Pre-Calculus Quarter 1, Weeks 9: Sigma Notation
Pre-Calculus Quarter 1, Weeks 9: Sigma Notation
Pre-Calculus Quarter 1, Weeks 9: Sigma Notation
QUARTER 1, WEEKS 9
SIGMA NOTATION
I. Learning Competency
Use the sigma notation to represent a series (STEM_PC11SMI-Ih- 3)
Apply the use of sigma notation in finding sums.
illustrate
II. Objectives
At the end of the lesson, the student should be able to:
1. write a series in sigma notation or summation notation ;
2. evaluate sums written in sigma notation;
3. determine the properties of sigma notation; and
4. calculate sums using sigma notation.
2i . The expression is read “the summation from 1 to 5 of 2i”. The number of terms of the
i 1
series is the difference between the upper bound and the lower bound plus one, (5 -1 =
4+1 = 5). Generally, we have
1
Evaluate a sum in sigma notation
The terms of the series above are generated by successively replacing the index of
summation with the consecutive integers from the first up to the last values of n. Thus, the
values of n for the series above are 1, 2, 3, … up to 5.
Illustrative Examples.
1. State the index and the number of terms in the following series
10 15 5
a. 3n
n 1
b. 3 2k
k 3
c. 3 2
j 3
j 1
Solutions:
a. Index is n
# of terms 10-1 = 9 +1 = 10
b. Index is k
# of terms 15-3 = 12 +1 = 13
c. Index is j
# of terms 5-3 = 2 +1 = 3
5
2. Write 3n 4 in expanded form and find the sum
n 1
Solution:
5
3 (n 1)7 7n 4
n 1 n 1
Similarly, just like in our previous discussion, I is the index of summation, m is the
lower bound, n is the upper bound and f(i) is a term. The sum of the first term o the
sequence with nth term an is
2
n
a
i 1
i a1 a2 ... an
To evaluate the sum of a series using sigma notation, we have to consider the
properties of sigma notation.
n n n
1. ai bi ai bi
i m i m i m
n n
2. ca
im
i c ai
i m
n n
3. c c(n m 1)
i m i m
n n
4.
im
f (i 1) f (i ) f (n 1) f (m)
im
4. i 3
i 1 2
Illustrative Examples.
3 = 3(12) = 36
i1
n n
b. Since the lower bound is not equal to 1, consider cai c ai
im i m
12
3 = 3(12-4 + 1) = 3(9) = 27
i 4
3
n
n n 1
c. Consider i
i 1 2
5
i=
5(5+1) 30
= = 15
2 2
i 1
n n n
that applies.
8 8 8
2i 3 = 2 i 3 = 2 [
8(8+1)
] − 3(8) = 72 – 24 = 48
2
i 3 i 1 i 1
2020
1
2. Evaluate ii 1
2
3i 2
Solution:
1 a b
Rewrite in the form such that a (n 2) b(n 1) 1
i 3i 2 2
n 1 n 2
2020 2020
1 1 1
Then, 2 =
i 1 i 3i 2 i 1 i 1 i2
It follows that the series
1 1 1 1 1 1 1 1 1 1
= . .....
2 3 3 4 4 5 2020 2021 2021 2022
1 1 505
=
2 2022 1011
10
3. Evaluate i(i 5)
i 1
Solution:
10 10 10(10 1) 2(10) 1 10(10 1)
i(i 5) = i
i 1 i 1
2
5i =
6
5
2
2310 110
= 5 385 275 110
6 2
ACTIVITY 1
8
2. k
k 3
2
k ______ ______
4
6
3. t (t 2)
t 0
______ ______
4
4. i
i 1
2
______ ______
6
5. n
n 5
2
n3 ______ ______
ACTIVITY 2.
Expand and Add
Direction. Write each expression in Expanded form and find the sum
6
1. (5i 1)
i 1
______________________________ _____
6
2. 2
k 0
k
______________________________ _____
4
3. (1) t
t 1
t
______________________________ _____
r
5
1
4. 32 ______________________________ _____
r 1 2
3
(1)k 1
5.
k 0 k 1
______________________________ _____
ACTIVITY 3
Express to Impress
1. 5 + 9 + 13 + 17+…..+ 45 __________
2. -1 + 2 -3 + 4 – 5 + 6 - ….+ 20 __________
3. 31 32 33 34 __________
5
1 1 1 1
4. 1 .... __________
2 3 4 100
1 1 1 1 1
5. __________
4 7 10 13 16
ACTIVITY 4.
Fun for Sum
Direction. Evaluate the following sums using properties of sigma notation.
20
1. (3i 4)
i 1
__________
5
2. k
k 0
2
3k 1 __________
4
t 1
3. t2
t 1
__________
20
4. 12
i12
__________
5
5. (2r 3)
r 1
2
__________
6
ENRICHMENT ACTIVITY
Fun with Telescoping Sum
Task 1: Fun with Telescoping Sum
Challenge
Use
telescoping
sum to
evaluate
the given
sigma
notation.