Sequences
The following figures are created with squares of side length of one and n is the number of squares in the
base. The figures are stair-step stacks of squares. P1 is the perimeter of the first figure, P2 is the perimeter
of the second figure, and so on. Use Pn-1 to represent the previous term.
According to this pattern, what is the
perimeter of a figure with a base of
(a) 7 squares?
(b) 70 squares?
P1 = 4
n=1
P3 = 12
n=3
P2 = 8
n=2
Is there a formula that can help?
P4 = 16
n=4
(a) 7-square base: You will notice that the perimeter increases by 4 with each subsequent figure. To find
the perimeter of the figure with a base of 5, you would +4 to the perimeter of the previous figure. Then
you +4 two more times. Perimeter of a figure with a base of 7 squares = 16 + 4 + 4 + 4 = 28.
Let’s look at this in a table format.
Figure
Perimeter, Pn
1
4
2
8
3
12
4
16
5
20
6
24
7
28
Pn-1 + 4
We can write a formula to find the perimeter of the figure with a base length of 5 and subsequent figures.
To write the pattern rule that you may remember from Algebra I: NEXT = NOW + 4, STARTING AT 4
Recursive formula This is just a different way of writing it:
Pn = Pn-1 + 4, if P1 = 4
where Pn = any term, Pn-1 = previous term, P1 = the first term, difference, d, = 4
(b) 70-square base: The pattern used above is not practical when you need a term that is not close to the
given terms. Let’s see if we can find a different pattern. You will notice that the perimeter is 4 times the
number of squares in the base. Perimeter of a figure with a base of 70 squares = 4(70) = 280
Figure
Perimeter, Pn
1
4
2
8
3
12
4
16
5
20
6
24
7
28
4n
We can write a formula to find the perimeter of the figure with a base length of n when n ≥ 1. You may
remember from Algebra I that it is a LINEAR FUNCTION.
Explicit formula Pn = 4n
Here is how to write the formula: Pn = P1 + d(n – 1) = 4 + 4(n – 1)
Sequences and Series
Sequence- an ordered list of numbers. Each number in a sequence is called a term or element, an. The
terms can be represented by a1, a2, a3, etc.
Examples of Sequences:
Finite Sequences
{1, 6, 11, 16, 21, 26}
{1, 4, 9, 16, 25}
Domain of sequence: {1, 2, 3, 4, 5, …}
Infinite Sequences
{-8, 10, 28, 56, 74, …}
{4, 8, 16, 32, 64, …}
Range of a sequence: the terms of the sequence
Series- when the terms of a sequence are added. Some sequences have a sum, others do not.
Examples of Series:
Finite Series
1 + 6 + 11 + 16 + 21 + 26
Infinite Series
-8 + 10 + 28 + 56 + 74 + …
Sum of a Series Sn represents the sum of a series
Example: Sn = 1 + 6 + 11 + 16 + 21 + 26 = 81
Note: -8 + 10 + 28 + 56 + 74 + … has no sum
Find the first five terms, the sum, and the 50th in the sequence defined by the general term an for each of the
following.
a. an  3n  2
In order to find the first five terms in the sequence
an  3n  2 , we plug n = 1, 2, 3, 4 and 5 into the
general term. So the first five terms in the sequence
are as follows:
a1  3 1  2  1
a2  3  2   2  4
a3  3  3  2  7
a4  3  4   2  10
a5  3  5   2  13
b. an 
n 1
n
In order to find the first five terms in the sequence
n 1
an 
, we plug n = 1, 2, 3, 4 and 5 into the
n
general term. So the first five terms are as follows:
11
2 1 1
3 1 2
a1 
0
a2 

a3 

1
2 2
3 3
4 1 3
5 1 4
a4 

4
a5 
4
1 2 3 4
Sequence: 0, , , , .
2 3 4 5
Sequence: 1, 4, 7, 10, 13
Sum
Sn = 1 + 4 + 7 + 10 + 13 = 35
The 50th term is found by plugging 50 in for n in our
general term: a50  3  50   2  148 .
1
2
3
4
2
3
4
5
Sum Sn = 0    

5

5
163
60
50  1
The 50th term, n = 50: a50 
50

49
50
.
Arithmetic Sequences
Sequences and series may have an apparent pattern, but others have a more random ordered list.
Sequences with patterns can be written using formulas. We will look at two specific types of patterns.
These sequences are called arithmetic and geometric.
Arithmetic Sequence EXAMPLE: 4, 7, 10, 13, 15, 18, …
difference = 7 – 4 = 3, or 10 – 7 = 3
To find subsequent terms, you add successive 3s: one 3 for the second term, two 3s for the third term,
three 3s for the fourth term, so 3(n – 1) for the nth term.
Recursive Formula
Explicit Formula
Use when you want to find the next few terms
Use when you want to find any term
an  an1  d , given a1
an  a1  d (n  1)
NEXT = PREVIOUS TERM + d, given the FIRST TERM
ANY TERM = FIRST + DIFFERENCE( TERM # - 1)
common difference, d - the same amount is added from one term to the next
To find the common difference, subtract consecutive terms: d = a2 – a1 or a3 – a2, etc.
The explicit formula of an arithmetic sequence is a linear function with slope d.
f ( x)  a1  d ( x  1) or f ( x)  dx  c
an = any term  a1 = first term, a2 = second term, etc.
an-1 = previous term
d = common difference  what you add to each term by to get the next term
n = term number
Examples of Arithmetic Sequences:
Finite: 11, 13, 15, 17, 19
[d = 2]
Infinite: 4, 4.2, 4.4, 4.6, 4.8, 5.0, …
[d = .2]
Terms of Arithmetic Sequences
Recursive Formula:
an  an1  d , given a1
Explicit Formula:
an  a1  d (n  1)
Find the first 4 terms of the sequence of numbers defined by the given recursive formula.
1.
an  an 1  5; a1  8
*previous term - 5
2. an  an 1  1 2 ; a1  7 *previous term + ½
a1  8
a1  7
a2  8  5  3
a2  ______  1  _______
2
a3  ______  1  _______
2
a4  ______  1  _______
2
a3  3  5  2
a4  2  5  7
8, 3, -2, -7
________, ________, ________, ________
Find the first 4 terms of the sequence of numbers defined by the given explicit rule.
3.
an  1.5n  4.5
a1  f (1)  1.5(1)  4.5  6
4.
an  5n
a1  f (1) 
a2  f (2)  1.5(2)  4.5  7.5
a2  f (2) 
a3  f (3)  1.5(3)  4.5  9
a3  f (3) 
a4  f (4)  1.5(4)  4.5  10.5
a4  f (4) 
6, 7.5, 9, 10.5
________, ________, ________, ________
For each arithmetic sequence, write the recursive and explicit formula. Find the 20 th term.
5. -34, -38, -42, -46, -50, …
6. -17, -6, 5, 16, 27, …
**You are given a1 = -34. Find d = -38-(-34) = -4
**Given a1 = -17. Find d: -6 – (-17) = 11
an  an1  4; a1  34
Explicit: an  34  4(n  1)  4n  30
Recursive:
Recursive: an  ____________; a1  ____
Explicit: an = ______________________
7. Write the explicit form of the arithmetic sequence with a3 = 2 and a7 = 22.
22  2 20
d

 5 ; Find the first term by subtracting d: a1 =a3 -5-5 = 2-5-5 = -8. an  8  5(n  1)  5n  13
73
4
Arithmetic Series and Sigma Notation
Arithmetic Sequence
4, 7, 10, 13, 15, 18
-34, -38, -42, -46, -50, …
Arithmetic Series
Type of Sequence
4+7+10+13+15+18
Finite
-34+-38+-42+-46+-50+ …
Infinite
Is there a sum?
Yes
No
What is the sum of the first 100 positive numbers 1, 2, 3, …, 98, 99? Solution: 49 sums of 100 + 50 = 4950
Sigma Notation (Summation Notation)
The Greek letter  is a mathematical symbol for finding the sum of the first n terms of a sequence ak.
n
a
k 1
k
 a1  a2  a3  a4  ...  an
Examples: Find each sum. You will need to find each individual term and add them.
5
1.
 2k  1 = 1 + 3 + 5 + 7 + 9 = 25
k 1
Find each term: f(1) = 1, f(2) = 3, f(3) = 5, f(4) = 7, f(5) = 9
Find the sum.
4
2.
  7k = 7  49  343  2401  2100
k 1
Find each term: f(1) = -7, f(2) = 49, f(3) = -343, f(4) = 2401
7
3.
a3  4(3)  2  14, a4  4(4)  2  18
 4k  2 a
k 3
5
 22, a6  26, a7  30
S5  14  18  22  26  30  110
Find the sum.
Sum of Arithmetic Series
Sum of a Finite Arithmetic Series
You can find the sum of a finite arithmetic series or a partial sum of n terms of an infinite arithmetic
series. We can use a formula for the sum of n terms. a1 is the first term and an represents the nth term.
Sn 
n(a1  an )
1
or n(a1  an )
2
2
Note: you cannot find the sum of all terms of an infinite arithmetic series.
Examples: Find the sum of each series. You will need the first term a1, the last term an, and n.
a1  f (1)  5  3(1)  2
10
4.
 5  3k
a10  f (10)  5  3(10)  25
k 1
1
Sn  (10)(2  25)  115
2

5. The first 7 terms of
 8i  6
i 1
a1  f (1)  8(1)  6  14
a7  f (7)  8(7)  6  62
Sn 
7(14  62)
 266
2
*This is a partial sum of an infinite series.
an  6  6(n  1)  6n  12
a1  f (1)  6
6. The first 20 terms of -6, 0, 6, 12, 18, …
a20  f (20)  6(20)  12  108
1
Sn  (20)(6  108)  1020
2
Applications
Party Planning : Suppose you are planning a grand
opening party at a local theater. Rather than count
all of the seats for planning purposes, you decide to
only count the first row of seats. You know that
there are 25 rows of seats, the first row has 22
seats, and each successive row has one additional
seat. What is the total number of people that can
attend your party?
Courtesty of idea go @ Freedigitalphotos.net
One Solution: This is an example of an arithmetic series in which each “term” (row of seats) increases by 1.
That means that d = 1. We need the sum of 25 rows, beginning with 22 seats in the first row and
increasing by 1 in each subsequent row: a1 = 22, d = 1, n = 25. Write the explicit formula an = 22 + 1(n 1)= n + 21. Find the number of seats in the last row: a25 = 25 +21 = 46 seats.
1
Find the sum: S n  (25)(22  46)  850 total seats
2
Purchasing an iPod: An NCVPS Algebra 2 student has $282 saved towards a new $399 IPod. Her savings
from a weekly babysitting can be modeled by the formula an  an1  40 . In how many weeks will she be
able to buy the iPod?
One Solution: Find the balance she needs to have enough to buy the iPod: $399 – 282 = $117. She is
saving $40 each week. 117/40 is approximately 3 weeks.
Bacteria Population: The number of bacteria at the end of 5 minutes in an experiment can be modeled by
the formula at  3.75t  37,060.25 . In the formula at is the number of bacteria at the end of t minutes.
There are 37,079 bacteria in the experiment at the end of 5 minutes. How many bacteria will be in the
experiment at the end of 8 minutes?
One Solution: This formula begins at t = 1, the bacteria increases by 3.75 every minute (slope), and at 5
minutes it is at 37,079. You need to find 3 additional minutes.
a5  37, 079; a6  3.75  37079; a7  3.75(2)  37079; a8  3.75(3)  37079
37,090.25 bacteria at 8 minutes