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Patterns
Inductive Reasoning
Inductive reasoning is making conclusions based
on patterns you observe. You can use inductive
reasoning to extend number patterns.
3, 9, 15, 21, …
Describe the pattern. Then find the next two
numbers in the pattern.
3 9 15 21
+6 +6 +6
The pattern is to add
6 to the previous term.
To find the next number, add 6 to 21 = 27.
The next term is 6 + 27 = 33.
Extending Number Patterns
1, 4, 9, 16, …
Describe the pattern. Then find the next two
numbers in the pattern.
1
4
9
16
The pattern is to square the consecutive integers.
12, 22, 32, 42
To find the next number, square the next two
consecutive integers.
52 = 25 and 62 = 36
Sequences
A number pattern is also called a sequence. Each
number in a sequence is a term of the sequence.
An arithmetic sequence is one where you add a
fixed number to the previous term. The fixed
number is called a common difference.
11, 23, 35, 47, …
Find the common difference of each arithmetic
sequence.
11 23 35 47
The common difference is 12.
+12 +12 +12
Arithmetic Sequence
An arithmetic sequence has the form
a1, a1 + d, a1 + 2d, ….
where a1 is the first term
and d is the common difference.
If a1 = 2 and d = 5,
then the sequence 2, 2 + 5, 2 + 2(5), ….
or 2, 7, 12, …. is arithmetic.
Arithmetic Sequence
If the sequence is arithmetic,
find the next two terms.
4, 17, 30, 43, 56, ….
The first term is a1 = 4 and d = 13,
then the sequence 4, 4 + 13, 4 + 2(13), ….
or 4, 17, 30, 43, 56…. is arithmetic.
The next two terms are 69 and 82.
Arithmetic Sequence
If the sequence is arithmetic, find the next two terms.
3, 5, 9, 15, 23, ….
The first term is a1 = 3.
a2 – a1 = 5 – 3 = 2
a 3 – a2 = 9 – 5 = 4
a4 – a3 = 15 – 9 = 6
a5 – a4 = 23 – 15 = 8
There is no common difference
so the sequence is not arithmetic.
Writing a rule for the
nth term of a sequence
Find the rule for the nth term of the
sequence –4, 1, 6, 11, 16….. then find a100.
The first term is a1 = –4.
a2 – a1 = 1–(–4) = 5
a 3 – a2 = 6 – 1 = 5
a4 – a3 = 11 – 6 = 5
a5 – a4 = 16 – 11 = 5
The first term of the sequence is a1 = –4 and the
common difference is d = 5.
an = a1 + (n – 1)d
an = –4 + (n – 1)5
Substitute 100 in for n. a100 = –4 + (100 – 1)5
a100 = 491
Fibonacci Rabbit Problem
Mathematician and businessman Leonardo Pisano Fibonacci
posed the following problem in his treatise Liber Abaci (pub. 1202):
How many pairs of rabbits will be produced in a year, beginning with
a single pair, if in every month, each pair bears a new pair
which becomes productive from the second month on?
Fibonacci Rabbit Problem
The way you find the total number of rabbits
for each month is to first find out how many pairs
of rabbits were newly born that month.
Then add that total to the number of rabbits you
had before the new ones were born.
So how many pairs of rabbits are newly born every month?
Since it takes two months for each new pair to give birth,
each pair of rabbits that was alive two months ago will
give birth to a new pair. In other words, the number of
new pairs in each month is equal to the number of
pairs alive two months ago.
Fibonacci Rabbit Problem
Next we need to find the number of rabbit pairs
that were alive before the new ones were born.
This is the number of pairs alive the month before.
In other words, to find the total number of pairs
of rabbits, you simply add together the number of
pairs that were alive in the preceding two months.
Do you know of any series of numbers which begins
with one and one and continues by adding the preceding
two numbers to get the next? Of course you do.
It's the Fibonacci sequence.
Fibonacci Sequence
The Fibonacci Sequence is the series of numbers:
0, 1, 1, 2, 3, 5, 8, 13, 21, 34, …..
The next number is found by adding
up the two numbers before it.
• The 2 is found by adding the two
numbers before it (1 + 1)
• Similarly, the 3 is just (1 + 2),
• And the 5 is just (2 + 3),
• and so on!
Example: the next number in the
sequence above would be (21 + 34) = 55
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