7.5 Use Recursive Rules with
Sequences and Functions
p. 467
• What is a recursive rule
for arithmetic sequences?
• What is a recursive rule
for geometric sequences?
• What is an iteration?
Explicit Rule
Recursive Rule
• Gives the beginning term(s) of a sequence
and a recursive rule that relates the given
term(s) to the next terms in the sequence.
• For example: Given a0=1 and an=an-1-2
• The 1st five terms of this sequence would
be: a0, a1, a2, a3, a4 OR
• 1, -1, -3, -5, -7
Recursive Equations
Example: Write the indicated rule for the
arithmetic sequence with a1=15 and d=5.
• Explicit rule
an=a1+(n-1)d
an=15+(n-1)5
an=15+5n-5
an=10+5n
• Recursive rule
(*Use the idea that you get
the next term by adding
5 to the previous term.)
Or an=an-1+5
So, a recursive rule would
be a1=15, an=an-1+5
Example: Write the indicated rule for the
geometric sequence with a1=4 and r=0.2.
• Explicit rule
an=a1rn-1
an=4(0.2)n-1
• Recursive rule
(*Use the idea that you get the next
term by multiplying the previous
term by 0.2)
Or an=r*an-1=0.2an-1
So, a recursive rule for the sequence
would be a1=4, an=0.2an-1
Example: Write the 1st 5 terms of
the sequence.
• a1=2, a2=2, an=an-2-an-1
1st
nd term
2
term
a3=a3-2-a3-1→a1-a2=2-2=0
a4=a4-2-a4-1→a2-a3=2-0=2
a5=a5-2-a5-1→a3-a4=0-2=-2
2, 2, 0, 2, -2
1
2
2
2
3
0
4
2
5
-2
Write the first six terms of the sequence.
a. a0 = 1, an = an – 1 + 4 b. a1 = 1, an = 3an – 1
SOLUTION
a. a0 = 1
b. a1 = 1
a1 = a0 + 4 = 1 + 4 = 5
a2 = 3a1 = 3(1) = 3
a2 = a 1 + 4 = 5 + 4 = 9
a3 = 3a2 = 3(3) = 9
a3 = a2 + 4 = 9 + 4 = 13
a4 = 3a3 = 3(9) = 27
a4 = a3 + 4 = 13 + 4 = 17
a5 = 3a4 = 3(27) = 81
a5 = a4 + 4 = 17 + 4 = 21
a6 = 3a5 = 3(81) = 243
Write the first six terms of the sequence.
a. 3, 13, 23, 33, 43, . . .
SOLUTION
The sequence is arithmetic with first term a1 =
3 and common difference d = 13 – 3 = 10.
an = an – 1 + d General recursive equation for a
n
= an – 1 + 10
Substitute 10 for d.
ANSWER
So, a recursive rule for the sequence is
a1 = 3, an = an – 1 + 10.
Write the first six terms of the sequence.
b. 16, 40, 100, 250, 625, . .
.
b. The sequence is geometric with first term
a1 = 16 and common ratio r = 40 = 2.5.
16
an = r an – 1
= 2.5an – 1
General recursive equation for an
Substitute 2.5 for r.
ANSWER
So, a recursive rule for the sequence is
a1 = 16, an = 2.5an – 1.
Write the first five terms of the sequence.
1. a1 = 3, an = an – 1 – 7
SOLUTION
a1 = 3
a2 = a1 –7 = 3 – 7 = – 4
a3 = a2 – 7
a3 = – 4 – 7 = – 11
a4 = a3 – 7 = – 11 – 7 = – 18
a5 = a4 – 7 = – 18 – 7 = – 25
ANSWER
3, –4, –11, –18, –25
Or think of it this way…
3
−4
−11
−18
−25
Write the first five terms of the sequence.
3. a0 = 1, an = an – 1 + n
SOLUTION
a0 = 1
a1 = a0 + 1 = 1 + 1 = 2
a 2 = a1 + 1 = 2 + 2 = 4
a3 = a2 + 3 = 4 + 3 = 7
a4 = a3 + 4 = 7 + 4 = 11
ANSWER
1, 2, 4, 7, 11
Write the first five terms of the sequence.
4. a1 = 4, an = 2an – 1 – 1
SOLUTION
a1 = 4
a2 = 2a1 – 1 = (2
a3 = 2a2 – 1 = (2
4) – 1 = 8 – 1 = 7
7) – 1 = 14 – 1 = 13
a4 = 2a3 – 1 = (2
13) – 1 = 26 – 1 = 25
a5 = 2a4 – 1 = (2
ANSWER
25) – 1 = 49
4, 7, 13 25, 49
Write a recursive rule for the sequence.
5. 2, 14, 98, 686, 4802, . . .
SOLUTION
The sequence is geometric with first term a1
= 2 and common ratio
a2
r=a =7
1
an = r an – 1
= 7 · an – 1
ANSWER
So, a recursive rule for the sequence is
a1 = 2, an = 7an – 1
Write a recursive rule for the sequence.
a.
1, 1, 2, 3, 5, . . .
SOLUTION
a. Beginning with the third term in the
sequence, each term is the sum of the two
previous terms.
ANSWER
So, a recursive rule is a1 = 1, a2 = 1, an = an – 2 + an – 1.
This sequence is the Fibonacci sequence.
By definition, the first two numbers in the Fibonacci
sequence are 0 and 1 (alternatively, 1 and 1), and each
subsequent number is the sum of the previous two.
0,1,1,2,3,5,8,13,21,34,55,89,144,…
Write a recursive rule for the sequence.
b.
1, 1, 2, 6, 24, . . .
SOLUTION
b.
Denote the first term by a0 = 1. Then note that
a1 = 1 = 1 a0, a2 = 2 = 2 a1, a3 = 6 = 3 a2, and
so on.
ANSWER
So, a recursive rule is a0 = 1, an = n an – 1.
This sequence lists factorial numbers.
Iterating Functions
Find the first three iterates x1, x2, and x3 of the
function f (x) = –3x + 1 for an initial value of x0 = 2.
SOLUTION
x1 = f (x0)
= f (2)
= –3(2) + 1
= –5
x2 = f (x1)
= f (–5)
= –3(25) + 1
= 16
x3 = f (x2)
= f (16)
= –3(16) + 1
= – 47
ANSWER
The first three iterates are – 5, 16, and – 47.
Find the first three iterates of the function for
the initial value.
11. f (x) = 4x – 3, x0 = 2
SOLUTION
x1 = f (x0)
= f (2)
=8–3
=5
x2 = f (x1)
= 4 (5) – 3
= 17
x3 = f (x2)
= 4 (17) – 3
= 68 – 3
= 65
ANSWER
The first three iterates are 5, 17, and 65.
7.5 Assignment:
p. 470, 3-27 odd, skip 21
Write a recursive rule for the
sequence 1,2,2,4,8,32,… .
• First, notice the sequence is neither arithmetic
nor geometric.
• So, try to find the pattern.
• Notice each term is the product of the previous
2 terms.
• Or, an-1*an-2
• So, a recursive rule would be:
a1=1, a2=2, an= an-1*an-2
Example: Write a recursive rule for
the sequence 1,1,4,10,28,76.
• Is the sequence arithmetic, geometric, or
neither?
• Find the pattern.
• 2 times the sum of the previous 2 terms
• Or 2(an-1+an-2)
• So the recursive rule would be:
a1=1, a2=1, an= 2(an-1+an-2)