1. SEQUENCE FUNCTION
 Sequence is a set of numbers written in a given order.
 Function whose domain is the set of natural numbers and whose range is a
subset of the real numbers
 Terms are the numbers or values in the sequence
Examples:
1. 1,2,3,4,5,6,...,n
2. 2,4,6,8,10,…,n
3. 5,10,15,20,…,n
A Sequence Function as a function whose domain is a subset of positive integers.
Example:
A ball is dropped from the height of 10 feet. Each time that it bounces, it reaches
a height that is half of its previous height. We can list the height to which the ball
bounces in order until it finally comes to rest.
After Bounce 2 4 6 5 8
Height (ft) 2 5 7 6 5
The numbers 2,4,6,8,10 form a sequence which is also an example of a sequence.
It can be in a form of ordered pairs and we let x as the after bounce and y as the height.
{(2,2),(4,5),(6,7),(5,6),(8,5)}
Finite Sequence is a function whose domain is the set of integers and has the last
term {1,2,3,…,n}. Infinite sequence is a list that continues without ends.
Example:
Identify whether the sequence is finite or infinite.
1. The first 5 natural numbers: 1, 2, 3 ,4, 5
2. The set of even whole numbers: 2, 4, 6, 8 , 10 ...
3. The multiples of 4 up to the number 40: 4, 8, 12, 16,…, 40.
4. The sequence of the multiples of 3: 3, 6, 9, 12...
Solution:
1 and 3 are examples of finite sequence while the examples in 2 and 4 are infinite
sequence. The 3 dots at the end of the infinite sequence indicates that the sequence goes
on without stopping.
FINDING THE TERM IN THE SEQUENCE
 Sequence is usually denoted by their general or nth term.
For example,
푡푛 =
푛 + 2
푛

2. The equation provides how we will be able to compute for the nth term of the
sequence. To get the value of a specific term , replace n with the number of the term.
For example:
To get the first 3 terms of 푡푛 = 푛+2
푛
, replace n with 2,4 and 6.
Solution:
푡푛 =
푛 + 2
푛
=
2 + 2
2
= 2
푡푛 =
푛 + 2
푛
=
4 + 2
2
= 3
푡푛 =
푛 + 2
푛
=
6 + 2
2
= 4
The domain of the sequence is the set {2,4,6}.
The range is denoted as {2,3,4}.
Example:
1. Find the first 5 terms of the sequence:
푡푛 = 2푛 − 3
The domain is the set {1, 2, 3, 4, 5}
Substituting this into 푡푛 = 2푛 − 3, we have:
푡푛 = 2(1) − 3 = -1
푡푛 = 2(2) − 3 = 1
푡푛 = 2(3) − 3 = 3
푡푛 = 2(4) − 3 = 5
푡푛 = 2(5) − 3 = 7
The first 5 terms of the sequence 푡푛 = 2푛 − 3 is {-1,1,3,5,7}
2. Find the first 3 terms of the sequence:
푡푛 = 푛2
The domain is {1,2,3}
Substituting this into 푡푛 = 푛2, we have:
푡푛 = 12 = 1
푡푛 = 22 = 4
푡푛 = 32 = 9

3. The first 3 terms of the sequence {1,4,9}
FINDING THE GENERAL TERM
When given the first few terms of the sequence, we can look at the pattern in order to
obtain the next terms or a general statement regarding the nth term.
Example:
1. Find the next term and describe the pattern.
1,4,9,16,25…
The next term is 36. The terms are all squares.
1st Term 12= 1
2nd Term 22= 4
3rd Term 32= 9
4th Term 42= 16
5th Term 52= 25
6th Term 62= 36
SERIES
 Series is the sum of the terms
t1 + t2 + t3 + t4 +,…, + tn
Note: The sum of the infinite sequence is an infinite series. A partial sum is the sum of
the first n terms. A partial sum is also called the sum of finite series, and is denoted as sn ,
where n denotes the number of terms in the sum.
For example:
If 4, 7, 10, 13, 16, 19…. Is a sequence, then 4 + 7 + 10 + 13 + 16 +19 is the
corresponding sum or series
If only we want to get the sum of only 4 terms in the sequence, it is a partial sum
written as S4 , which is equal to 4 + 7 + 10 + 13.
Example:
For the sequence 9, 13, 17, 21, 25, 29, 33, 37, find
1. S4
2. S1
Solution:
1. S4= 9+13+17+21 = 60
2. S1=9+13+17+21+25+29+33 = 147

4. SIGMA NOTATION
 Use the symbol Σ, a capital sigma of the Greek alphabet which corresponds to
the letter S.
 The symbol i ,is called the index of summation.
 Used as summation or the total sum of the sequence.
Example:
1
5
Σ 푖 2
푖=1
= 12 + 22 + 32 + 42 + 52
What we have observed here is that, the i =1 is located under the sigma symbol.
So, it indicates that the first term on the right-hand side is the value of i2 when i is equal
to 1. The next terms are the values of i2 when i is 2,3,4 and 5. We stop here because in the
upper sigma notation there is number 5.
The sigma notation can be defined by the equation:
푛
Σ 퐹(푖) = 퐹(푚) + 퐹(푚 + 1) + 퐹(푚 + 2) + ⋯ + 퐹(푛)
푖=푚
Where m and n are integers and m ≤ n. The right hand sign of the equation consist of the
sum of n – m + 1terms, the first of which is obtained by replacing i by m + 1 and so on
until the last term is obtained.
 The upper limit will be m
 The lower limit will be n
 The symbol i is the “dummy” symbol because any other symbol could be used
without changing the right side.
Example:
7
Σ 푖3
푖 =2
= 23 + 32 + 43 + 53 + 63 + 73
Sometimes the term of a sum involve subscripts . For instance the sum
푛1 + 푛2 + 푛3 + 푛4 + ⋯ + 푛푛
Can also be written in sigma notation
푛
Σ 푛푖
푖 =1

5. A. For each sequence, find the indicated partial sum.
1. 3,5,7,9,11…S5
2. 2,4,8,16,32…S7
3. 1,2,3,4,5,6,7,8,9,10,11,12,13,14,…,S6
4. 5,10,15,20,25,30,35,40,…,S7
B. Find the indicated term of the sequence.
1.푡푛 = 4푛 − 7: 푡5
2. 푡푛 = 4푛−2
3
: 푡6
3. 푡푛 = 5 + 2푛: 푡10
4. 푡푛 = 4푛
2
: 푡4
5. 푡푛 = 4(푛 − 7)2: 푡5
c. write the following in sigma notation
1. 2+4+6+8+10
2. 1+3+5+7+9+11+13+15+17
3. 52+62+72
4. 1+
1
2
+ 1
3
+ 1
4
+ 1
5
5. (n+1) + (n+2) +n+3)… (n+n)
PRINCIPLE MATHEMATICAL INDUCTION
Definition Let P(n) be a proposition on an integer variable n. Then P(n) is true for
all integers n  s if and only if the following two conditions are both satisfied :
(i) P(s) is true ,
ii) If P(k) , where k  s , is assumed to be true, then P(k 1) is true .

6. Example:
We use the mathematical induction to prove that
1+3+5+7+….+(2n – 1) =n2
STEP 1. We apply that the formula is true for n = 1. If n = 1 then the formula becomes
1=12
1=1
which is true.
STEP 2. Show that Pk is true.
1 + 3 + 5 + 7 + ⋯ + (2푘 − 1) = 푘2
If the equation 1 is true, then
1 + 3 + 5 + ⋯ + (2푘 − 1) + [2(푘 + 1) − 1] = (푘 + 1)2
This is also true.
Then we add 2k +1 – 1 to the left side of the Equation 1 and its equivalent 2k+1 to the
right side and we have:
1 + 3 + 5 + ⋯ + (2푘 − 1) + [2(푘 + 1) − 1] = 푘2 + (2푘 + 1)
1 + 3 + 5 + ⋯ + (2푘 − 1) + [2(푘 + 1) − 1] = (푘 + 1)2
Which is Equation 2.
Example:
Use mathematical induction to prove that n3 + 3n2 + 2n is divisible by 3 for n  1.
Basis case: n = 1 n3 + 3n2 + 2n = 13 + 3 ·12 + 2·1 = 1 + 3 + 2 = 6 and 3 | 6.
Hypothesis: Assume k3 + 3k2
(i.e. k3 + 3k2 + 2k = 3s for some integer s.)
Induction step:
(k + 1) 3 + 3(k + 1)2 + 2(k + 1) = (k3 + 3k2 + 3k + 1) + 3(k2 + 2k + 1) + 2(k + 1)
= (k3 + 3k2 + 2k) + k + 1 + 3k2 + 6k + 3 + 2k + 2
= 3s + 3k2 + 9k + 6 hypothesis and algebra
= 3s + 3(k2 + 3k + 2) factoring
= 3[s + k2 + 3k + 2] factoring
Therefore 3 | [(k + 1)3 + 3(k + 1)2 + 2(k + 1)].
Therefore, by the principle of mathematical induction, the statement is true for all n.
Example:
Use mathematical induction to prove that 2n < n! for n ≥ 4
Basis: n = 4 lhs 24 = 16 rhs 4! = 24 Since 16 < 24 the basis holds.

7. Hypothesis: Assume 2k < k! for k ≥ 4
Induction step:
Proof: 2k+1 = 2 • 2k
< 2 • k!
< (k + 1) • k! since k >=4
= (k + 1)!
Therefore, by the principle of mathematical induction, the statement is true for all n ≥ 4.