Semester 1 Mathematics STPM (Teacher's Edition)

146

Q2. = 3 2 −
−1 = 3( − 1)2 − ( − 1)

= − −1
= 3 2 − − [3( − 1)2 − ( − 1)]
= 3 2 − − (3 2 − 7 + 4)
= 6 − 4

Explicit formula: = 6 − 4
+1 = 6( + 1) − 4
= 6 + 2
= (6 − 4) + 6

Recursive formula: +1 = + 6 for n = 1, 2, 3, … and 1 = 2

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.

147

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

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 ∑ ∞ =0 3 ∑ ∞ =1 3 −1

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

148

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

Solution:
+ 2 + 3 + 4 + 5 + 6 = ∑ 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) +1(2 ).

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
234 50
Solution:
Solution:
1 ∙ 2 + 2 ∙ 3 + 3 ∙ 4+ . . . +19 ∙ 20
= 1. (1+1) + 2. (2+1) + 3. (3+1) + … 111 1
2 + 3 + 4 +. . . 50
+ 19. (19+1) =1+1 1 1 1 1
+ 2+1 + 3+1 + ⋯ + 49+1
∴ the ℎ term, = ( + 1)
Hence, the series may be written as ∴ the ℎ term, = 1 ,1 ≤ ≤ 49
∑ 1 =9 1 ( + 1)
+1 1
c) −3 + 4 − 5 + 6 +1
Solution: Hence, the series may be written as ∑ 4 =9 1
Ignoring the alternate sign, it can be seen that
the terms increase by 1. The rth term of the d) 8 + 9 + 10 + ⋯ +
series is given = + 2. 82+1 92+1 102+1 2+1

−3 + 4 − 5 + 6 Solution:
=(−1)1(1 + 2) + (−1)2(2 + 2) +
8 + 9 + 10 + ⋯ +
(−1)3(3 + 2) + (−1)4(4 + 2) 82+1 92+1 102+1 2+1
1+7 (2+27+)72+1+(3+37+)72+1 +7
∴ the ℎ term, = (−1) ( + 2) = (1+7)2+1 + + ⋯ + ( +7)2+1

Hence, the series may be written as ∴ the ℎ term, = +7
∑4 =1(−1) ( + 2) OR ∑ 5 =2(−1) +1( + 1) ( +7)2+1

Hence, the series may be written as

∑ =−17 +7 OR ∑ =8
( +7)2+1 2+1

149

e) + + +. . . + f) 9 − 27 + 81 − 243 + 729 − 2187 + 6561
× × × ( + )
Solution:
Solution:
9 − 27 + 81 − 243 + 729 − 2187 + 6561
1 + 1 + 1 +. . . + 1 = (−3)1+1 + (−3)2+1 + (−3)3+1 + (−3)4+1
2×4 3×5 4×6 ( +2)
1 1 1 1 +(−3)5+1 + (−3)6+1 + (−3)7+1
= 2×(2+2) + 3×(3+2) + 4×(4+2) + ⋯ + ×( +2)
∴ the ℎ term, = (−3) +1
∴ the ℎ term, = 1
( +2) Hence, the series may be written as
∑ 7 =1(−3) +1
Hence, the series may be written as

∑ =2 1 OR ∑ =−11 1
( +2) ( +1)( +3)

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

=0+2+4+6

359

=2 2

15

∴ 0+2+4+6; ∑3 =0 2 = 2 2
2 +1 15
359

150

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

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
2 +1
Solution:
∑ 7 =3(−2) −1 Solution:
= (−2)3−1 + (−2)4−1 + (−2)5−1 +
∑3 =0 2
(−2)6−1 + (−2)7−1 2 +1
= (−2)2 + (−2)3 + (−2)4 + (−2)5 + (−2)6 20 21 22 23
= 4 − 8 + 16 − 32 + 64 = 20+1 + 21+1 + 22+1 + 23+1
= 44
=1+2+4+8
4 − 8 + 16 − 32 + 64; 44
2359
= 27907

1+2+4+8 ; 2 77
90
2359

c) ∑ 5 =1(2 − 7) d) ∑6 =3 −5
−1
Solution:
∑5 =1(2 − 7) Solution:
=[2(1) − 7] + [2(2) − 7] + [2(3) − 7]
∑6 =3 −5
+[2(4) − 7] + [2(5) − 7] −1
= −5 − 3 − 1 + 1 + 3 = 3−5 + 4−5 + 5−5 + 6−5
= −5
3−1 4−1 5−1 6−1
−5 − 3 − 1 + 1 + 3 ; −5 = −1 − 1 + 0 + 1

35
= − 17

15

−1 − 1 + 0 + 1 ; − 17

3 5 15

151 f) ∑ 6 =2(3 + 2)2

e) ∑40( − 1)( − 3) Solution:
∑ 6 =2(3 + 2)2
Solution: = [3(2) + 2]2 + [3(3) + 2]2 + [3(4) + 2]2
∑40( − 1)( − 3)
= (0 − 1)(0 − 3) + (1 − 1)(1 − 3) +[3(5) + 2]2 + [3(6) + 2]2
= 64 + 121 + 196 + 289 + 400
+(2 − 1)(2 − 3) + (3 − 1)(3 − 3) = 1070
+(4 − 1)(4 − 3)
=3+0−1+0+3 64 + 121 + 196 + 289 + 400; 1070
=5

3 + 0 − 1 + 0 + 3; 5

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)

152

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

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

Solution:

1 = = 2
= 2 − = 6 − 2 = 4

The nth term is given by
= + ( − 1)

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

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

Solution:
1 = = −7
= −5 − (−7) = 2

The nth term is given by
= + ( − 1)

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

153

EXAMPLE 3

1 + 1 3 + 2 1 + ⋯ (11 ℎ term)
5 5

Solution:
1 = = 1

33
= 1 5 − 1 = 5

The nth term is given by
= + ( − 1)

∴ The 11 ℎ term is
3

11 = 1 + (11 − 1) (5) = 7

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:

1 = = 7 1 = = − 1
= 2 − = 4 − 7 = −3 = 2 − = 3 + 2 − ( − 1) = 2 + 3

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

∴ The 18 ℎ term is ∴ The 9 ℎ term is
18 = 7 + (18 − 1)(−3) = −44 9 = − 1 + 8(2 + 3) = 17 + 23

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

636 33

Solution: Solution:
1 1 = = 3

1 = = 6 22
21 1 = 2 − = 3 3 − 3 = 3

= 2 − = 3 − 6 = 2

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

∴ The 15 ℎ term is ∴ The 15 ℎ term is
1 11 21

15 = 6 + 14 (2) = 7 6 15 = 3 + 14 (3) = 12 3

154 f) 4 + 7 1 + 11 + ⋯ + (20 ℎterm)

e) −22 − 17 − 12 − ⋯ − (15 ℎ term) 2
Solution:
1 = = −22 Solution:
= −17 − (−22) = 5 1 = = 4

The nth term is given by 17
= + ( − 1) = 7 2 − 4 = 2

∴ The 15 ℎ term is The nth term is given by
15 = −22 + 14(5) = 48 = + ( − 1)

∴ The 20 ℎ term is
71

20 = 4 + 19 (2) = 70 2

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

155

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

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

156

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 ②

14 = 163
+ 13 = 163

②-①

8 = 96
= 12

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,

157

= + ( − 1) OR 20 = 20 (3 + )
2
820 = 20 (3 + )
= 3 + (20 − 1)(4)
2

= 79 = 79

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:
1 = = 7 1 = = + 2
= 2 − = 4 − 7 = −3 = − ( + 2 ) = −2
The nth term is given by The nth term is given by

= + ( − 1) = + ( − 1)
−98 = 7 + ( − 1)(−3) − 40 = + 2 + ( − 1)(−2 )

3 = 108 2 = 44
= 36 = 22

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

Solution: Solution:
1 = = 5 , = 2 − = 9 − 5 = 4 1 = = 1.3
= 2 − = 1.6 − 1.3 = 0.3
The nth term is given by The nth term is given by
6.7 = 1.3 + ( − 1)(0.3)
81 = 5 + ( − 1)4 0.3 = 5.7
4 = 80
= 19
= 20

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
1 = = 85, = 82 − 85 = −3
The nth term is given by Solution:
13 = 85 + ( − 1)(−3) The sum of the first 20 terms of an A.P. is
3 = 75 given by
= 25
20 1 7
The sum of the first 25 terms of an A.P. is 20 = 2 [2 (12) + 19 (12)]
given by 225
=2
25
25 = 2 [85 + 13] = 1225

158 d) −10 − 8 − 6 − ⋯ + 11 ℎterm

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

Solution: 11 = 2 [2(−10) + 10(2)]
40 =0

40 = 2 [2( − 4 ) + 39( − 2 )]
= 820 − 1720

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 ℎ 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, .

159

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]

ANSWER 1 Given that
Section 5 = 21 and 10 = 41
2.2 + 4 = 21 … (1)
+ 9 = 41 … (2)
Exercise (2) – (1) 5 = 20
2.2 = 4

Substituting = 4 into (1)
+ 4(4) = 21
+ 16 = 21
= 5

The ℎ term is = + ( − 1)
= 5 + ( − 1)4
= 1 + 4

2 Given that

The ℎ term is = 1 (5 − 3)
10
1 1
The 1 term is 1 = 10 (5 − 3) = 5

The 2 term is 2 = 1 (10 − 3) = 7
10 10

The common difference,
= 2 − 1
= 7 −1

10 5

=1

2

3(a) = 1 = 1
= 2(1)2 − 1
=1

3(b) 6 = 6 − 5
= [2(6)2 − 6] − [2(5)2 − 5]
= 21

160

3(c) = − −1
= [2( )2 − ] − [2( − 1)2 − ( − 1)]
= 4 − 3

Replacing by − 1

−1 = 4( − 1) − 3 = 4 − 7
− −1 = (4 − 3) − (4 − 7)

=4

Since − −1 is a constant, therefore the series is an A.P.

4 The integers from 1 to 200 are 1, 2, 3, 4, …,200

5(a) = 1, = 1, = 200, = 200
5(b)(i)

200 = 2 ( + )
= 200 (1 + 200)

2

= 20 100

The integers divisible by 5 are
5, 10, 15, 20, …, 200

= 5, = 5, = 200

= + ( − 1) = 200
5 + ( − 1)(5) = 200

= 40

40 = 40 (5 + 200) = 4100
2

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

= 20100 − 4100 = 16000

21 = 168
21 [2 + 20 ] = 168

2

+ 10 = 8 (shown)

2 + 3 = 50
+ + + 2 = 50

2 + 3 = 50 … (1)

5(b)(ii) Substitute = 8 − 10 into (1)
2(8 − 10 ) + 3 = 50

= −2
= 28
∴ 12 = 28 + 11(−2)
=6

21

∑ = 21 − 3

=4

= 168 − (28 + 50)
= 90

161

6(a) Given = 2 +
4 = 8

16 + 4 = 8 … (1)

6 = 24
36 + 6 = 24 … (2)

Solve (1) and (2)
= 1; = −2

(b) the nth term, .
= 2 − 2
= − −1
= 2 − 2 − [( − 1)2 − 2( − 1)]
= 2 − 3

ANSWER 1 a = u1
Section
2.2 b = u3 = a + 2d
c – 3 = u7 = a + 6d
STPM
Pass Year

a:b:c=4:3:2
a : a + 2d : a + 6d +3 = 4 : 3 : 2

4 4
+ 2 = 3 , + 6 + 3 = 2
3 = 4 + 8 ,
= 2 + 12 + 6
+ 8 = 0 , + 12 = −6
3
= 12
= − 2 ,

< 0

3
2 [2(12) + ( − 1) (− 2)] < 0

4 (51 − 3 ) < 0

3 (17 − ) < 0
4

n < 0 or n > 17

⸫ n = 18

162

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.

163

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 16 9

2 = (8)2 −2 = 3 (4)

or

2 = (130)2 −1 = 2 (25)
(− 23)2 −2 15

164

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]

24

Solution: Solution:

1 = = 1, 1 = = 1,
2 −3 3 9
= 1 = 1 = −3 / = 2 = −3 = −3 2 1 1
1 2
= = 2 =
1

7 = 1( 7−1) 12 = ( 12−1)
= 1[(−3)6] 1 11
= 729
= 1[(2) ]
1

= 2048

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

Solution:

1 = = 64, Solution:

= 2 = −16 = − 1 1 = = 1 ,
1 64 4 1

20 = ( 20−1) = = 1
1 19 1

= 64 (− 4)
= ( −1)
[1 ] −1
= 64(−1)19 (41)19 = 1
= −[43. 4−19]
1
= −1

= −(4−16)

= − 1
416

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

39

Solution: Solution:

1 = = 0.01 1 = = 1,
2 0.02
= 1 = 0.01 = 2 2 − 1 1
1 3 3
= = = −
1
16 = ( 16−1)
= 0.01(2)15 1 6−1
= 327.68 6 = 1[(− 3) ]

1
= − 243

165

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

1, 1 , 1 , … [ ] where 0 < < 1
2


Solution:

1 = = 1 ,

1
2 1
= 1 = = > 1
1



So,

− 1
= ( − 1 )
[( 1 1 )− −1
= 1 1
]

1 1 −
= −1 [(1 − )]

166

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]

24

Solution: Solution:

1 = = 1 , 1 = = 1 , 2 −3
1 1
2 1 1 = = = −3 < 1
1 2
= = 2 = < 1
1
So,
So,

1 − 12 1 − 7
12 = ( 1 − ) 7 = ( 1 − )

= 1 1 − −(2121)12) 4095 1 − (−3)7
( 1 = 2048 = 1 ( 1 − (−3) )

= 547

c) 1 −1+ 1− … [ 6] d) + 3, + 7, 4 − 2, … [ 2 ]
where < −3
3 9

Solution: Solution:

1 = = 1 , 1 1 = = + 3 ,
3
= 2 = − = − 1 < 1 = 2 = + 7 or = 3 = 4 − 2
1 3 1 + 3 2 + 7
1
So,
So,
+ 7 4 − 2
1 − 6 + 3 = + 7
6 = ( 1 − ) 3 2 − 4 − 55 = 0

= 1 (11−−((−−3131))6) 11
182 = 5 (ignore) or = − 3

= 243 the sequence become,

2 10 50
−3, 3 ,− 3 ,…

and

= −5

1 − 2
2 = ( 1 − )

2 1 − (−5)2
= − 3 [ 1 − (−5) ]

= − 1 (1 − 25 )

9

= 1 (25 − 1)

9

167

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 + 1 32 + ⋯ 1
100 10 000 000 000
= 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

168

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:

0. 5̇ = 5 + 5 + 1 5 + ⋯ 0. 7̇ 2̇ = 72 + 72 + 1 72 + ⋯
10 100 000 100 10 000 000 000

with = 5 and = 1 with = 72 and = 1
10 10 100 100

∞ = (150) = 5 ∞ = (17020) = 8
(1 − 110) 9 (1 − 1010) 11

∴ 0. 5̇ = 5 ∴ 0. 7̇ 2̇ = 8
9 11

c) 3.03̇ d) 3.28̇ 1̇ 4̇

Solution: Solution:

3.03̇ = 3 + 3 + 1 3 + 10 3 +⋯ 3.28̇ 1̇ 4̇ = 3 + 2 + 814 + 10 814
100 000 000 10 10 000 000 000
814
= 3 + ∞ + 10 000 000 000 + ⋯

with = 3 and = 1 2
100 10 = 3 + 10 + ∞
(1030)
∞ = (1 − 110) = 1 with = 814 and = 1
30 10 000 000
1
(10810400)
3.03̇ 1 ∞ = (1 − 1 0100) = 11
30 135
∴ = 3 +

1 ∴ 3.28̇ 1̇ 4̇ = 3 + 2 + 11
= 3 30 10 135

38
= 3 135

169

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
= − (−32)2

= − 81 = −20.25
4

∞ =
1−
20.25
∞ = 1−(−32) = −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.

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

170

b) 1 = 10 = 10 , 2 = 10 , 3 = 10
33 27 35 37
2 1 3 = 1
1 = 9 ,
2 9
10 19.
the series is a geometric series with = 27 and common ratio, =

Sum of the first terms,

= (1− )
1−
2107[1−(19) ]
=
= 5 1−91 (1) ]

[1 −
12 9

∞ = lim
= 5 [1
→∞ − (1) ]
9
lim
→∞ 12
=5
(1)12
as → 0 as → ∞
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)
(2) +1
(a) Express in the form , where is a constant.
5

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

STPM PAST YEAR

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]

171

ANSWER 1 2, − 1 , 1
Section 2(a) 2 8
2.2 = − 1
= 2 ,
Exercise 4
2.3

∞ = 1 −

28
∞ = 1 − (− 14) = 5

(1 − )
= 1 −
2[1−(−41) ]
= 1−(−41)

= 8 [1 − (− 1) ]

54

| ∞ − | < 10−5

| 8 − 8 [1 − (− 1 | < 10−5
5 5 4) ]

| 8 (− 41) | < 10−5
5

8 (41) < 10−5
5

1 < (5 × 10−5)

48

> −5.20412

−0.60206

> 8.644

⸫n=9

2 +1 2 −1
= (5) − (5)
2 +1 2 +1−2
= (5) − (5)
2 −2 2 +1
2 +1
= (5) − (5) (5)
2 −2 2 +1
= [1 − (5) ] (5)

21 2 +1
= − 4 (5)

2(b) 1 = − 21 (2)2 = − 21 , 2 = − 42 , 3 = − 84 ,

45 25 125 625
2 2 3 = 2
1 = 5 ,
2 5
21 2
The series is a geometric series with = − 25 and common ratio, = 5 .

Sum of the first terms,

172

(1 − )
= 1 −
21 [1 − (52) ]
− 25

= 2
5
1 −

7 2
= − 5 [1 − (5) ]

∞ = lim

→∞ 7 2

= lim − 5 [1 − (5) ]

→∞

= − 7 as (25) → 0 as → ∞
5

ANSWER 1 ∞ −
Section ∑ ( ) ( )
2.2
=
Exercise
2.3 = ( ) ( ) −

STPM
Past Year
= ( ) ( ) −
− ( ) − ( ) −

= ( ) ( ) =


Since is a constant, therefore the series is a geometric series.

−

Since = = , | | < showing that ∑∞ = ( ) ( ) − is a
−


convergent series.

∞ −
∑ ( ) ( ) = ∞
=


∞ = −
= ( )( )−

−( )
⁄
= ⁄


=

2(a) = − −
− = − − ( − ) = − − +

= − −
= ( − − ) − ( − − + )

= − + − −
= − ( − )

= ( − ) ( )

173

(b) + = [ − ( + )]
= ( − )( − )

= ( − )


=

+ = =




Since + is a constant, therefore the sequence is a geometric


sequence.

(c) + + + …
= ( − ) + ( − ) + ( − ) + ⋯
( − )
= − ( )
=

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 = 1 + 2 + 3 + ⋯ + ( − 1) + ….①

By rewriting this series backwards term by term,
∑ =1 = + ( − 1) + ( − 2) + ⋯ + 2 + 1 …..②

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

174

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 − 13
= ( + 1)3 − 1

Thus,

( + 1)3 − 1 = 3 ∑ 2 + 3 ∑ + ∑ 1

=1 =1 =1


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

1
4
∑ 3 = 2( + 1)2

=1 12

= [2 ( + 1)]

2

= (∑ )

=1

175

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

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


6
∑ 2 = ( + 1)(2 + 1)

=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:
a) 12 + 42 + 72 + 102 + ⋯
= ∑(3 − 2)2

=1

∑ =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
2 2

= 3 3 − 3 2 − 1

22

176

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



= ∑( + 1)( + 2)

=1

∑ =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 6 2 2

= 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)( + 3)

=1

∑ =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] + ⋯



= ∑(2 − 1)( + 1)2

=1

∑ =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

= 2 [ ( + 1)2 + ( + 1)(2 + 1) − 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

177

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

= [23 ( + 1)(2 + 1) + 2( + 1) + 1]
= (4 2 + 2 + 2 + 2 + 3)

33

= (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 =01(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

Q1. Find b) ∑ =1 ( + 3)( + 5)
a) ∑ =1( + 2)( + 3)

Solution: Solution:



∑( + 2)( + 3) ∑ ( + 3)( + 5)

=1 =1


= ∑ 2 + 5 + 6 = ∑ 3 + 8 2 + 15

=1 =1


= ∑ 2 + 5 ∑ + 6 ∑ 1 = ∑ 3 + 8 ∑ 2 + 15 ∑

=1 =1 =1 =1 =1 =1

= 2 ( + 1)2 + 8 ( + 1)(2 + 1)
= 6 ( + 1)(2 + 1) + 5 (2) ( + 1) + 6 4 (6)

= 1 (2 2 + 3 + 1) + 5 + 5 + 6]
[6 2 2 +15 (2) ( + 1)

= 1 2 + 3 + 26 = 1 ( )( + 1)[3 ( + 1) + 16(2 + 1) + 90]
[3 3] 12

= 1 ( 2 + 9 + 26) = 12 ( + 1)[3 2 + 3 + 32 + 16 + 90]

3

= 1 ( + 1)(3 2 + 35 + 106)

12

178

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

Solution: Solution:

10 16

∑ (2 2 + 3) ∑ ( + 1)( + 4)

=1 =8

10 10 16 16 16

= 2 ∑ 3 + 3 ∑ = ∑ 3 + 5 ∑ 2 + 4 ∑

=1 =1 =8 =8 =8

= 2 102 (10 + 1)2 + 3 10 (10 + 1) = 162 (16 + 1)2 − 72 (7 + 1)2
(4) (2) (4) (4)

16 7
+5 [ 6 (16 + 1)(32 + 1) − 6 (7 + 1)(14 + 1)]
= 6050 + 165

= 6215 +4 16 (16 + 1) − 7 (7 + 1)]
[2 2

= 17 712 + 6 780 + 432

= 24 924

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

Solution: Solution:
93 + 103 + 113 + … + 203 1 × 2 + 4 × 3 + 9 × 4 + ⋯ + 169 × 14

12 13

= ∑( + 8)3 = ∑ 2( + 1)

=1 =1

12 12 12 12 13 13

= ∑ 3 + 24 ∑ 2 + 192 ∑ + ∑ 512 = ∑ 3 + ∑ 2

=1 =1 =1 =1 =1 =1

= 122 (12 + 1)2 + 24 12 (12 + 1)(2(12) + 1) = 132 (13 + 1)2 + 13 (13 + 1)(13(2) + 1)
4 (6)
46
12
+192 ( 2 ) (12 + 1) + 512(12) = 8281 + 819

= 9100

= 6 084 + 15 600 + 14 976 + 6 144

= 42 804

179

c) 2 × 3 + 3 × 5 + 4 × 7 + ⋯ + 11 × 21 d) 1 × 2 × 4 + 2 × 3 × 6 + 3 × 4 × 8 + ⋯ +
15 × 16 × 32
Solution:
2 × 3 + 3 × 5 + 4 × 7 + ⋯ + 11 × 21 Solution:
1×2×4+2×3×6+3×4×8+⋯+
10 15 × 16 × 32

= ∑( + 1)(2 + 1) 15

=1 = ∑ ( + 1)(2 + 2)
10 10 10
=1
= 2 ∑ 2 + 3 ∑ + ∑ 1 15 15 15

=1 =1 =1 = 2 ∑ 3 + 4 ∑ 2 + 2 ∑

10 10 =1 =1 =1
= 2 ( 6 ) (10 + 1)(21) + 3 ( 2 ) (10 + 1) + 10
= 2 152 (15 + 1)2 + 4 15 (15 + 1)(31)
= 770 + 165 + 10 (4) (6)

= 945 15
+2 ( 2 ) (15 + 1)

= 28800 + 4960 + 240

= 34000

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:
11 + 12 + 13 + 14 + ⋯ to 2n terms 23 + 43 + 63 + 83 + ⋯ to n terms

2

= ∑( + 10) = ∑(2 )3

=1 =1

2 2

= ∑ + ∑ 10 = 8 ∑ 3

=1 =1 =1

2 = 8 2 ( + 1)2
= 2 (2 + 1) + 10(2 ) (4)

= (2 + 1) + 20 = 2 2( + 1)2

= 2 2 + 21

= (2 + 21)

180

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

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



= ∑ ( + 1)2 = ∑ ( + 3)( + 6)

=1 =1



= ∑ 3 + 2 ∑ 2 + ∑ = ∑ 3 + 9 ∑ 2 + 18 ∑

=1 =1 =1 =1 =1 =1

= 2 ( + 1)2 + 2 ( + 1)(2 + 1) = 2 ( + 1)2 + 9 ( + 1)(2 + 1)
(4) (6) 4 (6)

+18 (2) ( + 1)
+ 2 ( + 1)

= ( + 1)[3 ( + 1) + 4(2 + 1) + 6] = ( + 1)[ ( + 1) + 6(2 + 1) + 36]
12 4

= ( + 1)[3 2 + 11 + 10] = ( + 1)( 2 + 13 + 42)
12 4

1
= 12 ( + 1)( + 2)(3 + 5) = 4 ( + 1)( + 7)( + 6)

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)

181

EXAMPLE 1 1
Find ∑ =1 [1 − +11].

Solution:

1 11
∑ [ − + 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 2 1

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

=1−1

13

+1−1

35

+1−1

57

⋮⋮

+1 − 1

2 −1 2 +1

= 1 − 1
2 +1

= 2 [shown]

2 +1

182

Method 2

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

2

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

= (1) − ( + 1) [by the 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 2]
( +1)( +2) +1

=2−3+1

123

+2−3+1

234

+ 2 − 3 + 1
3 4 5

⋮⋮⋮

+ 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)

= 3 − +3
2 ( +1)( +2)

183

Method 2 + 4 2

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

= ∑ =1 [(2 − 2) − (1 − +1 2)]

+1 +1

= ∑ =1 [2 − +21] − ∑ =1 [ +11 − +12]

Let ( ) = 1 , Let ( ) = 1
+1


+ 4

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

= 2[ (1) − ( + 1)] − [ (1) − ( + 1)]

= 2 [11 − +1 1] − [21 − +1 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
2 + 1

= 2( + 1)2

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

184

2 + 1
= ∑ 2( + 1)2

=1


11
= ∑ [ 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

33

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

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

=1−1 = 1

3(1) 3(4)

+1 − 1 = 2
= 3
3(4) 3(7)

+ 1 − 1
3(7) 3(10)

⋮⋮

185

+ 1 − 1 =
3(3 −2) 3(3 +1)

11
= 3 − 3(3 + 1)

(3 + 1) − 1
= 3(3 + 1)

3
= 3(3 + 1)


= 3 + 1

∞1 lim 11
∑ = (3 − 1))
(3 − 2)(3 + 1) →∞ 3(3 +
=1

1
=3−0

1
=3

EXERCISE 2.4

1. a) Express 2 in partial fractions.
(2 +1)(2 +3)

b) Show that ∑ =1 2 = 1− 1 .
(2 +1)(2 +3)
3 2 +3

c) Find the sum of the series 1 +1 + 1 + 1 + ⋯+ 1 and find ∑∞ =1 1 .
3×5 7×9 9×10 37×39 (2 +1)(2 +3)
5×7

2. If ( ) = 1 , obtain and simplify ( + 1) − ( ) . Hence, find ∑ =1 1 .
( +1) ( +1)( +2)

3. Show that if ( ) = 1 then ( ) − ( − 1) = −4 . Hence, or otherwise,
(2 +1)(2 +3) (2 −1)(2 +1)(2 +3)

find the sum of the series 1 + 1 + 1 + 1 + ⋯ + 1 .
1×3×5 3×5×7 5×7×9 7×9×11 (2 −1)(2 +1)(2 +3)

186

STPM PAST YEAR

STPM MM 2013 (Section B)

Q1 a) Express 1 in partial fractions, and deduce that
( 2−1)

1 ≡ 1 [ ( 1−1) − ( 1+1)] . [4 marks]
( 2−1) 2

Hence, use the method of differences to find the sum of the first ( − 1) terms, −1 , of the

series

11 1 1
2 × 3 + 3 × 8 + 4 × 15 + ⋯ + ( 2 − 1) + ⋯ ,

and deduce . [6 marks]

b) Explain why the series converges to 1 , and determine the smallest value of such that
4
1
4 − < 0.0025 . [5 marks]

ANSWER 1(a) 2
Section (2 + 1)(2 + 3) ≡ 2 + 1 + 2 + 3
2.2
2 ≡ (2 + 3) + (2 + 1)
Exercise
2.4 Let = − 3 , 2 = (−2) ⇒ = −1
2

Let = − 1 , 2 = (2) ⇒ = 1
2

2 11
∴ (2 + 1)(2 + 3) ≡ 2 + 1 − 2 + 3

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

1 11
∴ ⇒ ( ) = 2 + 1 , ( + 1) = 2( + 1) + 1 = 2 + 3



= ∑ ( ) − ( + 1)

=1

= (1) − ( + 1)

= 1 − 1 (Shown)
3 2 +3

1(c) 111 1 1
3 × 5 + 5 × 7 + 7 × 9 + 9 × 10 + ⋯ + 37 × 39

1 18 2
= 2 ∑ (2 + 1)(2 + 3)
=1

187

11 1
= 2 [3 − 2(18) + 3]

2
= 13

∞1 11 1
∑ (2 + 1)(2 + 3) = 2 [3 − 2(∞) + 3]
=1

1
=6

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

11
( + 1) − ( ) = ( + 1)( + 2) − ( + 1)

− ( + 2)
= ( + 1)( + 2)

2
= − ( + 1)( + 2)

For 1 ≡ + +
( +1)( +2) +1 +2

1 ≡ ( + 1)( + 2) + ( )( + 2) + ( )( + 1)

Let = 0, 1 = (1)(2) ⇒ = 1
2

Let = −1, 1 = (−1)(1) ⇒ = −1

Let = −2, 1 = (−2)(−1) ⇒ =1

2

1 11 1
∴ ( + 1)( + 2) ≡ 2 − + 1 + 2( + 2)

11 1 12 1
2 − + 1 + 2( + 2) = 2 − 2( + 1) + 2( + 2)

11 1 1
= 2 − 2( + 1) − 2( + 1) + 2( + 2)

11 11
= [2 − 2( + 1)] − [2( + 1) − 2( + 2)]

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

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

1 1 1 11
∑ ( + 1)( + 2) = ∑ {[2 − 2( + 1)] − [2( + 1) − 2( + 2)]}
=1 =1

188



= ∑[ ( ) − ( + 1)] − ∑[ ( ) − ( + 1)]

=1 =1

= [ (1) − ( + 1)] − [ (1) − ( + 1)]

1111
= 2 − 2( + 1) − 4 + 2( + 2)

11 1
= 4 − 2( + 1) + 2( + 2)

1 22
= 4 [1 − + 1 + + 2]

1 − 1 2
= 4 [ + 1 + + 2]

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

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

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

= 1
(2 −1)(2 +1)

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

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

4
= − (2 − 1)(2 + 1)(2 + 3)

111 1
1 × 3 × 5 + 3 × 5 × 7 + 5 × 7 × 9 + 7 × 9 × 11 + ⋯

1
+ (2 − 1)(2 + 1)(2 + 3)

1
= ∑ (2 − 1)(2 + 1)(2 + 3)

=1

1 4
= − 4 ∑ − (2 − 1)(2 + 1)(2 + 3)

=1


1
= − 4 ∑[ ( ) − ( − 1)]

=1

1
= − 4 [ ( ) − (0)]

189

11 1
= − 4 [(2 + 1)(2 + 3) − (1)(3)]

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

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

ANSWER 1(a) 11
Section 2 − 1 ≡ ( − 1)( + 1)
2.2

Exercise ≡ − 1 + + 1
2.4 1 ≡ ( + 1) + ( − 1)

STPM Let = 1, 1 = (2) ⇒ = 1
Pass Year
2

Let = −1, 1 = (−2) ⇒ = − 1

2

11 1
∴ 2 − 1 ≡ 2( − 1) − 2( + 1)

11 11 1
( 2 − 1) ≡ [2( − 1) − 2( + 1)]

1 11 1
( 2 − 1) ≡ 2 [ ( − 1) − ( + 1)]

11 1 1 1
2 × 3 + 3 × 8 + 4 × 15 + ⋯ + ( 2 − 1) + ⋯ + ( 2 − 1)

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

−1 = 1 {2(11) − 1 = 2
2 2(3) = 3
= 4
+1 − 1

3(2) 3(4)

+1 − 1

4(3) 4(5)

+ 1 − 1 = − 1
( −1)( −2) ( −1)( )

+ 1 − ( 1+1)} =
( −1)

= 1 [21 − ( 1+1)]
2

11 1
−1 = 2 [2 − [( − 1) + 1][( − 1) + 2]]

∴ = 1 [21 − ( +1)1( +2)]
2

190

(b) As → ∞, 1 →0
( +1)( +2)

∴ → 1
4
The series converges to 1
4

1
4 − < 0.0025

11 1
4 − (4 − 2( + 1)( + 2)) < 0.0025

1 25
2( 2 + 3 + 2) < 10 000

2 + 3 + 2 > 200
2 + 3 − 198 > 0

= −3 ± √9 + 4 × 198
2
= −15.65 , 12.65 12.65
- 15.65

Since > 0 , ∴ > 12.65
∴ = 13

2.3 BINOMIAL Learning outcome:
EXPANSIONS a) Expand (a + b)n, where n ϵ Z+.
b) Expand (1 + x)n, where n ϵ Q, and identify the condition |x| < 1 for the validity of

this expansion.

c) Use binomial expansions in approximations.

Learning outcome: Expand (a + b)n, where n ϵ Z+.

n! = n(n - 1)(n - 2)… 3⸱2⸱1, with n ϵ ℤ+,

0! = 1

(n + 1)! = (n + 1)⸱n! ( )

The binomial coefficient, = ! ! for n, r ϵ ℤ+ and 0 ≤ r ≤ n .
( − )!



Binomial Theorem, ( + ) = ∑ =0( ) − , n ϵ ℤ+.

( + ) = + ( 1 ) −1 + ( 2 ) −2 2 + ( 3 ) −3 3 + … + ( ) − + ⋯ + ( 1) −1 +
−

EXAMPLE 1
Using the binomial theorem, find the expansion of (2x + 3y)4.

Solution:
(2 + 3 )4 = (2 )4 + (41) (2 )3(3 ) + (24) (2 )2(3 )2 + (43) (2 )(3 )3 + (3 )4

= 16 4 + 96 3 + 216 2 2 + 216 3 + 81 4

191

EXAMPLE 2

Using the binomial theorem, find the expansion of (a + b)4 .

Hence, find (b) ( − 2)4
(a) (2x + 3y)4
2

Solution:
( + )4 = ( )4 + (14) ( )3( ) + (42) ( )2( )2 + (34) ( )( )3 + ( )4

= 4 + 4 3 + 6 2 2 + 4 3 + 4

(a) (2x + 3y)4 = (2 )4 + 4(2 )3(3 ) + 6(2 )2(3 )2 + 4(2 )(3 )3 + (3 )4
= 16 4 + 96 3 + 216 2 2 + 216 3 + 81 4

(b) ( − 2)4 = ( )4 + 4 ( )3 (− 2) + 6 ( )2 (− 2)2 + 4 ( ) (− 2)3 + (− 2)4
2 22 2 2

= 4 − 3 2 + 3 2 4 − 2 6 + 8
16 2 2

EXAMPLE 3
Find the coefficient of x6 in the expansion of (3x2 + 4)7.

Solution:

( + ) = ∑ =0( ) −
where the (r + 1)th term = ( ) −

7

Thus (3 2 + 4)7 = ∑ =0(7 )(3 2)7− (4)
The term in x6 is when 2(7 – r) = 6, r = 4

Coefficient of x6 = (74) 3344 = 241920

The Expansion of (1 + x)n , n ϵ ℤ+.

(1 + ) = 1 + ( 1 ) + ( 2 ) 2 + ( 3 ) 3 + … + ( ) + ⋯ + ( 1) −1 +
−

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

2! !

EXAMPLE 1
Use the binomial theorem to expand (1 + x)5.

Solution:

(1 + )5 = 1 + 5 + 5∙4 2 + 5∙4∙3 3 + 5∙4∙3∙2 4 + 5
2! 3! 4!

= 1 + 5 + 10 2 + 10 3 + 5 4 + 5

192

EXAMPLE 2
Use the binomial theorem to expand (3 + p)5.

Solution:
(3 + p)5 = [3 (1 + 3 )]5

= 35 (1 + )5

3

= 35 [1 + 5 ( ) + 5∙4 ( )2 + 5∙4∙3 ( )3 + 5∙4∙3∙2 ( )4 + ( )5]
3 2! 3 3! 3 4! 3 3

= 35 [1 + 5 ( 3 ) + 10 ( 3 )2 + 10 ( 3 )3 + 5 ( 3 )4 + ( 3 )5]

= 243 + 405 + 270 3 + 90 4 + 15 4 + 5

EXAMPLE 3
Expand (1 + − 3 2)8 up to the term in y3.

Solution:
(1 + − 3 2)8 = [1 + ( − 3 2)]8

= 1 + 8( − 3 2) + 8∙7 ( − 3 2)2 + 8∙7∙6 ( − 3 2)3 + ⋯

2! 3!

= 1 + 8( − 3 2) + 28( 2 − 6 3 + ⋯ ) + 56( 3 + ⋯ ) + ⋯

= 1 + 8 + 4 2 − 112 3 + ⋯

WORKSHEET 2.11

Q1. Expand the following using binomial theorem.

a) (2 + 1)7



Solution:

(2 + 1)7 = (2 )7 + (71) (2 )6( −1) + (72) (2 )5( −1)2 + (37) (2 )4( −1)3 + (47) (2 )3( −1)4 +



(57) (2 )2( −1)5 + (67) (2 )( −1)6 + ( −1)7
= (128) 7 + 7(64) 5 + 21(32) 3 + 35(16) + 35(8) −1 + 21(4) −3 + 7(2) −5 + −7

= 128 7 + 448 5 + 672 3 + 560 + 280 −1 + 84 −3 + 14 −5 + −7

b) ( − √2)6

Solution:
( − √2)6 = 6 + (61) 5(−√2) + (26) 4(−√2)2 + (63) 3(−√2)3 + (46) 2(−√2)4 + (56) (−√2)5 +

(−√2)6
= 6 − 6√2 5 + 30 4 − 40√2 3 + 60 2 − 24√2 + 8

193

Q2. Find the term indicated in the binomial expansions of the following functions.

a) (3 + 4 )7, 4th term. b) ( + )8 , term containing 3.

Solution: Solution:
3+1 = (73) (3 )7−3(4 )3 5+1 = (58) ( )8−5( )5
= 35(81 4)(64 3) = 56 3 5
= 181440 4 3

c) (√ + 4√ )8, middle term. d) (2 3 − 1 )12, term independent of x.

Solution: Solution:
4+1 = (84) (√ )8−4(4√ )4
= 70( 2)(x) +1 = (1 2) (2 3)12− (− 1)
= 70 3


= (1 2) (2)12− (−1) 36−4

The term is independent of x if 36-4r =0,
so r=9.

Therefore, the required term is
9+1 = (192) (2)12−9(−1)9
= (220)(8)(-1)

= -1760

EXERCISE 2.5

1. Find the coefficient of 4 in the expansion of ( + 1)5( − 1 )3.



2. Expand (1 − 3 − 2)5 in ascending powers of up to the term in 4.
2

3. Expand (2 + 3 )4 as a series of ascending powers of x.
Write down the expansion of (2 − 3 )4.
Hence, find the value of (2 + 3√2)4 − (2 − 3√2)4.

194

STPM PAST YEAR

STPM MT 2014 (Section A)

Q1 Use the binomial expansions of (√3 + 2)6 and (√3 − 2)6 to evaluate (√3 + 2)6 + (√3 − 2)6 .

Hence, show that 2701 < (√3 + 2)6 < 2702 . [7 marks]

STPM MM 2020 (Section A) [2 marks]
Q2 Find the general term in the expansion of (2 + 1 2)12 . [2 marks]
[4 marks]
a) Find the term independent of x in the expansion.
b) Determine the coefficient of x3 in the expansion of (1 − 3)(2 + 1 2)12 .

ANSWER

Section 1. ( + ) ( − ) = [ + ( ) ( ) + ( ) + ( ) + ( ) +
2.3
( ) ] ) + (− ) + (− ) ]
Exercise
2.5 [ + ( ) (−

= ( + + + + + )( − + − )

To obtain term,

= ... + ( ) + (− )+ ( ) + ⋯

= − + = −

⸫ The coefficient of term is -2.

2. (1 − 3 − 2)5 = [1 − (23 + 2)]5
2
(3 10(3 10(3 5(3
= 1 − 5 + 2 ) + + 2)2 − + 2)3 + +
2 2 2 2
2)4 + ⋯

= 1 − 15 − 5 2 + 10 ( 9 2 + 4 + 3 3) − 10 ( 27 3 + 27 4 +
2 4 8 4
⋯ ) + 405 4 + ⋯
16
15 35 15 515
= 1 − 2 + 2 2 − 4 3 − 16 4

3. (2 + 3 )4 = (2)4 + 4(2)3(3 ) + 6(2)2(3 )2 + 4(2)(3 )3 + (3 )4
= 16 + 96 + 216 2 + 216 3 + 81 4

(2 − 3 )4 = 16 − 96 + 216 2 − 216 3 + 81 4

(2 + 3 )4 − (2 − 3 )4
= 16 + 96 + 216 2 + 216 3 + 81 4 − (16 − 96 + 216 2 − 216 3 +

81 4)
= 192 + 432 3

(2 + 3√2)4 − (2 − 3√2)4 = 192(√2) + 432(√2)3
= 1493.41

195

ANSWER

Section 1 (√3 + 2)6 = (√3)6 + (16) (√3)5(2) + (26) (√3)4(2)2 + (63) (√3)3(2)3
2.3 + (46) (√3)2(2)4 + (65) (√3)(2)5 + 26

STPM (√3 − 2)6 = 6 − (16) (√3)5(2) + (26) (√3)4(2)2 − (36) (√3)3(2)3
Past Year
(√3)

+ (46) (√3)2(2)4 − (65) (√3)(2)5 + 26

(√3 + 2)6 + (√3 − 2)6

= 2 [(√3)6 + (26) (√3)4(2)2 + (64) (√3)2(2)4 + 26]

= 2[27 + 15(9)(4) + 15(3)(16) + 64]

= 2702

(√3 − 2)6 = 2702 − (√3 + 2)6

1 < √3 < 2

−1 < √3 − 2 < 0
0 < (√3 − 2)6 < 1
−1 < −(√3 − 2)6 < 0
0 < 2702 − (√3 + 2)6 < 1
−2702 < −(√3 + 2)6 < −2701
2702 > (√3 + 2)6 > 2701

⸫ 2701 < (√3 + 2)6 < 2702 (shown)

2 1 2)12 12 (1 2) (2 )12− 1
( 2)
(2 + = ∑

The general =0 (r+1)th term = (1 2) (2 )12− ( 1 2)

term,

(a) The term independent of x is when ( )12− ( 1 2 = 0

)

12 − − 2 = 0

= 4

The term independent of x is
5 = (142) (2)12−4(1)4

= (495)(256)(1)

= 126720

(b) (1 − 3)(2 + 1 2)12 13 14
( 2) ( 2)
= (1 − 3) [… + (132) (2 )12−3 + (142) (2 )12−4 +. . . ]

= (1 − 3)[… + 112640 3 + 126720 +. . . ]

= ⋯ + 112640 3 − 126720 3 + ⋯

= ⋯ − 14080 3 + ⋯

The coefficient of x3 = -14080