Semester 1 Mathematics STPM (Student's Edition)

96

(iv) when 3 < ≤ 2 , cos > 0 , tan ≤ 0

2

sin
cos < √3

tan < √3

3
2 < ≤ 2

Combining the above results,

the set of values of x satisfying sin < √3 cos ,
4 < ≤ 2
is
3

EXAMPLE 2

Solve sin < 1 .
2

Solution:

1
sin = 2

5
= 6 , 6

5
∴ (0 , 6) ∪ ( 6 , 2 )

EXERCISE 1.14

1. Solve tan ≥ √3
2. Solve 2 cos ≤ 1 in the interval 0 ≤ ≤ 2 .
3. Solve the trigonometric inequality cos 2 > 3 sin + 2 for − < ≤ .
4. Sketch the graph of = sin 2 in the range 0 ≤ ≤ . Hence, solve the inequality

|sin 2 | < 1, where 0 ≤ ≤ 2 .

2

UNIT SAINS DAN MATEMATIK, SEKTOR PEMBELAJARAN, JABATAN PENDIDIKAN NEGERI PULAU PINANG

97

5. Sketch the graph of = 12 cos − 5 sin for 0 ≤ ≤ 2 , and determine −5 ≤
12 cos − 5 sin ≤ 0.

ANSWER 1 4
Section [3 , 2) ∪ [ 3 , 2 )
1.4
2 5
Exercise [3 , 3 ]
1.14

3 5
{ : − 6 < < − 6 }

4

y = sin 2x

5 y = 12 cos ̶ 5 sin
̶5 y= ̶5

1.176 ≤ ≤ 1.571 or 3.923 ≤ ≤ 4.317

UNIT SAINS DAN MATEMATIK, SEKTOR PEMBELAJARAN, JABATAN PENDIDIKAN NEGERI PULAU PINANG

98
CLONE STPM

1. Sketch the graphs of y = cos, y = sin 2 for  − ,  on the same coordinate axes. Hence
solve the inequality cos  sin 2 .

2. Sketch the graph of = |2 cos 2 | in the range of 0 ≤ ≤ . Hence, solve the inequality

|cos 2 |  1 , where 0 ≤ ≤ .
2

3. Sketch the graph of y = 24cos − 7sin for 0    2 and determine the range of  in

this interval satisfying the inequality −5  24cos − 7sin  0 .

ANSWER 1
Section
1.4

Clone
STPM

5
∈ { : − ≤ ≤ , 6 ≤ ≤ 2 , 6 ≤ ≤ }
2

or < 2 or 5 < < }
{ : 0 < < 6 3 <3 6

3

{ ∶ 1.287 ≤ ≤ 1.488 or 4.226 ≤ ≤ 4.428}

UNIT SAINS DAN MATEMATIK, SEKTOR PEMBELAJARAN, JABATAN PENDIDIKAN NEGERI PULAU PINANG

99

STPM PAST YEAR

1. Sketch the graph of 12 cos − 5 sin for 0 ≤ ≤ 2 and determine the range of values
of in this interval satisfying the inequality −5 ≤ 12 cos − 5 sin ≤ 0. [STPM 2015]

2. Sketch the graph of = sin 2 in the range 0 ≤ ≤ . Hence, solve the inequality

|sin 2 | < 1, where 0 ≤ ≤ . [STPM2013]

2

ANSWER 1

Section
1.4

STPM
Past
Year

The range of values of = {1.18 ≤ ≤ 1.57 , 3.92 ≤ ≤ 4.32}
2

y = sin 2x π
π/2

5 7 11
[0 , 12] ∪ (12 , 12) ∪ [ 12 , ]

UNIT SAINS DAN MATEMATIK, SEKTOR PEMBELAJARAN, JABATAN PENDIDIKAN NEGERI PULAU PINANG

100

2.1 Learning Outcome:
SEQUENCES a) Use an explicit formula and a recursive formula for a sequence
b) Find the limit of a convergent sequence

Sequences

Learning outcome: Use an explicit formula and a recursive formula for a sequence.

A sequence is a list of numbers, stated in a particular order, such that each number can be derived
from the previous number according to a certain rule.

Each number of a sequence is called a term.
1, 3, 5, 7, …

This is a sequence of odd numbers. If 1, 2, 3, … represent the terms, then

1 = 1,
2 = 3,
3 = 5,

⋮
= 2 − 1.

The sequence 1, 3, 5, 7, … can be generated using the formula = 2 − 1, where = 1, 2, 3, … .
Since can be expressed in terms of , we say that = 2 − 1, an explicit formula.

The above sequence of odd numbers can also be generated by a recursive formula = −1 + 2
with 1 = 1. To see this, observe the description below:

1 = 1,
2 = 3 = 1 + 2 = 1 + 2,
3 = 5 = 1 + 2 + 2 = 2 + 2,

⋮
= −1 + 2.

As we can see that the next term, can be generated by using the previous term, −1 plus 2.
That is why this type of formula is called recursive formula.

UNIT SAINS DAN MATEMATIK, SEKTOR PEMBELAJARAN, JABATAN PENDIDIKAN NEGERI PULAU PINANG

101

EXAMPLE 1
Write down the first six terms of each of these sequences.
+1 = 2 + 1, 1 = 4

Solution:
Given, 1 = 4,
then 2 = 2 1 + 1 = 2(4) + 1 = 9,

3 = 2 2 + 1 = 2(9) + 1 = 19,
4 = 2 3 + 1 = 2(19) + 1 = 39,
5 = 2 4 + 1 = 2(39) + 1 = 79,
6 = 2 5 + 1 = 2(79) + 1 = 159.
The first six terms of the sequence are 4, 9, 19, 39, 79, 159.

EXAMPLE 2
Write down the first six terms of each of these sequences.
+1 = 2 − 5, 1 = 2

Solution:
Given, 1 = 2,
then 2 = ( 1)2 − 5 = (2)2 − 5 = −1

3 = ( 2)2 − 5 = (−1)2 − 5 = −4
4 = ( 3)2 − 5 = (−4)2 − 5 = 11
5 = ( 4)2 − 5 = (11)2 − 5 = 116
6 = ( 5)2 − 5 = (116)2 − 5 = 13451
The first six terms of the sequence are 2, -1, -4, 11, 116, 13451.

EXAMPLE 3

Write down the first six terms of each of these sequences.

1 1 = 2
+1 = 1 − ,

Solution:

Given, 1 = 2, 1 1 1
1 2 2
then 2 = 1 − = 1 − =

11
3 = 1 − 2 = 1 − 1 = −1
2
11
4 = 1 − 3 = 1 − (−1) = 2
1 11
5 = 1 − 4 = 1 − 2 = 2
11
6 = 1 − 5 = 1 − 1 = −1
2
The first six terms of the sequence are 2, 1 , −1, 2, 1 , −1.
22

UNIT SAINS DAN MATEMATIK, SEKTOR PEMBELAJARAN, JABATAN PENDIDIKAN NEGERI PULAU PINANG

102

WORKSHEET 2.1

Write down the first six terms of each of these sequences.

a) +1 = 1, 1 = 5 b) +1 = 1 − 1, 1 = 3



Solution: Solution:

c) +1 = + −1, 1 = 2, 2 = 3 d) +1 = 2 −1 − 3 , 1 = 4, 2 = 3
Solution: Solution:

e) +1 = 2 + 1, 1 = 4 f) +1 = , 1 = 6, 2 = 3
Solution:
−1

Solution:

UNIT SAINS DAN MATEMATIK, SEKTOR PEMBELAJARAN, JABATAN PENDIDIKAN NEGERI PULAU PINANG

103

Learning outcome: Find the limit of a convergent sequence. Example

The Properties of Limits

Law

lim = lim 7 = 7

→ →4

lim = lim 3 = 23

→ →2

=8

lim ( ) = lim ( ) lim 4(3 2 − 50) = 4 lim(3 2 − 50)
→ → →5 →5

= 4[3(5)2 − 50]

= 100

lim[ ( ) ± ( )] = lim ( ) ± lim ( ) lim(5 + 3 2) = lim 5 + lim 3 2
→ → → →2 →2 →2

= 5(2) + 3(2)2

= 22

lim[ ( ) ∙ ( )] = lim ( ) ∙ lim ( ) lim [ 3(1 + 2)] = lim 3 ∙ lim (1 + 2)
→ → → →−1 →−1 →−1

= (−1)3 ∙ [1 + (−1)2]

= −2

( ) lim ( ) 2 + 3 lim( 2 + 3)
lim ( ) = ( ), − 1 →3
→ lim = − 1) ,
→ lim(
lim →3
→
→3
where lim ( ) ≠ 0
(3)2 + 3
→ = (3) − 1

=6

A sequence, is a convergent sequence if lim = , where is a constant.

→∞

UNIT SAINS DAN MATEMATIK, SEKTOR PEMBELAJARAN, JABATAN PENDIDIKAN NEGERI PULAU PINANG

104

Determine whether the following sequence converges or diverges. For those which are convergent,
determine the limiting value to which they are tending.

EXAMPLE 1 ≥ 1
2 + 1

= 3 − 2 ,

Solution:

2 + 1
lim = lim (3 2)
−
→∞ →∞

= lim 2 + 1
(3 − 2 )
→∞


= lim (2 + 1 )
(3 − 2 )
→∞

lim

→∞

2+0
=3−0

2
=3

Since 1 → 0 and 2 → 0 as → ∞. converges to 23.


EXAMPLE 2
5 − 2

= 2 + 1

Solution:

5 − 2
lim = lim (2 1)
+
→∞ →∞

= lim 5 − 2
( + 1 )
→∞
2

5−0
=2−0

5
=2

Since 2 → 0 and 1 → 0 as → ∞. converges to 52.


UNIT SAINS DAN MATEMATIK, SEKTOR PEMBELAJARAN, JABATAN PENDIDIKAN NEGERI PULAU PINANG

105

EXAMPLE 3
5 2 + 3

= 2 − 2

Solution: TIPS

lim = lim ( 5 2 + 3 ) If the highest power of n in the
numerator and denominator are the
→∞ →∞ 2 − 2 same, divide each term throughout by
this highest power terms as shown.
(5 2222−+ 32 2 2 )
= lim

→∞

= lim (15−+ 23 2)

→∞

5+0
=1−0

=5

Since 3 → 0 and 2 → 0 as → ∞. converges to 5.
2


EXAMPLE 4
1 + 3 × 2

= 5 − 7 × 2

Solution:

1 + 3 × 2
lim = lim (5 2 )
− 7 ×
→∞ →∞

1 + 3 × 2
(25 − 7
= lim 2 )
2 × 2
→∞

2

= lim 1 + 3
(25 − )
→∞
2 7

0+3
=0−7
3
= −7

Since 1 → 0 and 5 → 0 as → ∞. converges to − 3.
2 2
7

UNIT SAINS DAN MATEMATIK, SEKTOR PEMBELAJARAN, JABATAN PENDIDIKAN NEGERI PULAU PINANG

106

WORKSHEET 2.2

Determine whether the following sequence converges or diverges. For those which are convergent,
determine the limiting value to which they are tending.

a) = 2+1 b) = 3 + 1
( +1)
( +2)(2 +3)

Solution: Solution:

c) = (−1) +1 d) = (−1) +1
2 +1

Solution: Solution:

UNIT SAINS DAN MATEMATIK, SEKTOR PEMBELAJARAN, JABATAN PENDIDIKAN NEGERI PULAU PINANG

107

e) = 2−1 f) = 4 + (−0.1)
2+1 Solution:

Solution:

g) = 4 − 3 h) = (−0.1)
Solution: Solution:

UNIT SAINS DAN MATEMATIK, SEKTOR PEMBELAJARAN, JABATAN PENDIDIKAN NEGERI PULAU PINANG

108

EXERCISE 2.1

STPM PAST YEAR

STPM MT 2016 (Section A)

Q1. A convergent sequence is defined by +1 = 1 + 1 and 1 = 1 .
3
(1)
a) Write down each of the terms 2 , 3 and 4 in the form ∑ −=10 , and show that an
3

explicit formula for is given by = 3 [1 − (1) ]. [5 marks]

2 3

b) Determine the limit of as r tends to infinity. [2 marks]

STPM MT 2017 (Section A)

Q2. The sum of the first n terms of a sequence 1 , 2 , 3, … is given by = 3 2 − . Find an

explicit and a recursive formula for . [5 marks]

ANSWER

Section 1(a) 1 = 1

2.1 1
2 = 1 + 3
1 12
Exercise
2.1 3 = 1 + 3 + (3)
1 12 13
STPM 4 = 1 + 3 + (3) + (3)
Past Year

1(b) 3
2

2 Explicit formula:

= 6 − 4 for n = 1, 2, 3, … and 1 = 2
+1 = (6 − 4) + 6
Recursive formula: +1 = + 6

UNIT SAINS DAN MATEMATIK, SEKTOR PEMBELAJARAN, JABATAN PENDIDIKAN NEGERI PULAU PINANG

109

2.2 SERIES Learning Outcome:
a) Use the formulae for the nth term and for the sum of the first n terms of an

arithmetic series and of a geometric series.
b) Identify the condition for the convergence of a geometric series, and use the

formula for the sum of convergent geometric series.

c) Use the method of differences to find the nth partial sum of a series, and deduce the
sum of the series in the case when it is convergent.

Series

Learning outcome: Use the formulae for the nth term and for the sum of the first n terms of an
arithmetic series and of a geometric series.

A series is the sum of the terms of a sequence. Thus
1+3+5+7+9+...
2 + 4 + 6 + 8 + 10 + . . .
0+1+1+2+3+5+...

are examples of series.

The sum of the first n terms of a sequence is written as , where
= 1 + 2 + 3 + ⋯ + .

If the series terminates after a finite number of terms, it is called a finite series. For example,
1 + 3 + 5 + 7 + 9 + 11 + 13 + 15

is a finite series of eight terms.

An infinite series does not stop but continues indefinitely. Hence,
1 + 4 + 9 + 16 + 25 + . . . + 2 + . . .

is an infinite series.

The Greek letter Σ (sigma) can be used to express a series more concisely when the general term
(the nth term) is known. For example, the sum of the first five terms of the sequence

5, 8, 11, 14, . . . , 3n + 2 can be written as follows, using what is called sigma notation, or

summation notation:

5

∑(3 + 2)

=1

The letter r is called the index of summation. It is possible for the index summation to start at a
number other than 1.

An infinite series, for example 3 + 6 + 9 + 12 + 15 + . . .can be written as

∞

∑ 3

=1

UNIT SAINS DAN MATEMATIK, SEKTOR PEMBELAJARAN, JABATAN PENDIDIKAN NEGERI PULAU PINANG

110

Express each of the following series by using summation notation Σ.

EXAMPLE 1
1 + 4 + 9 + 16 + . . . + 100
Solution:
1 + 4 + 9 + 16 + . . . + 100
= 12 + 22 + 33 + 42+ . . . +102

∴ the ℎ term, = 2
Hence, the series may be written as ∑1 =0 1 2

EXAMPLE 2
1 + 3 + 9 + 27 + 81 + . . .

Solution:
1 + 3 + 9 + 27 + 81+. . .
= 30 + 31 + 32 + 33 + 34 + . . .

∴ the ℎ term, = 3 OR = 3 −1

∞ ∞

Hence, the series may be written as ∑ 3 ∑ 3 −1

=0 =1

(Notice that this is an infinite series. This is a sum of powers of 3. The symbol ∞ is used represent
infinity)

EXAMPLE 3
+ 2 + 3 + 4 + 5 + 6

Solution:

6

+ 2 + 3 + 4 + 5 + 6 = ∑

=1

.
EXAMPLE 4
2 − 4 + 6 − 8 + 10

Solution:
These are even integers with alternating sign. Therefore, the given series can be written as

5

∑(−1) +1(2 )

=1

.

UNIT SAINS DAN MATEMATIK, SEKTOR PEMBELAJARAN, JABATAN PENDIDIKAN NEGERI PULAU PINANG

111

WORKSHEET 2.3

Express each of the following series by using summation notation Σ.

a) 1 ∙ 2 + 2 ∙ 3 + 3 ∙ 4+ . . . +19 ∙ 20 b) 1 + 1 + 1 +. . . 1
Solution: 234 50

Solution:

c) −3 + 4 − 5 + 6 d) 8 + 9 + 10 + ⋯ +
Solution: 82+1 92+1 102+1 2+1

Solution:

e) + + +. . . + f) 9 − 27 + 81 − 243 + 729 − 2187 + 6561
× × × ( + ) Solution:

Solution:

UNIT SAINS DAN MATEMATIK, SEKTOR PEMBELAJARAN, JABATAN PENDIDIKAN NEGERI PULAU PINANG

112

Write down all the terms in each of these series and evaluate each sum.

EXAMPLE 1

5

∑(−1) 2 +1

=1

Solution:
∑5 =1(−1) 2 +1 = (−1)122 + (−1)223 + (−1)324 + (−1)4(2)5 + (−1)5(2)6

= −4 + 8 − 16 + 32 − 64
= −44

∴ −4 + 8 − 16 + 32 − 64 ; ∑5 =1(−1) 2 +1 = −44

EXAMPLE 2
3 2
∑ 2 + 1

=0

Solution:

3 2 2(0) 2(1) 2(2) 2(3)
∑ 2 + 1 = 20 + 1 + 21 + 1 + 22 + 1 + 23 + 1
=0

246
=0+3+5+9

2
= 2 15

∴ 0 + 2 + 4 + 6 ; ∑ 3 =0 2 = 2 2
3 5 9 2 +1 15

EXAMPLE 3

6

∑4

=1

Solution:

6

∑ 4 = 4 + 4 + 4 + 4 + 4 + 4 = 24

=1

∴ 4 + 4 + 4 + 4 + 4 + 4 ; ∑ 6 =1 4 = 24

UNIT SAINS DAN MATEMATIK, SEKTOR PEMBELAJARAN, JABATAN PENDIDIKAN NEGERI PULAU PINANG

113

WORKSHEET 2.4

Write down all the terms in each of these series and evaluate each sum.

a) ∑ 7 =3(−2) −1 b) ∑ 3 =0 2
Solution: 2 +1

Solution:

c) ∑ 5 =1(2 − 7) d) ∑6 =3 −5
Solution: −1

e) ∑04( − 1)( − 3) Solution:
Solution:
f) ∑6 =2(3 + 2)2
Solution:

UNIT SAINS DAN MATEMATIK, SEKTOR PEMBELAJARAN, JABATAN PENDIDIKAN NEGERI PULAU PINANG

114

Arithmetic Progression (A.P.)

An arithmetic progression (A.P.) is a sequence of numbers in which any term differs from the
previous term by a certain number called common difference.

• The common difference can be either positive or negative.
• The common difference of an A.P. is denoted by d.
• The first term of an A.P. is denoted by a.

Consider the sequence of numbers

2, 6, 10, 14, 18, . . .

This in an example of arithmetic progression where the common difference is 4.

Note that: 1 =
+1 = +

= +1 −

General Term of an Arithmetic Progression

Generally, the terms of an A.P. take the form

, + , + 2 , + 3 , . . .

and the nth term is given by

= + ( − 1)
Sum of the first n terms of an Arithmetic Progression

The sum of the first n terms of an A.P. is given by

= [2 + ( − 1) ] or = [ + ]
2 2

where is the last term.

Arithmetic Mean

If p, m and q are in an A.P., the common difference is

= − = −
2 = +

+
= 2

m is called the arithmetic mean of p and q.

UNIT SAINS DAN MATEMATIK, SEKTOR PEMBELAJARAN, JABATAN PENDIDIKAN NEGERI PULAU PINANG

115

For each of the following arithmetic series, write down the term indicated in brackets.

EXAMPLE 1
2, 6, 10, . . ., (12th term)

Solution:
1 = = 2
= 2 − = 6 − 2 = 4

The nth term is given by
= + ( − 1)

∴ The 12th term is
12 = 2 + (12 − 1)(4) = 46

EXAMPLE 2
−7 − 5 − 3−. . . (20th term)

Solution:
1 = = −7
= −5 − (−7) = 2

The nth term is given by
= + ( − 1)

∴ The 20th term is
20 = −7 + (20 − 1)(2) = 31

EXAMPLE 3

1 + 1 3 + 2 1 + ⋯ (11th term)
5 5

Solution:
1 = = 1

33
= 1 5 − 1 = 5

The nth term is given by
= + ( − 1)

∴ The 11th term is
3

11 = 1 + (11 − 1) (5) = 7

UNIT SAINS DAN MATEMATIK, SEKTOR PEMBELAJARAN, JABATAN PENDIDIKAN NEGERI PULAU PINANG

116

WORKSHEET 2.5

For each of the following arithmetic series, write down the term indicated in brackets.

a) 7, 4, 1, … , (18 ℎ term) b) − 1, 3 + 2, 5 + 5, . . . , (9 ℎ term)

Solution: Solution:

c) 1 , 2 , 7 , . . . , (15 ℎ term) d) 3 + 3 2 + 4 1 + ⋯ (15 ℎ term)

636 33

Solution: Solution:

e) −22 − 17 − 12 − ⋯ − (15 ℎ term) f) 4 + 7 1 + 11 + ⋯ + (20 ℎterm)
Solution:
2

Solution:

UNIT SAINS DAN MATEMATIK, SEKTOR PEMBELAJARAN, JABATAN PENDIDIKAN NEGERI PULAU PINANG

117

Find the number of terms in each of the following A.P.s

EXAMPLE 1
6,10, 14, … , 102

Solution:
1 = = 6
= 2 − = 10 − 6 = 4

The nth term is given by
= + ( − 1)
102 = 6 + ( − 1)4
24 = − 1
= 25

EXAMPLE 2
2 , 2 + 5, 2 + 10, . . . , 7

Solution:
1 = = 2
= 2 − = (2 + 5) − 2 = 5

The nth term is given by
= + ( − 1)
7 = 2 + ( − 1)5
5 = ( − 1)5
= − 1
= + 1

EXAMPLE 3
Find the sum of the arithmetic series 5 + 9 + 13 + ⋯ + 81

Solution:
1 = = 5
= 2 − = 9 − 5 = 4

The nth term is given by
= + ( − 1)
81 = 5 + ( − 1)4
19 = − 1
= 20

The sum of the first n terms of an A.P. is given by

= [ + ]
2
20
∴ 20 = 2 [(5) + 81]

= 860

UNIT SAINS DAN MATEMATIK, SEKTOR PEMBELAJARAN, JABATAN PENDIDIKAN NEGERI PULAU PINANG

118

EXAMPLE 4
Find the sum of the arithmetic series −15 − 9 − 3 − ⋯ to 10 ℎ term

Solution:
Method 1
1 = = −15
= 2 − = −9 − (−15) = 6

The 10th term is given by
= + ( − 1)
10 = −15 + (10 − 1)6

= 39

The sum of the first n terms of an A.P. is given by

= [ + ]
2
10
∴ 10 = 2 [(−15) + 39]

= 120

Method 2

1 = = −15
= 2 − = −9 − (−15) = 6

The sum of the first n terms of an A.P. is given by

= [2 + ( − 1) ]
2
10
10 = 2 [2(−15) + (10 − 1)6]
= 120

EXAMPLE 5
In an A.P., the 6th term is 67 and 14th term is 163. Find the common difference, the first term and the

nth term.

Solution:
Give that 6 = 67 and 14 = 163

6 = 67 ①
+ 5 = 67

14 = 163 ②
+ 13 = 163

②-①
8 = 96
= 12

UNIT SAINS DAN MATEMATIK, SEKTOR PEMBELAJARAN, JABATAN PENDIDIKAN NEGERI PULAU PINANG

119

Substitute = 12 into ①
+ 5(12) = 67

= 67 − 60
=7

The nth term is given by
= + ( − 1)
= 7 + ( − 1)12
= 12 − 5

EXAMPLE 6
The first term of an A.P. is 3. The sum of the first eight terms is 136 and the sum of the entire series is
820. Determine the common difference, the number of terms and the last term.

Solution:

= 3, 8 = 136 and = 820
8 = 2 [2 + ( − 1) ]
8
136 = 2 [2(3) + (8 − 1) ]

= 4

Since the sum of the first n terms is 820,
[2(3) + ( − 1)(4)] = 820

2

3 + 2 ( − 1) = 820
2 2 + − 820 = 0

(2 + 41)( − 20) = 0

= 20 [ cannot be negative]

The last term,

= + ( − 1) OR 20 = 20 (3 + )
2

= 3 + (20 − 1)(4) 820 = 20 (3 + )

2

= 79 = 79

UNIT SAINS DAN MATEMATIK, SEKTOR PEMBELAJARAN, JABATAN PENDIDIKAN NEGERI PULAU PINANG

120

WORKSHEET 2.6

Q1. Find the number of terms in each of the following A.P.s.

a) 7, 4, 1, . . . , −98 b) + 2 , , − 2 , . . . , − 40
Solution: Solution:

c) 5, 9, 13, . . . , 81 d) 1.3, 1.6, 1.9, … , 6.7
Solution: Solution:

UNIT SAINS DAN MATEMATIK, SEKTOR PEMBELAJARAN, JABATAN PENDIDIKAN NEGERI PULAU PINANG

121

Q2. Find the sum of all terms indicated in each of the following A.P.s.

a) 85 + 82 + 79+. . . +13 b) 1 + 2 + 5 + ⋯ + 20 ℎ term
Solution:
12 3 4

Solution:

c) ( − 4 ) + (2 − 6 ) + (3 − 8 ) d) −10 − 8 − 6 − ⋯ + 11 ℎterm
+ . . . + 40 ℎ term Solution:

Solution:

UNIT SAINS DAN MATEMATIK, SEKTOR PEMBELAJARAN, JABATAN PENDIDIKAN NEGERI PULAU PINANG

122
EXERCISE 2.2

1. Given that the fifth term of an arithmetic progression is 21 and its tenth term is 41, find the
common difference, first term and the nth term of this arithmetic progression.

2. The nth term of an arithmetic progression is 1 (5 − 3). Obtain the first two terms and also the
10

common difference of the arithmetic progression.

3. The sum of the first n terms of series is 2 2 − . Find
a) the first term,
b) the sixth term, and
c) the nth term.
Show that the series is an A.P. .

4. Find the sum of the integers from 1 to 200 which are not divisible by 5.

5. An arithmetic series has first term and common difference . The sum of the first 21 terms
is 168.
a) Show that + 10 = 8.
b) The sum of the second term and the third term is 50. The th term of the series is .
i) Find the value of 12 .
ii) Find the value of ∑2 1=4 .

6. The sum of the first four terms is 8 and the sum of the first six term is 24. If the sum of the first
nth term is given by = 2 + , determine
a) the values of p and q.
b) the nth term, .

STPM PAST YEAR

STPM MM 2017 (Section A)
Q1 Three non-zero numbers, a, b and c, are in the ratio of 4 : 3 : 2. If a, b and (c ̶ 3) are the first,

third and seventh terms of an arithmetic sequence respectively, find the smallest integer n such

that the sum of the first n terms is less than zero. [7 marks]

UNIT SAINS DAN MATEMATIK, SEKTOR PEMBELAJARAN, JABATAN PENDIDIKAN NEGERI PULAU PINANG

123 1 = 4
= 5
ANSWER The ℎ term is = 1 + 4
Section
2.2

Exercise
2.2

2 The 1 term is 1 = 1
5
7
The 2 term is 2 = 10

= 1
2

3(a) = 1

3(b) 6 = 21

3(c) − −1 = (4 − 3) − (4 − 7)
=4

The series is an A.P.

4 Therefore, the sum of the integers from 1 to 200 which is not divisible by 5
= 16000

5(a)

5(b)(i) ∴ 12 = 6

5(b)(ii) 21

∑ = 90

=4

6(a) = 1; = −2
(b) = 2 − 3

ANSWER 1 ⸫ n = 18
Section
2.2

STPM
Pass Year

UNIT SAINS DAN MATEMATIK, SEKTOR PEMBELAJARAN, JABATAN PENDIDIKAN NEGERI PULAU PINANG

124

Geometric Progression (G.P.)

A geometric progression (G.P.) is a sequence of numbers in which any term can be obtained from the
previous term by multiplying by a certain number called the common ratio.

• The common ratio is denoted by r.
• The first term of an G.P. is denoted by a.

Consider the sequence of numbers

2, 6, 18, 54. . .

This in an example of geometric progression where the common ratio is 3.

Note that:

1 = = 2
2 = (3) = 6
3 = (32) = 18

⋮
= ( −1)

General Term of a Geometric Progression

Generally, the terms of an G.P. take the form
, , 2, 3, . . .

and the nth term is given by
= −1

Sum of the First n terms of a Geometric Progression

The sum of the first n terms of an G.P. is given by

= (1− ) , for < 1 or = ( −1) , for > 1

1− −1

Geometric Mean

If p, m and q are in a G.P., the common ratio is


= =
2 =
= ±√

m is called the geometric mean of p and q.

UNIT SAINS DAN MATEMATIK, SEKTOR PEMBELAJARAN, JABATAN PENDIDIKAN NEGERI PULAU PINANG

125

Find the term indicated in the squared brackets in each of the following G.P.s.

EXAMPLE 1
2, 4, 8, … [ 10]

Solution:
1 = = 2 ,

= 2 = 4 = 2 or = 3 = 8 = 2
1 2 2 4

10 = 1( 10−1)
= 2(29)

= 1024

EXAMPLE 2
+ 3, + 7, 4 − 2, … [ 2 ]

Solution:
+ 7 2 −1

2 = ( + 3) ( + 3)
( + 7)2 −1

= ( + 3)2 −2

If student solve for , the values of are = 5, or = − 131. Then,

(12)2 −1
2 = (8)2 −2

16 9
= 3 (4)

or

2 = (130)2 −1
(− 23)2 −2

= 2 (25)
15

UNIT SAINS DAN MATEMATIK, SEKTOR PEMBELAJARAN, JABATAN PENDIDIKAN NEGERI PULAU PINANG

126

WORKSHEET 2.7

Find the term indicated in the squared brackets in each of the following G.P.s.

a) 1, −3, 9, … [ 7] b) 1, 1,1,… [ 12]
Solution:
24

Solution:

c) 64, −16, 4, … [ 20] d) 1 , 1 , 1 , … [ ]
Solution: 2

Solution:

e) 0.01, 0.02, 0.04, … , [16 ℎ term] f) 1, − 1 , 1 , … [6 ℎ term]
Solution:
39

Solution:

UNIT SAINS DAN MATEMATIK, SEKTOR PEMBELAJARAN, JABATAN PENDIDIKAN NEGERI PULAU PINANG

127

Find the sum of the term up to the term indicated in the squared brackets in each of the following
G.P.s.

EXAMPLE 1
2 + 4 + 8 + … [ 10]

Solution:
1 = = 2 ,

= 2 = 4 = 2 > 1
1 2

So,

10 − 1
10 = ( − 1 )

210 − 1
= 2( 2−1 )
= 2046

EXAMPLE 2 [ ] where 0 < < 1
11 1
, , 2 , …

Solution:

1 = = 1 ,


2 1 1
1
= = = > 1
1



So,

− 1
= ( − 1 )

= 1 [( 1 1 )− −1 1
]

1 1 −
= −1 [(1 − )]

UNIT SAINS DAN MATEMATIK, SEKTOR PEMBELAJARAN, JABATAN PENDIDIKAN NEGERI PULAU PINANG

128

WORKSHEET 2.8

Find the sum of the term up to the term indicated in the squared brackets in each of the following
G.P.s.

a) 1+ 1+ 1,… [ 12] b) 1 − 3 + 9 − … [ 7]
Solution:
24

Solution:

c) 1 − 1 + 1 − … [ 6] d) + 3, + 7, 4 − 2, … [ 2 ]
where < −3
39
Solution:
Solution:

UNIT SAINS DAN MATEMATIK, SEKTOR PEMBELAJARAN, JABATAN PENDIDIKAN NEGERI PULAU PINANG

129

Infinite Geometric Progression

Learning Outcome: Identify the condition for the convergence of a geometric series, and use the
formula for the sum of convergent geometric series.

The sum to infinity of a geometric progression (G.P.) is given by

∞ = ∞ −1 = 1
−
∑

=1

where −1 < < 1 or | | < 1.

Note that: When → ∞, → 0. So,

( (1 − 1−0
∞ = lim = lim )) = (1 ) =
1 − − 1 −
→∞ →∞

Express each of the following recurring decimals as a fraction or mixed number in its simplest form.

EXAMPLE 1 EXAMPLE 2

0. 3̇ 2̇ 6. 7̇

Solution: Solution:
0. 3̇ 2̇ = 0.323232 … 6. 7̇ = 6 + 0.7777 …

0. 3̇ 2̇ = 0.32 + 0.0032 + 0.000032 + ⋯ 6. 7̇ = 6 + 7 + 7 + 7 + ⋯
10 100 000
0. 3̇ 2̇ = 32 + 32 + 32 + ⋯ 1
100 10 000 000 000
1 = 6 + ∞

with = 32 and = 1 with = 7 and = 1
100 100 10 10

∞ = (13020) = 32 ∞ = (170) = 7
(1 − 1010) 99 (1 − 110) 9

⸫ 0. 3̇ 2̇ = 32 ⸫ 6. 7̇ = 6 + 7

99 9

7
= 69

UNIT SAINS DAN MATEMATIK, SEKTOR PEMBELAJARAN, JABATAN PENDIDIKAN NEGERI PULAU PINANG

130

WORKSHEET 2.9

Express each of the following recurring decimals as a fraction or mixed number in its simplest form.

a) 0. 5̇ b) 0. 7̇ 2̇

Solution: Solution:

c) 3.03̇ d) 3.28̇ 1̇ 4̇
Solution: Solution:

UNIT SAINS DAN MATEMATIK, SEKTOR PEMBELAJARAN, JABATAN PENDIDIKAN NEGERI PULAU PINANG

131

EXAMPLE 1
If x, 6 and x + 5 are three successive terms of a G.P. , find the possible values of x and the
corresponding values of the common ratio.
Given that x, 6 and x + 5 are the third, fourth and fifth terms of a G.P. and that the sum to infinity of
the series exists, determine the first term and the sum to infinity.

Solution:
6 + 5
= 6

2 + 5 − 36 = 0
( + 9)( − 4) = 0

x = − 9 or 4

For = −9, = − 2

3

For = 4, = 3

2

Given the sum to infinity of the series exists,

therefore = − 2 , since | | < 1.
3

Given that x is the third term,
3 = 2 = −9

9
= − (− 23)2

81
=− 4
= −20.25


∞ = 1 −

20.25
∞ = 1 − (− 23)

= −12.15

For the G.P., the first term is −20.25 and the sum to infinity is −12.15.

EXAMPLE 2

The th term, , of an infinite series is given by
1 2 +1 1 2 −1
= (3) + (3)

a) Express in the form 32 +1 , where is a constant.

b) Find the sum of the first terms of the series, and deduce the sum of the infinite series.

UNIT SAINS DAN MATEMATIK, SEKTOR PEMBELAJARAN, JABATAN PENDIDIKAN NEGERI PULAU PINANG

132

Solution:

a) = (1)2 +1 + (1)2 −1

3 3

= +1 1
32 −1
32 +1

= +1 1
32 +1−2
32 +1

= +1 1
32 +1∙3−2
32 +1

= +1 32
32 +1
32 +1

= 10
32 +1

b) 1 = 10 = 10 , 2 = 10 , 3 = 10
33 27 35 37

2 = 1 , 3 = 1

1 9 2 9

the series is a geometric series with = 10 and common ratio, = 19.
27

Sum of the first terms,

= (1− )
1−

= 1270[1−(19) ]

1−19
= 5 [1 − (1) ]

12 9

∞ = lim

→∞
= lim 5 [1 − (1) ]
→∞ 12 9

= 5 as (1) → 0 as → ∞
12 9

EXERCISE 2.3

1. The first three terms of a geometric series are 2, − 1 , 1 . Find the sum to infinity of this series.
2 8

Find also the smallest value of n such that the difference between the sum of the first n terms and

the sum to infinity is less than 10-5.

2. The th term, , of an infinite series is given by

2 +1 2 −1
= (5) − (5)

(a) Express in the form (2) +1 , where is a constant.

5

(b) Find the sum of the first terms of the series, and deduce the sum of the infinite series.

UNIT SAINS DAN MATEMATIK, SEKTOR PEMBELAJARAN, JABATAN PENDIDIKAN NEGERI PULAU PINANG

133

STPM MM 2016 (Section A)

Q1 Show that ∑ ∞ =1 (1) (1) −2 is a convergent series. Give a reason for your answer. [3 marks]

2 3

Hence, determine the sum of the convergent series. [2 marks]

STPM MT 2018 (Section A)

Q2 The sum of the first n terms of a sequence 1 , 2 , 3 , … is given by = 1 − 3−2 .
a) Show that = 8(3−2 ) .
[3 marks]

b) Express +1 in terms of , and deduce that the sequence is a geometric sequence.

[3 marks]

c) Find the sum of the infinite series 1 , 2 , 3 , … . [3 marks]

ANSWER 1 8
Section ∞ = 5
2.2
⸫n=9
Exercise
2.3

2(a) 21 2 +1
− 4 (5)

2(b) 7 2
= − 5 [1 − (5) ]

∞ = − 7 as (2) → 0 as → ∞
5
5

ANSWER 1 (12) (13) −2
Section −1 (12) −1 (31) −3
2.2 =

Exercise = (1) (1) = 1
2.3 23 6

STPM 9
Past Year ∞ = 5

2(a) +1 1 1
(b) 9 9

= =

(c) 1 + 2 + 3 + …

=1

UNIT SAINS DAN MATEMATIK, SEKTOR PEMBELAJARAN, JABATAN PENDIDIKAN NEGERI PULAU PINANG

134

Summation of Other Series

The types of series considered here may be classified as follow:
1. Summation of series using the natural number of series
2. Summation of series using the method of differences

Summation of Finite Series Involving Powers of Integers

Consider the series of the first positive integers
1 + 2 + 3 + ⋯ + ( − 1) + .

This series can also be written as



∑ = 1 + 2 + 3 + ⋯ + ( − 1) + … . ①

=1

By rewriting this series backwards term by term,



∑ = + ( − 1) + ( − 2) + ⋯ + 2 + 1 … . . ②

=1



① + ② ; 2 ∑ = ( + 1) + ( + 1) + ( + 1) + ⋯ + ( + 1) + ( + 1)

=1

= ( + 1) ←← ℎ ℎ

1
∑ = 2 ( + 1)

=1

Now, consider the series of the squares of the first positive integers, i.e



∑ 2 = 12 + 22 + 32 + ⋯ + ( − 1)2 + 2.

=1

By using the identity

( + 1)3 ≡ 3 + 3 2 + 3 + 1,

We have ( + 1)3 − 3 = 3 2 + 3 + 1.

By adding the terms of the identity one by one with values of from = to = 1, we get



∑[( + 1)3 − 3] = 3 ∑ 2 + 3 ∑ + ∑ 1

=1 =1 =1 =1

The LHS of this identity can be written as
[( + 1)3 − 3] + [ 3 − ( − 1)3] + ⋯ + (33 − 23) + (23 − 13) = ( + 1)3 − 1

Thus,

( + 1)3 − 1 = 3 ∑ 2 + 3 ∑ + ∑ 1

=1 =1 =1

UNIT SAINS DAN MATEMATIK, SEKTOR PEMBELAJARAN, JABATAN PENDIDIKAN NEGERI PULAU PINANG

135



3 ∑ 2 = ( + 1)3 − 1 − 3 ∑ − ∑ 1

=1 =1 =1

= ( + 1)3 − 1 − 3 ( + 1) −
2

= ( + 1)3 − ( + 1) − 3 ( + 1)
2

= 1 ( + 1)[2( + 1)2 − 2 − 3 ]
2

= 1 ( + 1)(2 2 + 4 + 2 − 2 − 3 )
2

= 1 ( + 1)(2 2 + )
2

= 1 ( )( + 1)(2 + 1)
2

Thus,

∑ 2 = 1 ( + 1)(2 + 1)
6
=1

Similarly, by using the identity
( + 1)4 − 4 = 4 3 + 6 2 + 4 + 1

we get

∑ 3 = 1 2( + 1)2
4
=1
12
= [2 ( + 1)]

2

= (∑ )

=1

Natural Numbers Series.

Natural numbers are 1, 2, 3, 4, 5,….

1. Sum of the first n natural numbers.

Sum of the first n natural numbers is the series
1 + 2 + 3 + 4 + 5 + … + n and can be written in form ∑ =1 .
This series is given by


∑ = 2 ( + 1)

=1

UNIT SAINS DAN MATEMATIK, SEKTOR PEMBELAJARAN, JABATAN PENDIDIKAN NEGERI PULAU PINANG

136

2. Sum of the squares of the first n natural numbers.

Sum of the squares of the first n natural numbers is the series
12 + 22 + 32 + 42 + 52 + ⋯ + 2 and can be written in form ∑ =1 2.

This series is given by

= ( + 1)(2 + 1)
6
∑ 2

=1

3. Sum of the cubes of the first n natural numbers.

Sum of the cubes of the first n natural numbers is the series
13 + 23 + 33 + 43 + 53 + ⋯ + 3 and can be written in form ∑ =1 3.

This series is given by

= 2 ( + 1)2
4
∑ 3

=1

EXAMPLE 1

Sum to n terms the following series:
a) 12 + 42 + 72 + 102 + ⋯

b) 2 × 3 + 3 × 4 + 4 × 5 + ⋯

c) 1 × 2 × 4 + 2 × 3 × 5 + 3 × 4 × 6 + ⋯
d) 1 × 22 + 3 × 32 + 5 × 42 + 7 × 52 + ⋯

Solution:
) 12 + 42 + 72 + 102 + ⋯

= ∑ =1(3 − 2)2

∑ =1(3 − 2)2 = ∑ =1(9 2 − 12 + 4)
= 9 ∑ =1 2 − 12 ∑ =1 + ∑ =1 4
= 9 × ( + 1)(2 + 1) − 12 × ( + 1) + 4

62

= 3 3 + 9 2 + 3 − 6 2 − 6 + 4

22

= 3 3 − 3 2 − 1

22

b) 2 × 3 + 3 × 4 + 4 × 5 + ⋯
= [(1 + 1)(1 + 2)] + [(2 + 1)(2 + 2)] + ⋯
= ∑ =1( + 1)( + 2)

∑ =1( + 1)( + 2) = ∑ =1( 2 + 3 + 2)
= ∑ =1 2 + 3 ∑ =1 + ∑ =1 2
= ( + 1)(2 + 1) + 3 × ( + 1) + 2

62

= 1 3 + 1 2 + 1 + 3 2 + 3 + 2

3 2 62 2

UNIT SAINS DAN MATEMATIK, SEKTOR PEMBELAJARAN, JABATAN PENDIDIKAN NEGERI PULAU PINANG

137

= 1 3 + 2 2 + 13

33

= 1 ( 3 + 6 2 + 13 )

3

c) 1 × 2 × 4 + 2 × 3 × 5 + 3 × 4 × 6 + ⋯

= (1 × 2 × 4) + (2 × 3 × 5) + (3 × 4 × 6) + ⋯

= 1(1 + 1)(1 + 3) + 2(2 + 1)(2 + 3) + ⋯
= ∑ =1 ( + 1)( + 3)

∑ =1 ( + 1)( + 3) = ∑ =1( 3 + 4 2 + 3 )

= ∑ =1 3 + 4 ∑ =1 2 + 3 ∑ =1

= 2 ( + 1)2 + 4 ( ( + 1)(2 + 1)) + 3 × ( + 1)
46 2

= ( +1) [ ( + 1) + 8 (2 + 1) + 6]

43

= ( +1) ( 2 + 19 + 26)

4 33

= 1 ( + 1)(3 2 + 19 + 26)

12

= 1 ( + 1)( + 2)(3 + 13)

12

d) 1 × 22 + 3 × 32 + 5 × 42 + 7 × 52 + ⋯
= [(2(1) − 1) × (1 + 1)2] + [2(2) − 1) × (2 + 1)2] + [(2(3) − 1) × (3 + 1)2] + ⋯
= ∑ =1(2 − 1)( + 1)2

∑ =1(2 − 1)( + 1)2 = ∑ =1(2 3 + 3 2 − 1)
= 2 ∑ =1 3 + 3 ∑ =1 2 − ∑ =1 1
= 2 × 2 ( + 1)2 + 3 × ( + 1)(2 + 1) −

46

= [ ( + 1)2 + ( + 1)(2 + 1) − 2]

2

= ( 3 + 4 2 + 4 − 1)

2

EXAMPLE 2
Using the standard results for ∑ =1 and ∑ =1 2, find the expression in terms of n for the sum of the
series 32 + 52 + 72 + 92 + ⋯ to n terms.
Simplifying your answer as far as possible. Hence, find the value of 32 + 52 + 72 + 92 + ⋯ + 812
and deduce the value of 412 + 432 + 452 + 472 + ⋯ + 812

Solution:
32 + 52 + 72 + 92 + ⋯ = ∑ =1(2 + 1)2

= ∑ =1(4 2 + 4 + 1)
= 4 ∑ =1 2 + 4 ∑ =1 + ∑ =1 1
= 4 × ( + 1)(2 + 1) + 4 × ( + 1) +

62

= [2 ( + 1)(2 + 1) + 2( + 1) + 1]

3

= (4 2 + 2 + 2 + 2 + 3)

33

UNIT SAINS DAN MATEMATIK, SEKTOR PEMBELAJARAN, JABATAN PENDIDIKAN NEGERI PULAU PINANG

138

= (4 2 + 4 + 11)

33

= (4 2 + 12 + 11)

3

Consider the A.P. 3 + 5 + 7 + 9 + ⋯ + 81,
= + ( − 1)
81 = 3 + ( − 1)2
= 40

Therefore, 32 + 52 + 72 + 92 + ⋯ + 812 = ∑4 =0 1(2 + 1)2
= 40 [4(40)2 + 12(40) + 11]

3

= 91 880

412 + 432 + 452 + 472 + ⋯ + 812
= (32 + 52 + 72 + 92 + ⋯ + 812) − (32 + 52 + 72 + 92 + ⋯ + 392)
= ∑ 4 =0 1(2 + 1)2 − ∑ 1 =9 1(2 + 1)2
= 91 880 − 10 659
= 81 221

WORKSHEET 2.10 b) ∑ =1 ( + 3)( + 5)
Solution:
Q1. Find
a) ∑ =1( + 2)( + 3)

Solution:

UNIT SAINS DAN MATEMATIK, SEKTOR PEMBELAJARAN, JABATAN PENDIDIKAN NEGERI PULAU PINANG

139 b) ∑ 1 =6 8 ( + 1)( + 4)
Solution:
Q2. Evaluate
a) ∑1 =01 (2 2 + 3)
Solution:

Q3. Evaluate each of the following series. b) 1 × 2 + 4 × 3 + 9 × 4 + ⋯ + 169 × 14
a) 93 + 103 + 113 + … + 203 Solution:

Solution:

UNIT SAINS DAN MATEMATIK, SEKTOR PEMBELAJARAN, JABATAN PENDIDIKAN NEGERI PULAU PINANG

140 d) 1 × 2 × 4 + 2 × 3 × 6 + 3 × 4 × 8 + ⋯
c) 2 × 3 + 3 × 5 + 4 × 7 + ⋯ + 11 × 21 +15 × 16 × 32
Solution:
Solution:

Q4. Find expressions in terms of n for the sum of the following series, to the number of terms stated,

simplifying the answers as far as possible.

a) 11 + 12 + 13 + 14 + ⋯ to 2 terms b) 23 + 43 + 63 + 83 + ⋯ to n terms

Solution: Solution:

UNIT SAINS DAN MATEMATIK, SEKTOR PEMBELAJARAN, JABATAN PENDIDIKAN NEGERI PULAU PINANG

141 d) 1 × 4 × 7 + 2 × 5 × 8 + 3 × 6 × 9
+ 4 × 7 × 10 + ⋯ to n terms
c) 1 × 22 + 2 × 32 + 3 × 42 + 4 × 52
+ ⋯ to n terms Solution:

Solution:

Method of Differences Other variations on the

Use the method of differences to find the nth partial sum of a series, formula are :
and deduce the sum of the series in the case when it is convergent.
∑[ ( ) − ( + 1)] =
The method of differences can be used to determine the sum of certain (1) − ( + 1)
series. If a series, Ur . has a function, f(r) such that ur = f(r+1) – f(r) for
every r then r=1, f(1) – f(2)
r=2, f(2) – f(3)
r=3, f(3) – f(4)

∑ = ∑[ ( + 1) − ( )] = ( + 1) − (1) ⁞
r=n-1, f(n-1) – f(n)
=1 =1 r=n, f(n) – f(n+1)

because all the middle terms cancel each other out. *******************
Find such a function can be difficult, and is not always possible. ∑[ ( ) − ( − 1)] =

( ) − (0)
r=1, f(1) – f(0)
r=2, f(2) – f(1)
r=3, f(3) – f(2)

⁞
r=n-1, f(n-1) – f(n-2)
r=n, f(n) – f(n-1)

UNIT SAINS DAN MATEMATIK, SEKTOR PEMBELAJARAN, JABATAN PENDIDIKAN NEGERI PULAU PINANG

142

EXAMPLE 1

Find ∑ =1 [1 − +1 1].



Solution:

∑ =1 [1 − 1] = 1 − 1
1 2
+1

+1 − 1

23

+1 − 1

34

⋮⋮

+1− 1

+1

=1− 1

+1


= + 1

EXAMPLE 2

Express 2 in partial fractions.
(2 −1)(2 +1)

Show that ∑ =1 2 = 2 and state whether or not the series is convergent.
(2 −1)(2 +1) 2 +1

Solution:

Let 2 = +
(2 −1)(2 +1) 2 −1 2 +1

2 = (2 + 1) + (2 − 1)

Let = 1 , 2 = 2 ⇒ = 1
2
1
Let = − 2 , 2 = −2 ⇒ = −1

Therefore, 2 = 1 − 1
(2 −1)(2 +1) 2 −1 2 +1

Method 1

∑ =1 2 = ∑ =1 (1 − 1)
(2 −1)(2 +1)
2 −1 2 +1

=1−1

13

+1 − 1

35

+1 − 1

57

⋮⋮

+1 − 1

2 −1 2 +1

=1− 1

2 +1

= 2 [shown]

2 +1

UNIT SAINS DAN MATEMATIK, SEKTOR PEMBELAJARAN, JABATAN PENDIDIKAN NEGERI PULAU PINANG

143

Method 2

Let ( ) = 1, then ( + 1) = 1 = 1
2( +1)−1 2 +1
2 −1

2

∑ (2 − 1)(2 + 1) = ∑[ ( ) − ( + 1)]
=1 =1

= (1) − ( + 1) [method of differences]

=1− 1

2 +1

= 2 [shown]

2 +1

As ⟶ ∞, 1 → 0. The summation approaches 1. Hence, the series is convergent with ∞ = 1.
2 +1

EXAMPLE 3

Express +4 in partial fractions.
( +1)( +2)

Hence, find the sum of the series

5 + 6 + 7 + ⋯ + +4
1×2×3 2×3×4 3×4×5 ( +1)( +2)

Show that the series is convergent and state the sum to infinity.

Solution:

Let +4 = + +
( +1)( +2) +1 +2

+ 4 = ( + 1)( + 2) + ( )( + 2) + ( )( + 1)

Let = 0, 4 = 2 ⇒ = 2

Let = −1, 3 = − ⇒ = −3

Let = −2, 2 = 2 ⇒ = 1

Therefore, +4 = 2 − 3 + 1
( +1)( +2) +1 +2

Method 1

∑ =1 +4 = ∑ =1 [2 − 3 + 1]
( +1)( +2) +1
+2

=2−3+1

123

+2 − 3 + 1

234

+2 − 3 + 1

345

⋮⋮⋮

+ 2 −3+ 1

−1 +1

+2− 3 + 1

+1 +2

=2−3+1+ 1 − 3 + 1

2 +1 +1 +2

=3− 2 + 1

2 +1 +2

= 3 − 2( +2)−( +1)
2 ( +1)( +2)

UNIT SAINS DAN MATEMATIK, SEKTOR PEMBELAJARAN, JABATAN PENDIDIKAN NEGERI PULAU PINANG

144

= 3 − +3
2 ( +1)( +2)

Method 2

+ 4 23 1
∑ ( + 1)( + 2) = ∑ [ − + 1 + + 2]
=1 =1

= ∑ =1 [(2 − 2) − (1 − 1 )]

+1 +1 +2

= ∑ =1 [2 − 2] − ∑ =1 [1 − 1]

+1 +1 +2

Let ( ) = 1 , Let ( ) = 1

+1



+ 4
∑ ( + 1)( + 2) = 2 ∑[ ( ) − ( + 1)] − ∑[ ( ) − ( + 1)]

=1 =1 =1

= 2[ (1) − ( + 1)] − [ (1) − ( + 1)]

= 2 [1 − 1 ] − [1 − 1 ]
1 +1 2 +2

=3− 2 + 1

2 +1 +2

= 3 − 2( +2)−( +1)
2 ( +1)( +2)

= 3 − +3
2 ( +1)( +2)

As ⟶ ∞, 1 → 0, 1 → 0 and the summation approaches 3.
+1 +2
Therefore, the series is convergent with ∞ = 32. 2

EXAMPLE 4

Given that is a positive integer and ( ) = 1 simplify ( ) − ( + 1) .
2
3 5 7
Hence, find the sum of the first terms of the series 12×22 + 22×32 + 32×42 + ⋯ .

Solution:

( ) − ( + 1)
11

= 2 − ( + 1)2
( + 1)2 − 2

= 2( + 2)2
2 + 2 + 1 − 2

= 2( + 1)2

UNIT SAINS DAN MATEMATIK, SEKTOR PEMBELAJARAN, JABATAN PENDIDIKAN NEGERI PULAU PINANG

145
2 + 1

= 2( + 1)2

12 3 22 + 22 5 32 + 32 7 42 + ⋯ + ℎ
× × ×

2 + 1
= ∑ 2( + 1)2

=1

1 1
= ∑ [ 2 − ( + 1)2]
=1



= ∑[ ( ) − ( + 1)]

=1

= (1) − ( + 1)
11

= 12 − ( + 1)2
1

= 1 − ( + 1)2
( + 1)2 − 1

= ( + 1)2
2 + 2

= ( + 1)2
( + 2)

= ( + 1)2

EXAMPLE 5

Express 1 in partial fraction.
(3 −2)(3 +1)

Hence, find ∑ =1 1 and deduce ∑ ∞ =1 1 .
(3 −2)(3 +1) (3 −2)(3 +1)

Solution:

Let 1 ≡ +
(3 −2)(3 +1) (3 −2) (3 +1)

1 = (3 + 1) + (3 − 2)

Let = − 1 ⇒ = − 1
3
3

Let = 2 ⇒ = 1
3
3

Therefore, 1 ≡ 1 − 1
(3 −2)(3 +1) 3(3 −2) 3(3 +1)

UNIT SAINS DAN MATEMATIK, SEKTOR PEMBELAJARAN, JABATAN PENDIDIKAN NEGERI PULAU PINANG