenceRelations.indd 240
12/09/15 4:08 AM
5 Answers
3 The sequence is −8, −13, −18, −23, −28, ...
EXERCISE 5.2
1 a This is a first-order recurrence relation.
b This is not a first-order recurrence relation.
2 a This is not a first-order recurrence relation.
b This is a first-order recurrence relation.
3 un = 4un − 1 + 3, u0 = 5
6
7
8
8 a un + 1 = 2un
9 a un + 1 = un + 2
E
c un + 1 = un − 7
c 23, 24, 25, 26, 27
G
d u n + 1 = u n − 0.5
d 7, −3, −13, −23, −33
e un + 1 = un + 4
9 a 1, 3, 9, 27, 81
PA
c 1, −4, 16, −64, 256
D
d −1, −2, −4, −8, −16
TE
10 a 1, 3, 7, 15, 31
b 5, 13, 37, 109, 325
10
11
11 a The first seven terms are 6, −5, 6, −5,
5, 6, −
−5,
5, 66..
EC
b The first seven terms are 1, 5, 25, 125, 625, 3125, 15 625.
12 C
R
13 E
O
R
13
14
C
1
16 −
8
17 D
12
1 1 3 1 3
1
, 2 , 6 , 20
4 4 4 4 4
4
b u n + 1 = u n + 2000, u 0 = 200 000; 200 000, 202 000,
204 000, 206 000, 208 000
c u n + 1 = u n − 7, u 0 = 100; 100, 93, 86, 79, 72
d u n + 1 = 2u n − 50, u 0 = $200; $200, $350, $650,
$1250, $2450
U
N
18 a u n + 1 = 3u
3u n , u0 = ; ,
15
16
EXERCISE 5.3
1 a un + 1 = un + 3,
b un + 1 = un − 4,
2 a un + 1 = un + 5,
b un + 1 = un + 11,
u1 = 1
u1 = 12
u1 = −24
u1 = 55
u1 = 1
u1 = 3
b un + 1 = 4un
b un + 1 = 3un
u1 = −3
u1 = −2
u1 = 12
u1 = 1
u1 = 2
u1 = −2
u1 = 6
u1 = 4
un + 1 = un + 4
g un + 1 = un − 5
h u n + 1 = u n + 6.5
B
a un + 1 = un − 1
u1 = −4
b un + 1 = un + 2
u1 = 3
c un + 1 = un + 3
u1 = −1
d un + 1 = un − 2
u1 = 4
A
a u n + 1 = u n − 2,
u1 = 12,
b u n + 1 = u n + 1,
u1 = 1,
a u n + 1 = 2u n
u1 = 5
b u n + 1 = 6u n
u1 = 1
c u n + 1 = −u n
u1 = −3
d u n + 1 = 4u n
u1 = −3
e u n + 1 = 3u n
u1 = 2
f u n + 1 = −u n
u1 = 5
g u n + 1 = 4u n
u1 = −2
h u n + 1 = −3u n u1 = 5
D
a u n + 1 = 3u n
u1 = 2
b u n + 1 = 4u n
u1 = −3
c u n + 1 = −u n
u1 = 0.5
d u n + 1 = 5u n
u1 = 3
e u n + 1 = 2u n
u1 = −5
f u n + 1 = −3u n u1 = 0.1
f
b −2, −10, −50, −250, −1250
15 −1
u1 = 3
b un + 1 = un + 7
b 5, 2, −1, −4, −7
14 7
O
u1 = 3
1
2000
b un +1 = − un, u1 = 20
2
7 a un + 1 = 5un
u1 = 2
O
5
6 a un + 1 = 5un,
PR
4
The sequence is 5, 23, 95, 383, 1535.
fn + 1 = 5fn − 6, f0 = −2
The sequence is −2, −16, −86, −436, −2186.
The second term is 2.
2
The fifth term is .
3
b, c, g, j
a 6, 8, 10, 12, 14
FS
The first-order recurrence relation is
un + 1 = un − 5, u1 = −8.
1
1
1
4 The sequence is 4 , 5, 5 , 6, 6 , ...
2
2
2
The first-order recurrence relation is
1
1
un +1 = un + , u1 = 4 .
2
2
5 a un + 1 = 5un,
u1 = 1
b un + 1 = −2
−2u
−
2 n, u1 = 4
n = 1, 2, 3, …
n = 1, 2, 3, …
Topic 5 RECURRENCE RELATIONS
c05RecurrenceRelations.indd 241
241
12/09/15 4:08 AM
7 un + 1 = 2un
17 A
u1 = −4,
8 un + 1 = 2un
b
c
d
e
c
Value of term
EC
n
b
0
−4
−8
−12
5 un + 1 = un − 2
6 un + 1 = un + 4
Value of term
n
u1 = 14
u1 = 6
n
n
1 2
n
4 5
b
un
20
16
12
8
4
0
Term number
1 2 3 4 5
Term number
n
un
100
80
60
40
20
n
1 2 3 4 5
Term number
un
100
0
−100
−200
−300
−400
−500
1 2
5
n
un
50
40
30
20
10
0
c
1 2 3 4 5 6
Term number
1 2 3 4 5
Term number
Term number
Value of term
1 2 3 4 5 6
Value of term
0
G
Value of term
TE
D
0
242
13 a
U
Value of term
un
36
33
30
27
24
21
18
15
12
9
6
3
un
8
4
0
R
un
500
400
300
200
100
n
O
R
Value of term
1 2 3 4 5
0
4
Value of term
c
un
14
12
10
8
6
4
2
0
3
1 2 3 4 5
1 2 3 4 5
Term number
PA
0
−1
−2
−3
−4
−5
−6
C
Value of term
2
12 a
n
N
Value of term
un
1
n
un
25
20
15
10
5
0
EXERCISE 5.4
1
1 2 3 4 5
Term number
FS
20 a
0
un
50
40
30
20
10
O
c
b
O
b
un
20
16
12
8
4
PR
e
19 a
11 a
E
d
10 B
Value of term
c
9B
Value of term
n = 1, 2, 3, …
n = 1, 2, 3, …
u n + 1 = 3u n ,
u1 = 2,
n = 1, 2, 3, …
u n + 1 = −u n , u1 = −1,
n = 1, 2, 3, …
u n + 1 = 2u n , u1 = 5,
n = 1, 2, 3, …
u n + 1 = u n + 6.5,
u1 = 4,
n = 1, 2, 3, …
u n + 1 = u n − 5,
u1 = 15,
n = 1, 2, 3, …
u n + 1 = u n + 4,
u1 = 1, n = 1, 2, 3, …
u n + 1 = 2u n , u1 = 4,
n = 1, 2, 3, …
u n + 1 = 4u n , u1 = −3, n = 1, 2, 3, …
u n + 1 = −6u n , u1 = 2,
n = 1, 2, 3, …
u n + 1 = 8u n , u1 = −0.1, n = 1, 2, 3, …
u n + 1 = −10u n , u1 = 3.5, n = 1, 2, 3, …
b u n + 1 = −7u n , u1 = 2,
Value of term
18 a u n + 1 = −u n ,
u1 = 1.5
u1 = 3
1 2 3 4 5
Term number
n
un
500
400
300
200
100
0
1 2 3 4 5
Term number
n
MATHS QUEST 12 FURTHER MATHEMATICS VCE Units 3 and 4
c05RecurrenceRelations.indd 242
12/09/15 4:08 AM
un
160
140
120
100
80
60
40
20
0
0
−5
−10
−15
−20
−25
un
100
75
50
25
Value of term
0
−25
−50
1 2 3 4 5
6
5
4
3
2
1
1 2 3 4 5
R
n
0
1 2 3 4 5
c un + 1 = un − 4
18 a u n + 1 = 2u n
b u n + 1 = −u n
c u n + 1 = −0.5u n
n
u1 = 2
u1 = 90
u1 = 8
u1 = 5
u1 = 6
u1 = 8
19 C
20 un + 1 = 3un
u1 = −1
21 a ii
b v
c iii
d iv
e vi
i
22 i matches to d.
U
0
b u n + 1 = u n − 20
f
N
un
150
120
90
60
30
n
O
R
un
80
60
40
20
EC
un
25
20
15
10
5
1 2 3 4 5
n
E
un
10
8
6
4
2
PA
D
n
TE
1 2 3 4 5
0
d
1 2 3 4 5
17 a u n + 1 = u n + 3
0
c
n
un
15
12
9
6
3
0
b
d
C
15 a
0
1 2 3 4 5
Term number
n
3 4 5
G
Value of term
un
10
8
6
4
2
0
1
c un
n
Term number
c
un
5
b
n
4 5
FS
1 2
n
1 2 3 4 5
O
0
Term number
b
un
250
200
150
100
50
O
−20
16 a
PR
Value of term
14 a
ii matches to b.
iii matches to a.
1 2 3 4 5
n
iv matches to c.
Topic 5 RECURRENCE RELATIONS
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243
12/09/15 4:09 AM