PRE-U STPM MATHEMATICS (T) TERM 1

CONTENTS







Chapter
1 FUNCTIONS 1
• • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • •
1.1 Functions 2
1.2 Polynomial and Rational Functions 21
1.3 Exponential and Logarithmic Functions 43
1.4 Trigonometric Functions 56

Chapter
2 SEQUENCES AND SERIES 92
• • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • •
2.1 Sequences 93
2.2 Series 99
2.3 Binomial Expansions 119


Chapter
3 MATRICES 136
• • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • •
3.1 Matrices 137
3.2 Systems of Linear Equations 160


Chapter
4 COMPLEX NUMBERS 172
• • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • •
4.1 Complex Numbers 173


Chapter
5 ANALYTIC GEOMETRY 192
• • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • •
5.1 Analytic Geometry 193


Chapter
6 VECTORS 225
• • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • •
6.1 Vectors in Two and Three Dimensions 226
6.2 Vector Geometry 246

STPM Model Paper (954/1) 266

Answers 268




iii





iii-Content STPM.indd 3 3/28/18 4:18 PM

STPM Scheme of Assessment







Term of Paper Code and Type of Test Mark Duration Administration
Study Name (Weighting)
First Term 954/1 Written test 60
Mathematics Section A (26.67%)
(T)
Paper 1 Answer all 6 questions of 45
variable marks. Central
1
Section B 1— hours assessment
2
Answer 1 out of 2 15
questions
All questions are based on
topics 1 to 6.
Second 954/2 Written test 60
Term Mathematics Section A (26.67%)
(T)
Paper 2 Answer all 6 questions of 45
variable marks. Central
1
Section B 1— hours assessment
2
Answer 1 out of 2 15
questions.
All questions are based on
topics 7 to 12.

Third Term 954/3 Written test 60
Mathematics Section A (26.67%)
(T)
Paper 3 Answer all 6 questions of 45
variable marks. Central
1
Section B 1— hours assessment
2
Answer 1 out of 2 15
questions.
All questions are based on
topics 13 to 18.

First, 954/4 Coursework 180 School-based
Second Mathematics 3 assignments, each based (20%) assessment for
and Third (T) on topics 1 to 6, topics 7 Through-out school candidates
Terms Paper 4 to 12 and topics 13 to 18. the three Assessment
terms by appointed
assessor for private
candidates










ii-IBC Scheme STPM.indd 1 3/29/18 12:39 PM

CHAPTER
2 SEQUENCES


AND SERIES









2
Subtopic Learning Outcome

2.1 Sequences (a) Use an explicit formula and a recursive formula for a sequence.
(b) Find the limit of a convergent sequence.


2.2 Series (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 a 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.


2.3 Binomial (a) Expand (a + b) , where n ∈ Z .
+
n
expansions (b) Expand (1 + x) , where n ∈ Q, and identify the condition |x| , 1 for the
n
validity of this expansion.
(c) Use binomial expansions in approximations.



Bilingual Keywords

approximation – penghampiran
arithmetic – aritmetik
binomial expansion – kembangan binomial
converge – menumpu
explicit – tak tersirat
geometric – geometri
limit – had
sequence – jujukan
series – siri
recursive – rekursif















02 STPM Math T T1.indd 92 3/28/18 4:21 PM

Mathematics Term 1 STPM Chapter 2 Sequences and Series

2.1 Sequences


Sequences

Consider the set of numbers 3, 7, 13, 21, …, 111, … Sequences
This set of numbers can be written as INFO
1 + 1 × 2, 1 + 2 × 3, 1 + 3 × 4, 1 + 4 × 5, …, 1 + 10 × 11, …
th
The n term, u , can be written as
n
u = 1 + n(n + 1), n  Z +
n
For the set of numbers 1, — , — , — , …, the n term, v , can be written as
1 1 1
th
3 5 7 n 2
1
v = 2n – 1 , n  Z +
n
+
A set of numbers u , u , u , …, u which is arranged with each term u as a function f(n), n  Z , is known
1
n
3
2
n
as a sequence. A sequence is usually denoted by {u }.
n
If the n term of a sequence is given, say u = f(n), n  Z , then we can find the successive terms of
th
+
n
this sequence. For example, if u = — , n  Z , then the sequence is 1, — , — , — , …
1
1 1 1
+
n n 4 9 16
2
1
We say that this sequence is defined by the explicit formula, u = — , n  Z .
+
n n 2
There is a type of sequence where the value of each term is related to its preceding terms, with the initial terms
being given. For example, if the first two terms of a sequence are u = 2 and u = 6, and u n + 2 = u + u ,
n
1
2
n + 1
then we can find the successive terms of the sequence, i.e.
u = u + u = 6 + 2 = 8,
1
3
2
u = u + u = 8 + 6 = 14,
2
3
4
u = u + u = 14 + 8 = 22
4
5
3
and so on. Hence, the sequence obtained is 2, 6, 8, 14, 22, 36, … We say that this sequence can be defined by
the recursive formula, u n + 2 = u n + 1 + u , with u = 2 and u = 6.
2
1
n
Example 1
th
Write down the first five terms of the sequence with its n term
n
u = n + 1 , n  Z +
n
n
Solution: u = n + 1 , n  Z +
n
1
Hence, u = 1 + 1 = 1
2
1
2
u = 2 + 1 = 3 2
2
3
u = 3 + 1 = 4 3
3
4
u = 4 + 1 = 5 4
4
5
u = 5 + 1 = 6 5
5
Hence, the first five terms of the sequence are 1 , 2 , 3 , 4 and 5 .
2 3 4 5 6
93
02 STPM Math T T1.indd 93 3/28/18 4:21 PM

Mathematics Term 1 STPM Chapter 2 Sequences and Series

Example 2

+
If a sequence with u n + 1 = 2u , n  Z , and u = 1 is given, find the first five terms of this sequence. State
n
1
the value of the n term in terms of n.
th
Solution: u n + 1 = 2u , n  Z .
+
n
u = 1
1
— 1
Hence, u = 2u = 2 2
1
2
1
1
— + —
u = 2u = 2 2 4
2
3
1
1
— + — + —
2 u = 2u = 2 1 8
2
4
3
4
1
1
1
1
— + — + — + —
4
2
8
u = 2u = 2 16
4
5
Hence, the first five terms of the sequence are
1
1
1
1
1
1
— 1 — + — 1 — + — + — 1 — + — + — + — 1
8
4
2
4
2
8
2
1, 2 , 2 2 4 , 2 and 2 16
— 1 — 3 — 7 15
—
4
8
16
2
or 1, 2 , 2 , 2 and 2 .
1
1
1
1
— + — + — + … + ––––––
2
4
8
th
The n term, u = 2 2 n – 1
n
A sequence with terms which are repeated after a certain fixed number of terms is known as a periodic sequence.
In trigonometry, the range of values for the graphs of sine and cosine functions are between –1 and 1. This
range repeats itself after an interval of 2π radians. For example,
+
the sequence with u = cos nπ , n  Z , has the terms 0, –1, 0, 1, … which repeats itself after four terms.
2
n
Therefore, this sequence is a periodic sequence.
n
Consider the sequence for u = (–1) , n  Z , which has the terms –1, 1, –1, 1, … This sequence oscillates
+
n
finitely between –1 and 1. Therefore, this sequence is also a periodic sequence.
n
The sequence for u = (–10) , n  Z , has terms –10, 100, –1000, 10 000, …, and oscillates infinitely between
+
n
–∞ and ∞. Therefore, this sequence is not a periodic sequence.

Convergent and divergent sequences
Consider the sequence
n
1.1, 1.01, 1.001, 1.0001, …, 1 + (0.1) , …
The n term of this sequence is
th
1
u = 1 + 1 2 n
n
10
1
1
When n → ∞, 1 2 n → 0 and 1 + 1 2 n → 1.
10
10
We write
1
n
lim u = lim 1 + 1 2 4 = 1.
3
x → ∞ n n → ∞ 10
We say that this sequence converges to the value 1.
lim
u exists, then this sequence is called a convergent
In general, if u is the n term of a sequence and n → ∞ n
th
n
sequence and the value of the limit of u is known as the limiting value or limit of sequence.
n
A sequence which is not convergent is called a divergent sequence.
94
02 STPM Math T T1.indd 94 3/28/18 4:21 PM

Mathematics Term 1 STPM Chapter 2 Sequences and Series

Example 3


Determine if each of the following sequences is convergent or divergent.
(a) 3, 3, 3, …, 3, …
1
1
1 2
(b) 1, – , 1 , – , .... , (–1) n – 1 1 , …

2 3 4 n
2
3
(c) 1, r, r , r , …, r n – 1 , … if
(i) |r| , 1, (ii) |r|  1.
Solution: (a) u = 3

lim u = lim 3 = 3.
n → ∞ n n → ∞ 2
The sequence is convergent.
1 2
(b) u = (–1) n – 1 1
n
n
lim u = lim 5 (–1) n – 1 1
1 26
n → ∞ n n → ∞ n
1
= (–1) n – 1 lim 1 2
n → ∞ n
= (–1) n – 1 · 0
= 0
The sequence is convergent.
(c) u = r n – 1
n
lim u = lim (r n – 1 )
n → ∞ n n → ∞
lim u = 0 if |r| , 1
n → ∞ n
and lim u = ∞ if |r|  1.
n → ∞ n
The sequence is convergent if |r| , 1 and divergent if |r|  1.



Properties of the limits of sequences

lim
lim
v = B, then
u = A and
If n → ∞ n n → ∞ n
(a) lim (u ± v ) = lim u ± lim v = A ± B. (b) lim (u · v ) = lim u · lim v = A · B.
n → ∞ n n n → ∞ n n → ∞ n n → ∞ n n n → ∞ n n → ∞ n
lim
u
u
n
(c) lim 1 2 = n → ∞ n = A , provided B ≠ 0.
n → ∞ v lim B
n v
n → ∞ n
Example 4
1
If u and v are the n terms of two sequences where u = 3 and v = respectively, find the limit of each
th
n
n
n
n
n
sequence when n → ∞. Hence, find lim 1 3n + 1 2 .
n → ∞ n
Solution: lim u = lim 3 = 3.
n → ∞ n n → ∞
1
lim v = lim 1 2 = 0

n → ∞ n n → ∞ n
95


02 STPM Math T T1.indd 95 3/28/18 4:21 PM

Mathematics Term 1 STPM Chapter 2 Sequences and Series

lim 1 3n + 1 2 = lim 1 3 + 1 2
n → ∞ n n → ∞ n
1
lim
lim
= n → ∞ 3 + n → ∞ 1 2
n
= 3 + 0
= 3

The summation notation ∑

Suppose we want to find the sum of the series of numbers
2 1 + 2 + 3 + … + 100

One simpler way of representing the sum of a series such as this is by using the Greek alphabet ∑, read as
100
“sigma”. For example, we write ∑ r to represent the sum 1 + 2 + 3 + … + 100, where the r term, u , is r, i.e.
th
r = 1 r
100 100
∑ u = ∑ r = 1 + 2 + 3 + … + 100.
r = 1 r r = 1
200
Similarly, we write ∑ to represent the sum 101 + 102 + 103 + … + 200, i.e.
r = 101
200
∑ r = 101 + 102 + 103 + … + 200.
r = 101
Notice that
101 + 102 + … + 200 = (100 + 1) + (100 + 2) + … + (100 + 100)
So, we can also write
200 100
∑ r = ∑ (100 + r).
r = 101 r = 1
In general, if u is the r term of a series, then
th
r
n
∑ u = u + u + u + … + u
r = 1 r 1 2 3 n
Notice that if k is a constant,
n
∑ (ku ) = ku + ku + … + ku n
r
1
2
r = 1 = k(u + u + … + u )
n
2
1
n
= k ∑ u
r = 1 r
When u = u = u = … = u = k, then
2
3
1
n
n
∑ k = k + k + … + k
r = 1 = nk
n
∑ (u + v ) = (u + v ) + (u + v ) + … + (u + v )
r = 1 r r 1 1 2 2 n n
= (u + u + … + u ) + (v + v + … + v )
n
n
1
1
2
2
n n
= ∑ u + ∑ v
r = 1 r r = 1 r
n n n
Similarly, ∑ (u – v ) = ∑ u – ∑ v
r = 1 r r r = 1 r r = 1 r
96
02 STPM Math T T1.indd 96 3/28/18 4:21 PM

Mathematics Term 1 STPM Chapter 2 Sequences and Series

Example 5

Express each of the following series by using the summation notation ∑.
(a) 1 + 4 + 9 + 16 + … + 100
(b) 1 · 2 + 2 · 3 + 3 · 4 + … + 19 · 20

2
2
2
2
2
Solution: (a) 1 + 4 + 9 + 16 + … + 100 = 1 + 2 + 3 + 4 + … + 10
th
∴ the r term, u = r 2 10
r
Hence, the series may be written as ∑ r .
2
r = 1
(b) The r term, u = r(r + 1). 19
th
r
Hence, the series may be written as ∑ r(r + 1). 2
r = 1
Example 6
Rewrite each of the following series by using the summation notation ∑.
(a) 2 + 5 + 10 + 17 + … + 401 (b) 1 + 1 + 1 + … + 1
2 3 4 50
Solution: (a) 2 + 5 + 10 + 17 + … + 401
2
2
2
2
= (1 + 1) + (2 + 1) + (3 + 1) + (4 + 1) + … + (20 + 1)
2
th
2
The r term, u = r + 1, (1  r  20)
r
20
Hence, the series may be written as ∑ (r + 1).
2
r = 1
(b) 1 + 1 + 1 + … + 1
2 3 4 50
= 1 + 1 + 1 + … + 1
1 + 1 2 + 1 3 + 1 49 + 1
1
th
The r term, u = r + 1 , 1  r  49.
r
49
Hence, the series may be written as ∑ 1 .
r = 1 r + 1
Example 7
Write down the first three terms and the last term of the following series.
20 10
(a) ∑ r(r + 2) (b) ∑ (–1) r + 1 r
2
r = 1 r = 1
Solution: (a) The r term, u = r(r + 2)
th
r
Thus u = 1(1 + 2) = 3
1
u = 2(2 + 2) = 8
2
u = 3(3 + 2) = 15
3
The last term, u = 20 (20 + 2) = 440
20
(b) The r term, u = (–1) r + 1 r
2
th
r
2 1
Thus u = (–1) 2 = 2
1
u = (–1) 2 = –4
3 2

2
u = (–1) 2 = 8
4 3
3
The last term, u = (–1) 2 = –1024
11 10
10
97
02 STPM Math T T1.indd 97 3/28/18 4:21 PM

Mathematics Term 1 STPM Chapter 2 Sequences and Series

Exercise 2.1


th
1. In each of the sequences below, write down the next two terms and state the n term.
(a) 5, 10, 15, 20, … (b) 4, 7, 10, 13, … (c) 2, 5, 8, 11, …
3
(d) 2, 4, 8, 16, … (e) 3, – , 3 , – 3 , … (f) 1 , 3 , 5 , 7 , …
5 25 125 3 5 7 9
,
(g) 1, 5 , 25 125 , … (h) 1 , 1 , 1 , 1 , … (i) 1, –2, 3, –4, …

2 4 8 2 6 12 20
2. Write down the first four terms of each of the following sequences.
(a) 5 n 6 (b) 5 (–1) n + 1 6 (c) 5 (2x) n – 1 26
2 n + 1 n! (2n – 1)
n 2n – 1
(–1) x
n 6
(d) 5 (2n – 1)(2n + 1) 6 (e) 5 cos nx
nx
3. Find the n term of each of the following sequences.
th
1
(a) – , 3 , – 5 , 7 , – 9 , … (b) 1, 0, 1, 0, 1, … (c) 2 , 0, 3 , 0, 4 , …
5 8 11 14 17 3 4 5
4. Find the next five terms of each of the sequences below, defined by the recursive formula given.
n
(a) u n + 1 = nu – n, u = 3. (b) u n + 1 = 2u – (–1) , u = 1.
n
1
n
1
(c) u n + 2 = 3u n + 1 – 2u , u = 1, u = 3.
2
n
1
5. Determine if each of the following sequences are convergent or divergent. For those that are convergent,
find their limits.
(a) 1 + 1 , 1 + 1 , 1 + 1 , … (b) 1 + 1 , 2 + 1 , 3 + 1 , …
2 3 4 2 3 4
(c) 2 , 2 , 2 , … (d) 3 – 1 , 3 – 2 , 3 – 3 , …
1 + 1 1 + 1 1 + 1 2n 3n 4n
2 3 4
6. Evaluate
2
3
1
(a) lim 1 2 (b) lim 1 2 2 (c) lim n + 2 2
n → ∞ n n → ∞ n – 1 n → ∞ n
2
1
(d) lim n + 2 2 (e) lim 1 2n + 2n 2 (f) lim 1 1 + 2 × 5 n n 2
n → ∞ n – 1 n → ∞ 5n – n n → ∞ 4 + 3 × 5
2
7. Write each of the following series by using the ∑ notation.
(a) 1 + 8 + 27 + 64 + … + 1 000 (b) 1 + 3 + 5 + 7 + … + 99
(c) 1 + 1 + 1 + 1 + … + 1 (d) 1 – 1 + 1 – 1 + 1 – 1
2 4 8 512 3 9 27 81 243
(e) 11 + 8 + 5 + 2 – 1 – 4 – 7 (f) 5 + 7 + 11 + 19 + 35 + 67
3
2
(g) –x + 2x – 3x + … + 10x 10 (h) 1 + 2 + 3 + … + 8
2 · 3 3 · 4 4 · 5 9 · 10
8. Write down the terms of each of the following series without the ∑ notation.
10 8 1 5
(a) ∑ (r – r) (b) ∑ (c) ∑ (3r + 2)
2
r = 1 r = 1 r 2 r = 1
9 10 7
(d) ∑ (–1) r – 1 (3r) (e) ∑ (8 – r) (f) ∑ (r + 2)(r + 4)
r = 1 r = 1 r = 1
5 5
2
(g) ∑ (6 – r) 2 (h) ∑ (2r – 3r – 5)
r = 1 r = 1
98





02 STPM Math T T1.indd 98 3/28/18 4:21 PM

Mathematics Term 1 STPM Chapter 2 Sequences and Series

2.2 Series


Arithmetic series

Consider the sequence of numbers
2, 5, 8, 11, …, 29
Each term (except the first term) in this sequence can be obtained by adding a fixed number 3 to the term
before it. Hence, the above sequence can also be written as
2, [2 + 1(3)], [2 + 2(3)], [2 + 3(3)], …, [2 + 9(3)].

A sequence such as this is called an arithmetic progression and the fixed number is called the common difference. 2
For the arithmetic progression
10, 7, 4, 1, –2, …, –17

the common difference is 7 – 10, i.e. –3.
Hence, the arithmetic progression can also be written as
10, [10 + 1(–3)], [10 + 2(–3)], [10 + 3(–3)], …, [10 + 9(–3)].
If the first term of an arithmetic progression is a and the common difference is d, then this arithmetic progression
can be represented by
a, (a + d), (a + 2d), (a + 3d), …, [a + (n – 1)d].
th
and the n term is
u = a + (n – 1)d
n


Example 8


Given that the fifth term of an arithmetic progression is 21 and its tenth term is 41, find the common
th
difference, first term and the n term of this arithmetic progression.
Solution: Let a be the 1 term and d the common difference.
st
So, the 5 term is a + 4d,
th
i.e. a + 4d = 21 ………… 
th
The 10 term is a + 9d,
i.e. a + 9d = 41 ………… 
 – : 5d = 20
d = 4
Substituting d = 4 into :
a + 16 = 21
a = 5

Hence, the 1 term is 5 and the common difference is 4.
st
th
The n term is u = a + (n – 1)d
n
= 5 + (n – 1)4
= 1 + 4n


99





02 STPM Math T T1.indd 99 3/28/18 4:21 PM

Mathematics Term 1 STPM Chapter 2 Sequences and Series

Example 9

th
The n term of an arithmetic progression is 1 (5n – 3). Obtain the first two terms and also the common
difference of the arithmetic progression. 10
1
Solution: The n term, u = 10 (5n – 3)
th
n
1
The 1 term, u = 10 (5 – 3)
st
1
= 1
5
2 The 2 term, u = 10 (5 × 2 – 3)
1
nd
2
= 7
10
The common difference, d = u – u 1
2
= 7 – 1
10 5
= 1
2


Sum of a finite arithmetic series

When the terms of an arithmetic progression are added up, we will obtain an arithmetic series.
Consider the following arithmetic series made up of 10 terms, i.e.
S 10 = 3 + 6 + 9 + 12 + … + 27 + 30 ………… 
By rewriting the terms of the series backwards, we obtain
S = 30 + 27 + 24 + 21 + … + 6 + 3 ………… 
10
Adding,  + : 2S = 33 + 33 + 33 + 33 + … + 33 + 33
10
= 33 × 10 (since the series has 10 terms)
= 330
∴ S = 165
10
th
For any arithmetic series with first term a, common difference d and n term l, the series can be written as
S = a + (a + d) + (a + 2d) + … + (l – d) + l ………… 
n
By rewriting the terms of the series backwards, we obtain
S = l + (l – d) + (l – 2d) + … + (a + d) + a ………… 
n
 + : 2S = (a + l) + (a + l) + (a + l) + … + (a + l) + (a + l)
n
= n(a + l) since the series has n terms
n
∴ S = — (a + l).

n 2
Since the n term, l = a + (n – 1) d,
th
S = — [a + a + (n – 1) d]

n
n 2 Arithmetic

n
i.e. S = — [2a + (n – 1) d] Sequence
n 2 VIDEO
Both these formulae derived can be used to find the sum of the first n terms of an arithmetic series.

100





02 STPM Math T T1.indd 100 3/28/18 4:21 PM

Mathematics Term 1 STPM Chapter 2 Sequences and Series

Example 10

Find the sum of the integers which can be divided by 6 exactly and which lie in between 50 and 150.
st
Solution: The 1 term divisible by 6 is 54.
The last term divisible by 6 is 144.
The required series is
S = 54 + 60 + 66 + … + 144
n
This is an arithmetic progression with a = 54 and d = 60 – 54 = 6.
The last term, l = a + (n – 1)d = 144
Hence, 54 + (n – 1)6 = 144
6n = 96 2
n = 16
Hence, the sum of this series is
S = 16 (a + l)
2
16
= 16 (54 + 144)
2
= 8 × 198
= 1584


Example 11

In an arithmetic progression, the sum of the first ten terms is 520 and the 7 term is twice the 3 term.
th
rd
Find the first term, a, and the common difference, d.
Solution: The sum of the first ten terms is
S = 10 (2a + 9d) = 520
10
2
2a + 9d = 104 ………… 
rd
th
The 7 term is a + 6d and the 3 term is a + 2d.
a + 6d = 2(a + 2d)
a = 2d ………… 
Substituting  into : 4d + 9d = 104
13d = 104
d = 8
a = 16
st
Hence, for the given arithmetic progression, the 1 term is 16 and the common
difference is 8.


Example 12
n
r
Find the sum of the first 10 terms of the series ∑ ln 3 , and the smallest value of n such that the sum of
the first n terms exceeds 1000. r = 1

r
Solution: By substituting r = 1, 2, 3, … into ln 3 , we have
n
2
r
3
∑ ln 3 = ln 3 + ln 3 + ln 3 + … + ln 3 n
r = 1
= ln 3 + 2 ln 3 + 3 ln 3 + … + n ln 3.
101





02 STPM Math T T1.indd 101 3/28/18 4:21 PM

Mathematics Term 1 STPM Chapter 2 Sequences and Series

This series is an arithmetic series with 1st term, a = ln 3 and common difference
d = ln 3, i.e.
a = d = ln 3.
10
∑ ln 3 = ln 3 + 2 ln 3 + 3 ln 3 + … + 10 ln 3
r
r = 1 = (1 + 2 + 3 + … + 10) ln 3
= 10 (1 + 10) ln 3
2
= 55 ln 3
n
and ∑ ln 3 = (1 + 2 + 3 + … + n) ln 3
r
r = 1 n
= (1 + n) ln 3.
2
2 We have to find the smallest value of n such that
n
(1 + n) ln 3  1000
2
n(1 + n)  2000
ln 3
n(1 + n)  1821
By using the calculator, we find that
if n = 42, n(1 + n) = 42 × 43 = 1806 , 1821
if n = 43, n(1 + n) = 43 × 44 = 1892  1821
n
Hence, the smallest value of the integer n such that ∑ ln 3  1000 is 43.
r
r = 1
Example 13

th
The n term of an arithmetic progression is 32 and its first term is 2. If the sum of the first n terms is 357,
th
find the value of n. If the smallest term in the series which exceeds 100 is the k term, find the value of k.
st
Solution: The 1 term, a = 2
th
The n term, a + (n – 1)d = 32 d = common difference
2 + (n – 1) d = 32
(n – 1) d = 30 ………… 
st
Sum of the 1 n terms is 357.
Thus n [2a + (n – 1) d] = 357
2
i.e. n (4 + 30) = 357
2 17n = 357
n = 21
Substitute n = 21 into ,
20d = 30
d = 3
2
th
The k term, u = a + (k – 1) d
k
= 2 + (k – 1) · 3
2
= 1 (3k + 1)
2
For 1 (3k + 1)  100
2
3k + 1  200
3k  199
k  66.3
Hence, the smallest term in the series which exceeds 100 is the 67 term, i.e. k = 67.
th


102





02 STPM Math T T1.indd 102 3/28/18 4:21 PM

Mathematics Term 1 STPM Chapter 2 Sequences and Series

Example 14

The sum of the first n terms of an arithmetic progression is given by S = pn + qn .
2
n
Given that S = 6 and S = 11,
3
5
(a) find the values of p and q,
th
(b) deduce, or otherwise, an expression for the n term and the value of its common difference.
Solution: (a) S = pn + qn 2
n
When n = 3, S = 3p + 9q = 6
3
p + 3q = 2 ………… 
When n = 5, S = 5p + 25q = 11
5
p + 5q = 11 …………  2
5
 – : 2q = 1
5
q = 1
10
Substituting q = 1 into ,
10
p + 3 = 2
10
3
p = 2 – 10
= 17
10
Hence, p = 17 and q = 1 .
10 10
(b) S = pn + qn 2
n
Thus S n – 1 = p(n – 1) + q(n – 1) 2
The n term, u = S – S n – 1
th
n
n
= pn + qn – p(n –1) – q(n – 1) 2
2
= pn + qn – pn + p – q(n – 2n + 1)
2
2
= p + 2nq – q
1
= 17 + 2n 1 2 – 1
10 10 10
= 1 (n + 8)
5
The common difference, d = u – u n – 1
n
= 1 (n + 8) – 1 (n – 1 + 8)
5 5
= 1 (n + 8) – 1 (n + 7)
5 5
= 1
5
Check:
u = 1 (2 + 8) = 10
2
5
5
u = 1 (1 + 8) = 9
1
5
5
d = u – u 1
2
= 10 – 9
5 5
= 1
5
103





02 STPM Math T T1.indd 103 3/28/18 4:21 PM

Mathematics Term 1 STPM Chapter 2 Sequences and Series
Arithmetic mean
If a, b and c are three consecutive terms of an arithmetic progression, the term b is called the arithmetic mean
of a and c.
The common difference = b – a = c – b
∴ 2b = a + c
b = 1 (a + c)
2

The arithmetic mean of two numbers, p and q, is 1 (p + q).
2

2
Exercise 2.2


th
1. For each of the following arithmetic series, write down the term indicated in brackets, and the n term.
th
(a) 7 + 11 + 15 + … (7 term) (b) 18 + 11 + 4 + … (6 term)
th
2
1
th
(c) –7 – 5 – 3 – … (20 term) (d) 3 + 3 + 4 + … (15 term)
th
3 3
2. Find the sum of the terms of each of the following series.
(a) 5 + 9 + 13 + … + 81 (b) 85 + 82 + 79 + … + 13
1
3
(c) –22 – 17 – 12 – … + 68 (d) 1 + 1 + 2 + … + 18 2
5 5 5
3. Find the sum of each of the following arithmetic series.
1
th
th
(a) 3 + 8 + 13 + … to 18 term (b) 4 + 7 + 11 + … to 20 term
2
th
th
(c) 21 + 18 + 15 + … to 12 term (d) –15 – 9 – 3 – … to 10 term
th
th
4. Find the sum of the arithmetic series –11 – 7 – 3 + 1 + … from the 11 term to the 20 term.
5. Find the sum of the odd numbers between 0 and 500 which are divisible by 7.
6. Find the sum of the numbers between 1 and 200 inclusive, which are not divisible by 6.
7. The first and last terms of an arithmetic series are 29 and 179 respectively. If the total number of terms
is 25, find the common difference and the sum of the series.
8. The second and seventh terms of an arithmetic series are –5 and 10 respectively. Find the eighth term
and the smallest value of n such that the sum of n terms exceeds 500.
th
nd
9. In an arithmetic series, the sum of the first 15 terms is 615, and the 13 term is 6 times the 2 term.
Find the first three terms.
2
10. The sum of the first n terms of a series is 3n + n. Show that the series is an arithmetic progression, and
find the first term and common difference.
11. In an arithmetic progression, the sum of the first 2n terms is equal to the sum of the following n terms.
If the first term is 12 and common difference is 3, find the value of n.
2
12. Show that the sum of the odd integers from 1 till (2n – 1) is n . Find the smallest value of n such that
the sum of n terms exceeds 4 000.
13. Find the arithmetic mean of
(a) 3 and 27 (b) 3 and –27 (c) 1 and 1 (d) log 3 and log 27
3 27 10 10



104





02 STPM Math T T1.indd 104 3/28/18 4:21 PM

Mathematics Term 1 STPM Chapter 2 Sequences and Series
Geometric series

Consider the sequence of numbers
1, 2, 4, 8, 16, …
Each term (except the first term) in the sequence is obtained by multiplying the previous term with a fixed
number 2.
Thus, this sequence can also be written as
4
1, 1 × 2, 1 × 2 , 1 × 2 , 1 × 2 , …
2
3
This type of sequence is called a geometric progression and the fixed number is called common ratio.
If a geometric progression has first term 3 and common ratio –2, then the terms are
2
3, 3(–2), 3(–2) , 3(–2) , … 2
3

or 3, –6, 12, –24, …
If the first term of a geometric progression is a and its common ratio is r, then the geometric progression may
be represented as
2
a, ar, ar , …, ar n – 1
th
with its n term,
u = ar n – 1
n

Sum of a finite geometric series
When the terms of a geometric progression are added up, we will obtain a geometric series.
Consider the following geometric series which is made up of 10 terms, with the first term 1 and common ratio
5, i.e.
2
3
S = 1 + 1(5) + 1(5) + 1(5) + … + 1(5) 9
10
3
2
9
S = 1 + 5 + 5 + 5 + … + 5 ………… 
10
10
4
2
3
 × 5: 5S = 5 + 5 + 5 + 5 + … + 5 ………… 
10
10
 – : (5 – 1) S = 5 – 1
10
10
S = 5 – 1
5 – 1
10
= 1 (5 – 1)
10
4
For any geometric series with first term a and common ratio r ≠ 1, the sum of the first n terms, S , can be
n
written as
3
S = a + ar + ar + ar + … + ar n – 1 ……… 
2
n
2
n
 × r: rS = ar + ar + ar + ar + … + ar ……… 
4
3
n
n
 – : (r – 1)S = ar – a
n
n
S = a(r – 1) , for r  1
r – 1
n
Geometric
n
or S = a(1 – r ) , for r , 1 VIDEO Sequence
1 – r
n
105
02 STPM Math T T1.indd 105 3/28/18 4:21 PM

Mathematics Term 1 STPM Chapter 2 Sequences and Series

Example 15

The 5 term of a geometric series is 18 and the 7 term is 54. Given that the sum of the first 10 terms of
th
th
this series is positive, find the 1 term, the common ratio and the sum of the first 10 terms.
st
Solution: Let a be the 1 term and r the common ratio.
st
4
th
the 5 term, ar = 18 ………… 
6
th
and the 7 term, ar = 54 ………… 
2
 ÷ : r = 54 = 3
18
r = ± 3
2 Substituting r = ± 3 into ,
a(9) = 18
a = 2
n
From the formula S = a(r – 1) ,
r – 1
n
10
when r = 3 , S = 2[( 3) – 1]
10
3 – 1
= 2(243 – 1)
3 – 1
= 484
3 – 1
10
When r = – 3 , S = 2[1 – (– 3) ]
10
1 – (– 3 )
2(1 – 243)
=
1 + 3
= – 484
3 + 1
Since it is given that S  0, thus
10
484
a = 2, r = 3 and S = 3 – 1 .
10
Example 16

The first term of a geometric series is 5 and the common ratio is 1.5. Find the number of terms needed
such that the sum of the series exceeds 200.
Solution: Given that a = 5 and r = 1.5
The sum of the 1 n terms is
st
5[(1.5) – 1]
n
n
S = 1.5 – 1 = 10[(1.5) – 1]
n
For S to exceed 200,
n
n
10[(1.5) – 1]  200
n
(1.5) – 1  20
n
(1.5)  21
n log 1.5  log 21
10
10
log 21
n  10
log 1.5
10
= 7.5
Hence, the smallest value of n is 8, i.e. 8 terms are needed such that its sum exceeds 200.
106


02 STPM Math T T1.indd 106 3/28/18 4:21 PM

Mathematics Term 1 STPM Chapter 2 Sequences and Series
Geometric mean
If a, b and c are three consecutive terms of a geometric progression, the term b is the geometric mean of a and c.
b
c
The common ratio = — = —
a b
2
∴ b = ac
b = (ac)
The geometric mean of two numbers, p and q, is (pq).



Example 17 2

If (x + 1), 2 2 and (3x – 2) are three consecutive terms of a geometric progression, find the integral value of x.

Solution: (x + 1), 2 2 and (3x – 2) forms a geometric progression.

Thus, 2 2 is the geometric mean of (x + 1) and (3x – 2)
i.e. 2 2 = (x + 1)(3x – 2)
8 = (x + 1)(3x – 2)
2
8 = 3x + x – 2
2
3x + x – 10 = 0
(3x – 5)(x + 2) = 0
x = 5 or –2
3
Hence, the integral value of x is –2.



Exercise 2.3


th
1. For each of the following geometric series, write down the term indicated in brackets and the n term.
th
(a) 1 + 1 + 2 + … (8 term) (b) 162 + 54 + 18 + … (6 term)
th
2
4
1
th
th
(c) 200 – 50 + 12 + … (5 term) (d) – – 2 – 1 – … (7 term)
2 9 3
2. Find the number of terms in each of the geometric series below, and also the sum of the series.
(a) 1 + 1 + … + 64 (b) 9 – 6 + … + 256
4 2 729
(c) 100 + 50 + … + 25 (d) 2 – 8 + … – 14 2150
16 3 2187
3. Find the sum of each of the following geometric series.
th
th
(a) 100 + 20 + … to 8 term (b) 4 – 2 + … to 10 term
th
th
k
(c) 2 – 6 + … to n term (d) a + a k + 2 + … to n term
4. The terms of a geometric series are positive, with the first term 80. If the sum of the first three terms is
185, find the common ratio of the series.
5. If two numbers, m and n, are such that m, n and 10 form an arithmetic progression, whereas n, m and
10 form a geometric progression, find the values of m and n.



107





02 STPM Math T T1.indd 107 3/28/18 4:21 PM

Mathematics Term 1 STPM Chapter 2 Sequences and Series
6. Find the geometric mean of each of the following numbers.

(a) 3 and 27 (b) 1 and 1 (c) 10 and 10 27
3
3 27
7. Given that the geometric mean of 4p – 3 and 9p + 4 is 6p – 1, find the values of p.
8. The second and fifth terms of a geometric series are 405 and –120 respectively. Find the seventh term
and the sum of the first seven terms of the series.
9. In a geometric progression, the second term exceeds the first term by 20 and the fourth term exceeds the
second term by 15. Find the possible values of the first term.

1
1
3
2 10. Find the sum of the first n terms of the series 12 + + + … . Find also the number of terms required
4
4
such that the sum exceeds 100.
11. A geometric series has first term 16 and common ratio 3 . If the sum of the first n terms exceeds 60,
find the smallest value of n. 4
Sum of an infinite geometric series

When the number of terms of a geometric series is infinite, it is called an infinite geometric series.
Consider the infinite geometric series
1
1 + 1 + 1 + … + 1 2 n – 1 + …
2 4 2
The sum of the first n terms is
1
n
3
1 1 – 1 2 4
S = 2
n
1 – 1
2
1
n
3
= 2 1 – 1 2 4
2
1
As n increases, 1 2 n → 0 and S → 2.
n
2
lim
S = 2.
We say the limit of S as n → ∞ exists and equals 2, and we write n → ∞ n
n
This series is called a convergent series with a sum of 2. Notice that the common ratio is 1 , 1.
2
We now consider the infinite geometric series
1 + 2 + 4 + … + 2 + …
n
The sum of this series is infinite.
This series is called a divergent series.
In general, for a geometric series with first term a and common ratio r,
n
S = a(1 – r )
n
1 – r
= a – 1 a 2 r n
1 – r 1 – r




108





02 STPM Math T T1.indd 108 3/28/18 4:21 PM

Mathematics Term 1 STPM Chapter 2 Sequences and Series
lim
S
S = n → ∞ n
∞
= lim 3 a – 1 a 2 4
n
r
n → ∞ 1 – r 1 – r
= a – 1 a 2 lim r n
1 – r 1 – r n → ∞
n
If |r|  1, lim r → ∞, and the series is divergent.
n → ∞
n
If |r| , 1, lim r → 0, and the series is convergent with sum to infinity S = a .
n → ∞ ∞ 1 – r
2
When |r| , 1, the geometric series 1 + a + ar + ar + … is convergent, 2
a
with the sum to infinity S = 1 – r .
∞


Example 18

Find the sum of each of the following series.
(a) 18 – 6 + 2 – … (b) 1 – 1 + 1 – …
4 16

1
1
Solution: (a) 18 – 6 + 2 + … = 18 + 18 – 1 2 + 18 – 1 2 2 + …
3
3
This is a geometric series with 1 term a = 18.
st
Common ratio r = – 1
3
u
|r| = – 1 u = 1 , 1
3
3
a
Sum to infinity S = 1 – r
∞
18
=
1
1 – – 1 2
3
= 54
4
= 13 1
2
1
1
(b) 1 – 1 + 1 – … = 1 + 1 – 1 2 + 1 – 1 2 2 + …
4 16 4 4
This is a geometric series with 1 term a = 1.
st
Common ratio r = – 1
4
1
|r| = |– | = 1 , 1
4 4
a
Sum to infinity S = 1 – r
∞
1
=
1
1 – – 1 2
4
= 4
5
109





02 STPM Math T T1.indd 109 3/28/18 4:21 PM

Mathematics Term 1 STPM Chapter 2 Sequences and Series

Example 19

Express the recurring decimal 0.327 as an infinite geometric series. Hence, express 0.327 as a fraction in
its simplest form.
(Note: The recurring decimal 0.327 = 0.327327327…)

Solution: 0.327 = 0.327327327…
= 0.327 + 0.000327 + 0.000000327 + …
= 0.327 + 0.327 × 10 + 0.327 × 10 + …
–3
–6
–3
This is an infinite geometric series with a = 0.327 and r = 10 = 0.001.
2 Hence, 0.327 = 0.327 a
1 – 0.001 Using S = –––––
∞
1 – r

= 0.327
0.999
= 327
999
= 109
333

Example 20

Find the sum of the series 1 + 3x + 9x + 27x + … , stating the range of values of x for which the result
2
3
is valid.
2
3
Solution: 1 + 3x + 9x + 27x + …
is a geometric series with a = 1 and r = 3x.
1
Hence, the sum to infinity is S = 1 – 3x .
∞
The result is valid if |3x| , 1,
i.e. |x| , 1
3
1
or – , x , 1
3 3


Example 21

The first term of a geometric series is 2 and the common ratio is 0.95. The sum of the first n terms of this
series is S and the sum of this series is S . Find the smallest value of n such that S – S , 1.
n
n
∞
∞
Solution: The first term, a = 2
Common ratio, r = 0.95 n
a
Using the formulas S = a(1 – r ) and S = 1 – r ,
∞
1 – r
n
S – S , 1
∞
n
n
⇒ a – a(1 – r ) , 1
1 – r 1 – r
a – a + ar n , 1
1 – r 1 – r 1 – r
ar n , 1
1 – r
110




02 STPM Math T T1.indd 110 3/28/18 4:21 PM

Mathematics Term 1 STPM Chapter 2 Sequences and Series

Substituting the values a = 2 and r = 0.95.
2(0.95) n , 1
1 – 0.95
n
(0.95) , 0.025
n log (0.95) , log (0.025)
10
10

n(–0.0223) , –1.6021
n  –1.6021 = 71.8
– 0.0223
i.e. the smallest value of n such that S – S , 1 is 72.
n
∞
2
Exercise 2.4

1. Determine whether each of the following series is convergent or otherwise.
(a) 2 + 2 + 2 + … (b) 18 + 15 + 12 + … (c) 40 – 20 + 10 – …
3 3 2
1
3
(d) 4 + 8 + 16 + … (e) k + 2k + 3k + … (f) 7 – 3 + 1 – …
5 15 45 2 4
2. Find the sum of each of the following geometric series.
(a) 6 + 2 + 2 + … (b) 1 – 1 + 1 – …
3 2 4
(c) 10 + 1 + 0.1 + … (d) 45 – 30 + 20 – …

3. Express each of the following recurring decimals as an infinite geometric series or as the sum of a constant
and an infinite geometric series. Hence, express each of the decimals as a fraction in its simplest form.
· · · · · · · ·

(a) 0.48 (b) 0.072 (c) 0.5813
· · · · · ·

(d) 0.3354 (e) 0.9218
4. Find the sum of each of the following infinite geometric series and state the range of values of x for which
the result is valid.
2
(a) 1 + 2x + 4x + … (b) 1 – 1 x + 1 x – …
2
2 4
2
3
(c) 3 – 6x + 12x – … (d) x + 1 x + 1 x + …
2
3 9
5. The sum of the first n terms of a geometric series is 2 n (3 – 1). Obtain the first three terms and the sum
n
to infinity of the series. 3
6. Find the value of r such that the sum of the series
2
1 + r + r + … + r n – 1 + …
is twice the sum of the series
2
3
1 – r + r – r + …
7. A ball rebounds to a height five-eighths of its previous height above the ground. If a ball is dropped from
a height of 3 metres, find the total distance travelled by the ball before it comes to rest.
1
8. The first three terms of a geometric series are 2, – and 1 respectively. Find the sum to infinity of the
2 8
series. Find 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


111





02 STPM Math T T1.indd 111 3/28/18 4:21 PM

Mathematics Term 1 STPM Chapter 2 Sequences and Series
1
th
9. Find the sum to the n term of the geometric series 108 + 60 + 33 + … . If s is the smallest number
3
which exceeds this sum for all values of n, find the value of s. Find also the smallest value of n such that
the sum of the series exceeds 99% of the value of s.
10. Write down the first four terms of a geometric series with sum 4 and first term 1. Find the value of r
such that the series 3
2
r + r 2 + r 2 + …
2 2
1 + r 2 (1 + r )
is convergent, and also the non zero value of the sum.
2
Summation of finite series involving powers of integers
Consider the series of the first n positive integers
1 + 2 + 3 + … + (n – 1)+ n.
This series can also be written as
n
∑ r = 1 + 2 + 3 + … + (n – 1) + n ………… 
r = 1
By rewriting this series backwards term by term,
n
∑ r = n + (n – 1) + (n – 2) + … + 2 + 1 ………… 
r = 1
n
 + : 2 ∑ r = (n + 1) + (n + 1) + (n + 1) + … + (n + 1) + (n + 1)
r = 1
= n(n + 1) Since the series has n terms
n
∑ r = 1 n(n + 1)
r = 1 2
Now, consider the series of the squares of the first n positive integers, i.e.
n
2
2
2
2
2
2
∑ r = 1 + 2 + 3 + … + (n – 1) + n .
r = 1
By using the identity
(r + 1)  r + 3r + 3r + 1,
3
2
3
3
2
we have (r + 1) – r = 3r + 3r + 1.
3
By adding the terms of the identity one by one with values of r from r = n to r = 1 , we get
n n n n
∑ [(r + 1) – r ] = 3 ∑ r + 3 ∑ r + ∑ 1
3
3
2
r = 1 r = 1 r = 1 r = 1
The LHS of this identity can be written as
3
3
3
3
3
3
[(n + 1) – n ] + [n – (n – 1) ] + … + (3 – 2 ) + (2 – 1 ) = (n + 1) – 1 3
3
3
3
3
= (n + 1) – 1
n n n
Thus, (n + 1) – 1 = 3 ∑ r + 3 ∑ r + ∑ 1
3
2
r = 1 r = 1 r = 1
n n n
3
2
3 ∑ r = (n + 1) – 1 – 3 ∑ r – ∑ 1
r = 1 r = 1 r = 1
3
= (n + 1) – 1 – 3 n(n + 1) – n
2
= (n + 1) – (n + 1) – 3 n(n + 1)
3
2
112
02 STPM Math T T1.indd 112 3/28/18 4:21 PM

Mathematics Term 1 STPM Chapter 2 Sequences and Series

2
= 1 (n + 1)[2(n + 1) – 2 – 3n]
2
2
= 1 (n + 1)(2n + 4n + 2 – 2 – 3n)
2
2
= 1 (n + 1)(2n + n)
2
= 1 n(n + 1)(2n + 1)
2
n
2
Thus ∑ r = 1 n(n + 1)(2n + 1)
r = 1 6
Similarly, by using the identity
3
4
4
2
(r + 1) – r = 4r + 6r + 4r + 1, 2
n 1
we get ∑ r = n (n + 1) 2
3
2
r = 1 4
= 3 1 2 n(n + 1) 4 2
n 2
1
= ∑ r 2
r = 1
Sum of the first n positive integers is
n 1
∑ r = n(n + 1).
r = 1 2
Sum of the squares of the first n positive integers is
n
∑ r = 1 n(n + 1)(2n + 1)
2
r = 1 6
Sum of the cubes of the first n positive integers is
n 1
2
3
2
∑ r = n (n + 1) .
r = 1 4
Example 22
Evaluate
100 50 25
(a) ∑ r (b) ∑ r 2 (c) ∑ r 3
r = 1 r = 1 r = 1
Solution: (a) By using
n 1
∑ r = n(n + 1)
r = 1 2
and substituting n = 100,
100 1
∑ r = (100)(101)
r = 1 2
= 5050
(b) By using
n 1
2
∑ r = n(n + 1)(2n + 1)
r = 1 6
and substituting n = 50,
50 1
∑ r = (50)(51)(101)
2
r = 1 6
= 42 925


113





02 STPM Math T T1.indd 113 3/28/18 4:21 PM

Mathematics Term 1 STPM Chapter 2 Sequences and Series

(c) By using
n 1
∑ r = n (n + 1) 2
2
3
r = 1 4
and substituting n = 25,
25
3
2
∑ r = 1 (25) (26) 2
r = 1 2
= 105 625

2 Example 23

2
2
Find the sum of the series 1 + 3 + 5 + 7 + … to n terms.
2
2
2
2
2
2
Hence, find the value of 1 + 3 + 5 + … + 99 .
2
2
2
Deduce the value of 51 + 53 + … + 99 .
Solution: The sum of the squares of the odd integers is
2
2
2
2
1 + 3 + 5 + 7 + …
The r term is (2r – 1) 2
th
n
2
Hence, the sum of the series is ∑ (2r – 1) .
r = 1
n n
2
∑ (2r – 1) = ∑ (4r – 4r + 1)
2
r = 1 r = 1
n n n
2
= 4 ∑ r – 4 ∑ r + ∑ 1
r = 1 r = 1 r = 1
4
4
= 4 3 n (n + 1)(2n + 1) – 4 3 n (n + 1) + n
2
6
= n [2(n + 1)(2n + 1) – 6(n + 1) + 3]
3
2
= n (4n + 6n + 2 – 6n – 6 + 3)
3
2
= n (4n – 1)
3
50 th
2
2
2
2
1 + 3 + 5 + … + 99 = ∑ (2r – 1) 2 n term, 2n – 1 = 99
r = 1 n = 50
2
= 50 [4(50) – 1]
3
= 50 × 9999
3
= 166 650
2
2
2
2
2
2
2
2
2
51 + 53 + … + 99 = (1 + 3 + … + 99 ) – (1 + 3 + … + 49 )
50 25
= ∑ (2r – 1) – ∑ (2r – 1) 2
2
r = 1 r = 1
2
= 166 650 – 25 [4(25) – 1]
3
= 166 650 – 20 825
= 145 825
114
02 STPM Math T T1.indd 114 3/28/18 4:21 PM

Mathematics Term 1 STPM Chapter 2 Sequences and Series

Example 24

th
Find the r term of the series
1 · 2 · 3 + 2 · 3 · 4 + 3 · 4 · 5 + …
n n
2
By using the results for ∑ r and ∑ r , find the sum of the first n terms of the above series.
r = 1 r = 1
Solution: The series is
1 · 2 · 3 + 2 · 3 · 4 + 3 · 4 · 5 + …
th
The r term is r(r + 1)(r + 2).
st
The sum of the 1 n terms of the series is
n n 2
2
3
∑ r(r + 1)(r + 2) = ∑ (r + 3r + 2r)
r = 1 r = 1
n n n
3
= ∑ r + 3 ∑ r + 2 ∑ r
2
r = 1 r = 1 r = 1
4
4
2
= n 2 (n + 1) + 3 3 n (n + 1)(2n + 1) + 2 3 n (n + 1)
4 6 2
= n (n + 1)[n(n + 1) + 2(2n + 1) + 4]
4
2
= n (n + 1)(n + 5n + 6)
4
= n (n + 1)(n + 2)(n + 3)
4
Summation of series using the method of differences
Consider the sum of the terms of a series such as
u + u + u + … + u n
3
2
1
th
Suppose that the r term, u , can be expressed in the form f(r) – f(r – 1), where f(r) is a function of r.
r
n n
Then ∑ u = ∑ [f(r) – f(r – 1)]
r = 1 r r = 1
= f(n) – f(n – 1)
+ f(n – 1) – f(n – 2)
+ f(n – 2) – f(n – 3)
+ … Substituting the values of
+ f(3) – f(2) r, term by term, with r = n,
n – 1, n – 2, …, 3, 2, 1
+ f(2) – f(1)
+ f(1) – f(0)
= f(n) – f(0)
th
This means that if the r term, u , of a series can be expressed as the difference of a function of r and a function
r
n
of (r – 1), then the sum of the series, ∑ u , can be determined.
r = 1 r
This method of finding the sum of a series is called the method of differences.
n
If u = f(r) – f(r – 1), then ∑ u = f(n) – f(0)
r
r
r = 1

115





02 STPM Math T T1.indd 115 3/28/18 4:21 PM

Mathematics Term 1 STPM Chapter 2 Sequences and Series

Example 25

Show that
r(r + 1)(r + 2)(r +3) – (r – 1)r(r + 1)(r + 2) = 4r(r + 1)(r + 2).
n
Hence, find ∑ r(r + 1)(r + 2).
r = 1
Solution: r(r + 1)(r + 2)(r + 3) – (r – 1)r(r + 1)(r + 2)
= r(r + 1)(r + 2)[(r + 3) – (r – 1)]
= 4r(r + 1)(r + 2)
2 Let f(r) = r(r + 1)(r + 2)(r + 3)
Then f(r – 1) = (r – 1)r(r + 1)(r + 2)
f(r) – f(r – 1) = r(r + 1)(r + 2)(r + 3) – (r – 1)r(r + 1)(r + 2)
= 4r(r + 1)(r + 2)
n
For the series ∑ r(r + 1)(r + 2), u = r(r + 1)(r + 2).
r = 1 r
f(r) – f(r – 1) = 4u r
or u = 1 [f(r) – f(r – 1)]
4
r
n n
∴ ∑ u = 1 ∑ [f(r) – f(r – 1)]
r = 1 r 4 r = 1
= 1 [f(n) – f(0)]
4
= 1 [n(n + 1)(n + 2)(n + 3) – 0]
4
n
Thus, ∑ r(r + 1)(r + 2) = 1 n(n + 1)(n + 2)(n + 3).
r = 1 4


Note: Compare the method of differences shown in Example 25 with the method shown in Example 24.
The method of differences can also be used if an expression can be expressed in partial fractions, as shown in
the following example.



Example 26

n
Express 1 in partial fractions. Hence, find ∑ 1 .
r(r + 1) r = 1 r(r + 1)
Solution: Let 1 ≡ A + B
r(r + 1) r r + 1
∴ 1 ≡ A(r + 1) + Br
Let r = 0 : 1 = A
Let r = –1 : 1 = –B
B = –1
Thus 1 = 1 – 1
r(r + 1) r r + 1



116





02 STPM Math T T1.indd 116 3/28/18 4:21 PM

Mathematics Term 1 STPM Chapter 2 Sequences and Series

n n
∑ 1 = ∑ 1 1 – 1 2
r = 1 r(r + 1) r = 1 r r + 1
n
= – ∑ 1 1 – 1 2
r = 1 r + 1 r
Let f(r) = 1 , f(r – 1) = 1
r + 1 r
n n
Thus ∑ 1 = – ∑ [f(r) – f(r – 1)]
r = 1 r(r + 1) r = 1
= –[f(n) – f(0)]
1
= – 1 n + 1 – 1 2 2

= 1 – 1
n + 1
= n
n + 1



Example 27


Show that 1 – 1 = – 2 .
(r + 1)(r + 2) r(r + 1) r(r + 1)(r + 2)

n
Using the method of differences, find ∑ 1 .
r = 1 r(r + 1)(r + 2)

Solution: 1 – 1 = r – (r + 2)
(r + 1)(r + 2) r(r + 1) r(r + 1)(r + 2)
= – 2
r(r + 1)(r + 2)

1
Thus 1 = – 3 1 – 1 4
r(r + 1)(r + 2) 2 (r + 1)(r + 2) r(r + 1)
1
1
Let u = r(r + 1)(r + 2) and f(r) = (r + 1)(r + 2) .
r
1
∴ u = – [f(r) – f(r – 1)].
r
2
n n
1
and ∑ u = – ∑ [f(r) – f(r – 1)]
r = 1 r 2 r = 1
1
= – [f(n) – f(0)]
2
1
= – 3 1 – 1 4
2 (n + 1)(n + 2) 2
n
3

∑ 1 = 1 1 – 1 4 .
r = 1 r(r + 1)(r + 2) 2 2 (n + 1)(n + 2)


117





02 STPM Math T T1.indd 117 3/28/18 4:21 PM

Mathematics Term 1 STPM Chapter 2 Sequences and Series

Example 28


Find the r term, u , of the series 2 · 1! + 5 · 2! + 10 · 3! + 17 · 4! + … + (n + 1) · n!
2
th
r
By expressing u as the difference of two functions of r, find the sum of the above series.
r
Solution: For the series
2 · 1! + 5 · 2! + 10 · 3! + 17 · 4! + … + (n + 1) · n!,
2
th
2
the r term, u = (r + 1) · r!
r
= [r(r + 1) – (r – 1)] · r!
2 = r(r + 1)r! – (r – 1) · r!
= r(r + 1)! – (r – 1)r! (r + 1)r ! = (r + 1)!


Let f(r) = r(r + 1)! and f(r – 1) = (r – 1)r!
Then u = f(r) – f(r – 1)
r
n n
and ∑ u = ∑ [f(r) – f(r – 1)]
r = 1 r r = 1
= f(n) – f(0)
= n(n + 1)! – 0
n
2
∴ ∑ (r + 1) r! = n(n + 1)!
r = 1

Exercise 2.5


1. Evaluate
4 12 8
(a) ∑ (r + 3r) (b) ∑ (100 – r ) (c) ∑ (4r + 5)
2
3
r = 1 r = 10 r = 3
5 90 8 rπ 5
r
(d) ∑ (e) ∑ sin (f) ∑ (–1) (1 + 2 r + 1 )
r = 1 r r = 1 3 r = 0
n n n n
2
3
2. Using the results for ∑ r, ∑ r and ∑ r , find the value of ∑ u for each of the following cases.
r = 1 r = 1 r = 1 r = 1 r
2
3
(a) u = 2r – r + 1 (b) u = r (r + 2)
r
r
(c) u = (r + 2)(r + 3)(2r – 1) (d) u = (r + 1)(r + 3)(r + 5)
r
r
3. For each of the following series, write down its r term. Hence, find the sum up to the n term.
th
th
(a) 1 · 4 + 2 · 7 + 3 · 10 + … (b) 1 · 5 + 2 · 6 + 3 · 7 + …
2
2
2
(c) 1 · 3 + 2 · 4 + 3 · 5 + … (d) 1 · 4 · 7 + 4 · 7 · 10 + 7 · 10 · 13 + …
(e) 1 + 1 + 1 + … (f) 1 · 2 · 3 + 2 · 3 · 4 + 3 · 4 · 5 + …
1 · 2 · 3 2 · 3 · 4 3 · 4 · 5
2
2
2
2
4. Given that 1 + 2 + 3 + … + n = 1 n(n + 1)(2n + 1), find the sum of the series
6
2
2
2
2 + 4 + 6 + … + 50 2
Using the above result, find the value of
2
2
2
2
1 + 3 + 5 + … + 49 .
3
3
3
3
5. Given that 1 + 2 + 3 + … + n = 1 n (n + 1) , find the sum of the first twenty terms of the series
2
2
4
2 + 16 + 54 + 128 + 250 + …
118
02 STPM Math T T1.indd 118 3/28/18 4:21 PM

Mathematics Term 1 STPM Chapter 2 Sequences and Series
6. Find the sum of each of the following series.
n 2n
(a) ∑ (3r – 1) 2 (b) ∑ r(r + 2)(r + 5)
r = 1 r = 1
2n 3n
2
(c) ∑ r(2r + 3) (d) ∑ r(r + 1)
r = n r = 1
n n
2
3
3
7. Using the identity r – (r – 1) ≡ 4r – 6r + 4r – 1 and ∑ r = 1 n(n + 1)(2n + 1), find ∑ r .
4
4
2
r = 1 6 r = 1
th
8. Using the method of differences, find the sum of the first n terms of the series whose r term, u , are as
r
follows.
2
1
(a) u = (r + 1)(r + 2) (b) u = (r + 2)(r + 3) 2
r
r
2r – 1
1
(c) u = r(r + 3) (d) u = r(r + 1)(r + 2)
r
r
9. Find the sum to n terms of the series.
2
2
2
2
(a) 1 · 5 + 5 · 9 + 9 · 13 + 13 · 17 + …
(b) 1 + 1 + 1 + 1 + …
1 · 3 · 4 2 · 4 · 5 3 · 5 · 6 4 · 6 · 7
n
10. If f(r) = 1 , simplify f(r) – f(r + 1). Hence, find ∑ r .
r! r = 1 (r + 1)!
n 2n
2
11. If ∑ u = 3n + 2n, find u . Hence, find ∑ u .
r = 1 r r r = n + 1 r
12. If f(r) = 1 , simplify f(r) – f(r + 1). Hence, find the sum of the first n terms of the series
r 2 3 5 7 …
2
2
1 · 2 2 + 2 · 3 2 + 3 · 4 2 +
2
13. Find the numbers A, B and C such that
1 + r ≡ A(r + 2)(r + 1) + B(r + 1) + C for all values of r.
2
Hence, prove that
n
∑ (1 + r ) r! = n(n + 1)!
2
r = 1
2.3 Binomial Expansions
n
The 1 2 and n! notation
r
Consider the product of the first 50 positive integers as follows:
50 × 49 × 48 × … × 3 × 2 × 1.
The product of these integers is a huge number. To simplify and for ease of writing, the product of these numbers
may be written as 50! (which is read as “factorial fifty”).
Hence,
10! = 10 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1
= 3 628 800



119





02 STPM Math T T1.indd 119 3/28/18 4:21 PM

Mathematics Term 1 STPM Chapter 2 Sequences and Series
In general, n! is defined as

+
n! = n(n – 1)(n – 2) … 3 · 2 · 1, with n  Z ,
and 0! = 1.
Notice that
(n + 1)! = (n + 1)[n(n – 1)(n – 2) … 3 · 2 · 1]
= (n + 1) · n!


Example 29

Evaluate 10! .
2 8!
Solution: 10! = 10 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1
8! 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1
= 10 × 9
= 90
Alternative method:
10! = 10 × 9! By using (n + 1)! = (n + 1) × n!
8! 8!
= 10 × 9 × 8!
8!
= 90


Example 30

Evaluate 18! .
15! 4!
Solution: 18! = 18 × 17 × 16 × 15!
15! 4! 15! × 4!
= 18 × 17 × 16
4 × 3 × 2 × 1
= 204


Example 31

Rewrite each of the following products in factorial form.
(a) 45 × 44 × 43 × 42 × 41 (b) 10 × 9 × 8
5 × 4 × 3
Solution: (a) 45 × 44 × 43 × 42 × 41
= 45 × 44 × 43 × 42 × 41 × 40 × 39 × … × 3 × 2 × 1
40 × 39 × … × 3 × 2 × 1
= 45!
40!

(b) 10 × 9 × 8 = 10 × 9 × 8 × 7! × 2!
5 × 4 × 3 5 × 4 × 3 7! 2!
= (10 × 9 × 8 × 7!) × 2!
(5 × 4 × 3 × 2!) × 7!
= 10! 2!
5! 7!



120





02 STPM Math T T1.indd 120 3/28/18 4:21 PM

Mathematics Term 1 STPM Chapter 2 Sequences and Series
In problems on permutations and combinations, the number of combinations of choosing r objects from n
n
different objects is given by 1 2 . For example, the number of ways of choosing 3 books out of 10 different
r
10
books is 1 2 , where
3
10
10!
1 2 = 3!7! = 120 ways
3
n
The notation 1 2 is called the binomial coefficient as it is found in binomial expansions.
r
n
The binomial coefficient 1 2 is defined as
r
n
n!
1 2 = (n – r)! r! for n, r  Z and 0  r  n.
+
r
Notice that 2
n
n!
1 2 = (n – 0)! 0! = 1 since 0! = 1
0
n
n!
1 2 = (n – 1)! 1! = n(n – 1)! = n
1
(n – 1)!
n
n!
1 2 = (n – 2)! 2! = n(n – 1)(n – 2)! = n(n – 1)
2
2!
(n – 2)! 2!
n
n!
1 2 = (n – 3)! 3! = n(n – 1)(n – 2)(n – 3)! = n(n – 1)(n – 2)
3!
(n – 3)! 3!
3
So, in general,
n
n!
1 2 = (n – r)! r! = n(n – 1)(n – 2)(n – 3) … (n – r + 1) , r  n.
r
r!
Example 32
Show that
n
n
n
n
=
(a) 1 2 1 n – r 2 (b) 1 2 1 r + 1 2 1 n + 1 2
+
=
r
r
r + 1
Solution: (a) From the definition,
n
n!
1 2 = (n – r)! r!
r
n
n!
and 1 n – r 2 = [n – (n – r)]! (n – r)!
= n!
r! (n – r)!
= n!
(n – r)! r!
n
n
Hence, 1 2 = 1 n – r 2 .
r
(b) From the definition,
n
n
n!
n!
+
1 2 1 r + 1 2 = (n – r)! r! + [n – (r + 1)]! (r + 1)!
r
= n! + n!
(n – r)(n – r – 1)! r! (n – r – 1)! (r + 1) r!
= n! 3 1 + 1 4
(n – r – 1)! r! n – r r + 1
121
02 STPM Math T T1.indd 121 3/28/18 4:21 PM

Mathematics Term 1 STPM Chapter 2 Sequences and Series

= n! 3 r + 1 + n – r 4
(n – r – 1)! r! (n – r)(r + 1)
= (n + 1) n!
(n – r)(n – r – 1)! (r + 1)r!
= (n + 1)!
(n – r)! (r + 1)!
(n + 1)!
1 n + 1 2 = [(n + 1) – (r + 1)]! (r + 1)!
r + 1
= (n + 1)!
(n – r)! (r + 1)!
n
n
2 Hence, 1 2 1 r + 1 2 1 n + 1 2
=
+
r
r + 1
n
Both the above results (a) and (b) are very useful in evaluating 1 2 , especially when the difference between
r
n and r is small.
For any positive integers n and r, n  r,
n
n
(a) 1 2 1 n – r 2
=
r
n
n
=
(b) 1 2 1 r + 1 2 1 n + 1 2
+
r + 1
r
Example 33
Evaluate
7
18
15
15
12
12
+
(a) 1 2 (b) 1 2 (c) 1 2 1 2 (d) 1 2 1 2
+
3
4
4
15
5
7
7
7!
Solution: (a) 1 2 = (7 – 3)! 3!
3
= 7!
4! 3!
= 7 × 6 × 5 × 4!
4! × 3 × 2 × 1
= 35
18
18!
(b) 1 2 = (18 – 15)! 15!
15
= 18 × 17 × 16 × 15!
3! × 15!
= 18 × 17 × 16
3 × 2 × 1
= 816
15
15
16
=
(c) 1 2 1 2 1 2 By using the result n + 1 2
+
2 1
1 2 1
n
n
4
5
5
=
+
= 16! r r + 1 r + 1
11! 5!
= 16 × 15 × 14 × 13 × 12 × 11!
11! × 5 × 4 × 3 × 2 × 1
= 4368
122
02 STPM Math T T1.indd 122 3/28/18 4:21 PM

Mathematics Term 1 STPM Chapter 2 Sequences and Series

12
12
12
12
=
+
(d) 1 2 1 2 1 2 1 12 – 7 2
+
7
4
4
12
12
+
= 1 2 1 2
5
4
13
= 1 2
5
= 13!
8! 5!
= 13 × 12 × 11 × 10 × 9 × 8!
8! × 5 × 4 × 3 × 2 × 1
= 1287
2
Exercise 2.6
1. Evaluate
(a) 5! (b) 8! (c) 10! (d) 8!
5!
(e) 10! (f) 18! (g) 20! (h) 16!
7! 15! 17! 4! 10! 6!
(i) 9! (j) 10!
2! 3! 4! (5!)

2. Rewrite in factorial form
(a) 10 × 9 × 8 (b) 12 × 11
(c) 22 × 21 × 20 (d) 18 × 17 × 16
5 × 4 × 3
(e) 21 × 20 × 19 (f) n(n – 1)(n – 2)
8 × 7 × 6
(g) (n + 2)(n + 1)n(n – 1) (h) (2n + 5)(2n + 4)(2n + 3)
(i) n(n – 1)(n – 2) (j) (n – 5)(n – 6)(n – 7)(n – 8)
3 × 2 × 1 5 × 4 × 3
3. Factorise
(a) 10! + 9! (b) 2(8!) – 7!
(c) 3(7!) + 4(9!) (d) (n + 1)! + n!
2
(e) (n + 1)! – (n – 1)! (f) n (n + 1)! + 2n(n!)
(g) (n + 1)! + n! + (n – 1)! (h) 8! + 7!
4! 5! 3! 2!
(i) 10! – 8! + 7! (j) n! + (n – 1)!
7! 3! 5! 4! 3! 2! r! (r + 1)!

4. Evaluate
8
15
10
20
(a) 1 2 (b) 1 2 (c) 1 2 (d) 1 2
0
17
4
3
8
10
8
5
6
13
5
12
÷
(e) 1 21 2 (f) 1 21 2 (g) 1 2 1 2 (h) 1 2 1 2
÷
3
4
3
4
2
2
3
5
10
16
9
16
–
+
(i) 1 2 1 2 (j) 1 2 1 2
5
4
4
12
123
02 STPM Math T T1.indd 123 3/28/18 4:21 PM

Mathematics Term 1 STPM Chapter 2 Sequences and Series
The binomial theorem

When a binomial expression such as (a + b) is raised to the power of n (where n is a small positive integer),
we obtain a binomial expansion such as follows:
0
When n = 0, (a + b) = 1
n = 1, (a + b) = a + b
1
2
2
n = 2, (a + b) = a + 2ab + b 2
n = 3, (a + b) = a + 3a b + 3ab + b 3
2
2
3
3
2 2
4
3
4
3
n = 4, (a + b) = a + 4a b + 6a b + 4ab + b 4
5
n = 5, (a + b) = a + 5a b + 10a b + 10a b + 5ab + b 5
4
3 2
5
2 3
4
The coefficients of each binomial expansion form an array which is known as a Pascal triangle, as shown
2 below:
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
Each number in the Pascal triangle is the sum of two numbers adjacent to it in the previous line. For example,
10 = 4 + 6. This means that the triangle can be extended to obtain the coefficients of the binomial expansion
n
of (a + b) for higher values of n. This way of expansion is rather tedious. An alternative method, known as
the binomial theorem, can be used to expand (a + b) for any n  Z .
n
+
Notice that we can arrange the above binomial expansions in the following way:
(a + b) = a + 3a b + 3 · 2 ab + b 3
3
2
3
2
1 · 2
3
4
4
3
2 2
(a + b) = a + 4a b + 4 · 3 a b + 4 · 3 · 2 ab + b 4
1 · 2 1 · 2 · 3
5
3 2
4
5
(a + b) = a + 5a b + 5 · 4 a b + 5 · 4 · 3 a b + 5 · 4 · 3 · 2 ab + b 5
4
2 3
1 · 2 1 · 2 · 3 1 · 2 · 3 · 4
In general, if n is a positive integer, we have
n – 2 2
n
3
(a + b) = a + na n – 1 b + n(n – 1) a b + n(n – 1)(n – 2) a n – 3 b + … + b .
n
n
1 · 2 1 · 2 · 3
n
n
n
However, since n = 1 2 , n(n – 1) = 1 2 , n(n – 1)(n – 2) = 1 2 , …
1
2
3
1 · 2 · 3
1 · 2
n
n
we can expand (a + b) by using the 1 2 notation,
r
n
n
n
n
n
3
n
n
n
b +
i.e. (a + b) = a + 1 2 a n – 1 b + 1 2 a n – 2 2 1 2 a n – 3 b + … + 1 2 a n – r b + … + 1 n –1 2 ab n – 1 + b ,
r
r
1
2
3
or in short,
n n
n
(a + b) = ∑ 1 2 a n – r r +
b , n  Z .
r = 0 r
The above result is called the binomial theorem.
124
02 STPM Math T T1.indd 124 3/28/18 4:21 PM

Mathematics Term 1 STPM Chapter 2 Sequences and Series
n
The expansion of (1 + x) , n  Z +
n
From the binomial theorem for (a + b) , when n is a positive integer with a = 1 and b = x, we have a finite series
n
n
n
n
n
2
n
r
(1 + x) = 1 + 1 2 x + 1 2 x + … + 1 2 x + … + 1 n –1 2 x n – 1 + x
2
1
r
n
r
= 1 + nx + n(n – 1) x + … + n(n – 1) … (n – r + 1) x + … + nx n – 1 + x .
2
2! r!
n
n
In general, if we want to expand (a + b) , it would be easier if we change it to the form (1 + x) as shown below:
3 1
n
(a + b) = a 1 + b 24 n
a
1
= a 1 + b 2 n
n
a
n
n
= a (1 + x) , where x = b . 2
a
Example 34
Use the binomial theorem to expand (1 + x) .
7
7
7
7
7
7
7
3
6
4
2
5
7
Solution: (1 + x) = 1 + 1 2 x + 1 2 x + 1 2 x + 1 2 x + 1 2 x + 1 2 x + x 7
5
1
2
4
6
3
4
3
6
= 1 + 7! x + 7! x + 7! x + 7! x + 7! x + 7! x + x 7
2
5
6! 1! 5! 2! 4! 3! 3! 4! 2! 5! 1! 6!
6
3
2
= 1 + 7x + 7 · 6 x + 7 · 6 · 5 x + 7 · 6 · 5 x + 7 · 6 x + 7x + x 7
4
5
1 · 2 1 · 2 · 3 1 · 2 · 3 1 · 2
3
= 1 + 7x + 21x + 35x + 35x + 21x + 7x + x 7
6
4
2
5
Example 35
4
Using the binomial theorem, find the expansion of (3x + 4y) .
4
4
4
Solution: (3x + 4y) = (3x) + 1 2 (3x) (4y) + 1 2 (3x) (4y) + 1 2 (3x)(4y) + (4y) 4
3
4
3
4
2
2
2
3
1
4 4
4 4
3
3
3
= 3 x + 4·3 ·4x y + 6·3 ·4 x y + 4·3·4 xy + 4 y
2 2 2 2
3
3
4
3
= 81x + 432x y + 864x y + 768xy + 256y 4
2 2
Example 36
4
15
Find the coefficient of x in the expansion of (2x – 1) .
Solution: From the binomial theorem,
n n
n
b
(a + b) = ∑ 1 2 a n – r r
r = 0 r
n
th
b
where the (r + 1) term = 1 2 a n – r r
r
15 15
15
Thus (2x – 1) = ∑ 1 2 (2x) 15 – r (–1) r
r = 0 r
4
The term in x is when 15 – r = 4, i.e. r = 11.
15
Coefficient of x = 1 2 2 (–1) 11
4
4
11
= – 15 × 14 × 13 × 12 × 2 4
1 × 2 × 3 × 4
= –21 840

125
02 STPM Math T T1.indd 125 3/28/18 4:21 PM

Mathematics Term 1 STPM Chapter 2 Sequences and Series

Example 37

2 7
Expand (1 + x + 2x ) up to the term in x .
3
2
Solution: Rewriting (1 + x + 2x ) as a binomial,
2
2 7
(1 + x + 2x ) = [1 + (x + 2x )] 7
2 3
= 1 + 7(x + 2x ) + 7 · 6 (x + 2x ) + 7 · 6 · 5 (x + 2x ) + …
2 2
2
1 · 2 1 · 2 · 3
We stop the expansion here
2 4
because the next term (x + 2x )
4
contains x and higher powers of x.
2 = 1 + 7(x + 2x ) + 21(x + 4x ) + 35x + …
3
2
2
3
3
2
= 1 + 7x + 35x + 119x + … Ignore x and higher powers of x.
4
Exercise 2.7
1. Use the binomial theorem to expand
2 6
7
4
(a) (p + q) (b) (m – n) (c) (2 + k )
1

(d) (2a – b) 5 (e) 1 x + 1 2 3 (f) 1 y – 2y 2 8
x
2. In each of the following expansions, find the term as stated.
th
10
8
(a) (1 + x) , 5 term (b) (2 – 3x) , term in x 2
2 7
4 6
12
th
(c) (2a + b) , 10 term (d) (p – 3q ) , term in p q
(e) 1 x – 1 2 6 , constant term (f) 1 x + 1 2 9 , term in x 1 3 .
2
x
x
3. Expand the expression
1
1 2x + 1 2 2 5 + 2x – 1 2 2 5 ,
simplifying the terms. x x
n
2
4. The coefficient of x is four times the coefficient of x in the expansion of (1 + x) . Find the value of n.
3
n
2
1
5. In the binomial expansion of 1 + 1 x , the coefficients of the fourth and fifth terms are equal. Find the
value of n. 3
8
5
4
6. The coefficient of x in the binomial expansion of (1 + 5x) is the same as the coefficient of x in the
7
expansion of (a + 5x) . Find the value of a.
n
7. If the first three terms in the expansion of (1 + ax) in ascending powers of x are 1 – 4x + 7x , find n
2
and a.
8. Find the first four terms of each of the following expansions, in ascending powers of x.
2 7
(a) (1 + x) (b) (1 + x – x )
7
3
2 8
9. Expand (1 + 2x + 3x ) in ascending powers of x up to and including the term in x .
6
10. Find the first three terms, in ascending powers of x, of the expansion (1 – 3x)(1 + 2x) .
11. Find the coefficient of the terms in x as indicated, in the following expansions.
2
(a) (1 + x )(2 – 3x) , term in x (b) (1 – 3x – 2x )(1 + x ) , term in x 20
2 20
3
7
2
1
5
(c) x x – x 2 2 2 12 , term in x 4 (d) 1 x + 1 2 2 (1 – x) , term in x 2
x
126
02 STPM Math T T1.indd 126 3/28/18 4:21 PM

Mathematics Term 1 STPM Chapter 2 Sequences and Series
n
The expansion of (1 + x) , n  Q
+
n
We have seen that the binomial expansion of (1 + x) , where n ∈ Z , is a finite series with (n + 1) terms, i.e.
n
(1 + x) = 1 + nx + n(n – 1) x + n (n – 1)(n – 2) x + … + x .
2
3
n
2! 3!
However, if n is any rational number, i.e. n ∈ Q, then the expansion is an infinite series, i.e.
(1 + x) = 1 + nx + n(n – 1) x + n(n – 1)(n – 2) x + … + n(n – 1) … (n – r + 1) x + …
3
2
n
r
2! 3! r!
This series is called the binomial series, and is valid if |x| , 1, i.e. –1 , x , 1. It is used to find approximations,
up to a degree of accuracy very close to its actual value. 2
n
n!
Note: The notation 1 2 = (n – r)! r! is not applicable if n is not a positive integer.
r
Example 38

Expand each of the following expressions as an ascending series in x, up to the term in x . State the range
4
of x such that the expansion is valid.
(a) 1 1 + 1 x 2 — 1 3 (b) (1 – 2x) –3
2

Solution: By using the binomial expansion
2
3
(1 + x) = 1 + nx + n(n – 1) x + n(n – 1)(n – 2) x
n
2! 3!
+ n(n – 1)\(n – 2)(n – 3) x + …
4
4!
(a) Substitute x = 1 x and n = 1 , we get
2 3
1 1 – 1 2 1 1 – 1 21 1 – 2 2
1
1


1
—
2
2
2
3
1 2
3
2
1 1 + 1 x = 1 + 1 1 x + 3 3 2! 1 1 x + 3 3 3! 3 1 1 x

2
2
3 2
2
1 1 – 1 21 1 – 2 21 1 – 3 2
1

2
+ 3 3 3 3 1 1 x + …
4
4! 2
1 – 2 2 1 – 2 21 – 5 2
1
1
2
2
= 1 + 1 x + 3 3 1 1 x + 3 3 3 1 1 x
3
2
6 2 4 6 8
1 – 2 21 – 5 21 – 8 2
1
3 3 3 3 1
2
4
+ 1 x + …
24 16
= 1 + 1 x – 1 x + 5 x – 5 x + …
2
3
4
6 36 648 1944
1
The expansion is valid if | x| , 1, i.e. |x| , 2 or –2 , x , 2.

2
127
02 STPM Math T T1.indd 127 3/28/18 4:21 PM

Mathematics Term 1 STPM Chapter 2 Sequences and Series

(b) Substitute x = –2x and n = –3, we get

(1 – 2x) = 1 + (–3)(–2x) + (–3)(–3 – 1) (–2x) + (–3)(–3 – 1)(–3 – 2) (–2x)
3
2
–3
2! 3!
4
+ (–3)(–3 – 1)(–3 – 2)(–3 – 3) (–2x) + …
4!

2
= 1 + 6x + (–3)(–4) (4x ) + (–3)(–4)(–5) (–8x )
3
2 6
+ (–3)(–4)(–5)(–6) (16x ) + …
4
24
4
2
3
= 1 + 6x + 24x + 80x + 240x + …
2
The expansion is valid if |–2x| , 1, i.e. 2|x| , 1,
1
or |x| , 1 or – , x , 1 .
2 2 2
Example 39


10
10
3
Expand (1 + x) up to and including the term in x . Hence, obtain an approximation for (1.01) and (0.99) .
10
2
3
Solution: (1 + x) = 1 + 10x + 10 · 9 x + 10 · 9 · 8 x + …
10
1 · 2 1 · 2 · 3
3
2
= 1 + 10x + 45x + 120x + …
Substitute x = 0.01 into the expansion,
3
2
10
(1 + 0.01) = 1 + 10(0.01) + 45(0.01) + 120(0.01) + …
= 1 + 0.1 + 0.0045 + 0.000120 + …
= 1.10462
= 1.1046 (correct to 4 decimal places)

Substitute x = –0.01 into the expansion,

3
2

10

(1 – 0.01) = 1 + 10(–0.01) + 45(–0.01) + 120(–0.01) + …
= 1 – 0.1 + 0.0045 – 0.000120 + …
= 0.90438
= 0.9044 (correct to 4 decimal places)
10
10
Note: By using the calculator, we find that (1.01) = 1.1046 and (0.99) = 0.9044 (both correct to 4 decimal
places). This shows that the accuracy of the approximation does not differ much even if we were to
10
expand (1 + x) up to x or more. Hence, usually three or four terms should be sufficient for a good
5
approximation.




128





02 STPM Math T T1.indd 128 3/28/18 4:21 PM

Mathematics Term 1 STPM Chapter 2 Sequences and Series

Example 40


Expand 1 + 2x as a series in ascending powers of x, up to the term in x . By substituting x = 1 , find
2
1 – 2x 100
an approximation for 51, stating the number of significant figures your result is accurate to.

— 1
1 + 2x (1 + 2x) 2
Solution: 1 – 2x = — 1
(1 – 2x) 2

— 1 – — 1
2
= (1 + 2x) (1 – 2x) 2 2
1 – 1 2
1
3
2
= 1 + 1 (2x) + 2 1·2 2 (2x) + … 4

2
1 – 1 21 – 3 2
3
1
2
× 1 + – 1 2 (–2x) + 2 1·2 2 (–2x) + … 4
2

1
= 1 + x – 1 x + … 21 1 + x + 3 x + … 2
2
2
2
2
2
= 1 + 2x + 2x + … Ignore x and higher powers of x.
3
When x = 1 , substituting into the expansion
100
1 + 2
100 = 1 + 2 1 1 2 + 2 1 1 2 2 + …
1 – 2 100 100
100
102 = 1 + 1 + 1
98 50 5000
51 = 5101
49 5000
1 51 = 5101
7 5000
51 = 35 707
5000
= 7.1414 (correct to 5 significant figures)



Example 41


1
1
– — 1 3 1 – — —
4
1
Expand (1 + x) 4 in ascending powers of x up to the term in x . Prove that 1 + 2 4 = 5 .
2
2 80
1
– — 1 1 —
4
Using your expansion for (1 + x) 4 and x = 80 , find an approximation for 5 , giving your answer correct
to five decimal places.
129




02 STPM Math T T1.indd 129 3/28/18 4:21 PM

Mathematics Term 1 STPM Chapter 2 Sequences and Series



4
– — 1 1 1 – 1 21 – 5 2
4
1
2
Solution: (1 + x) 4 = 1 + – 4 2 x + 1·2 x + …

2
= 1 – 1 x + 5 x – …
4 32
1
3
1
1 2
4
4

1 + 1 2 – — 1 = 3 81 – —
2 80 2 80
1
—
1 2
= 3 80 4

2 81
1
2 = 3 1 5 × 2 4 2 —
4
2 3 4
= 3 · 2 (5) — 1 4
2 3
— 1
= 5 4
– — 1 1
By using the expansion of (1 + x) 4 and x = ,
80
1
1
1
5
1
1 2
1 1 + 80 2 – — = 1 – 1 1 2 + 32 80 2
4

4 80
= 1 – 1 + 1
320 40 960
= 40 833
40 960
1
— 3 40 833
4
Hence, 5 = ×
2 40 960
= 1.49535 (correct to 5 decimal places)
Exercise 2.8
3
1. Expand each of the following expressions as a series in ascending powers of x up to the term in x .
1
–2
(a) (1 + x) (b) (1 – x) –1 (c) (1 + 2x) 3
(d) (2 + x) –2 (e) (3 + 2x) –1
3
2. Expand each of the following expressions as a series in ascending powers of x up to the term in x . State
the range of values of x in each case for which the expansion is valid.
1
2
(a) 1 1 + 1 2 x 2 – — (b) (1 – 2x) — 1 3 (c) (2 + x) –1
1 x – 1
(d) (e) (1 – x) 1 + x (f)
1 + x x + 1
x + 2 1
2 –1
(g) (h) (i) (1 + x + x )
1 – 3x (x – 2)(1 + 2x)
130





02 STPM Math T T1.indd 130 3/28/18 4:21 PM

Mathematics Term 1 STPM Chapter 2 Sequences and Series
4
4
3. Use the expansion of (x + y) to evaluate (1.03) , correct to four decimal places.
5
5
4. Use the expansion of (2 – x) to evaluate (1.98) , correct to five decimal places.
15
3
5. Obtain the expansion of (1 + 2x) in ascending powers of x up to the term in x . Hence, evaluate (1.002) ,
15
correct to five decimal places.
– — 1
2
6. Find the first four non-zero terms of the expansion of (1 + 2x ) 2 in ascending powers of x.
1
– —
7. Obtain the first four terms of the expansion of (1 – x) 2 in ascending powers of x. Deduce the value of
0.9, correct to four decimal places.
1
—
8. By substituting x = 0.08 into (1 + x) and its expansion, find 3 , correct to four significant figures. 2
2
2
9. By substituting x = 1 into (1 – x) – — 1 and its expansion, find 10 , correct to five significant figures.
10
–2
10. Expand (2 – x) as a series in ascending powers of x, up to the term in x .
4
Deduce the value of 1 , correct to three significant figures.
(1.8) 2
— 1 1
2
11. Expand (1 + 2x) in ascending powers of x, up to the term in x . By substituting x = , find an
3
approximation for 5 , giving your answer correct to three decimal places. 8
3
12. Expand 1 – 3x in ascending powers of x, up to the term in x . State the range of values of x for which
1 + 4x
the expansion is valid.
2 5
13. Expand (1 – x – 2x ) in ascending powers of x, up to the term in x . By substituting
4
5
x = 0.01, estimate the value of (0.9898) , correct to six decimal places.

Summary


1. A series S = u + u + u + … + u is said to be convergent if there exists a finite number a, such that
1
n
n
3
2
lim S = a.
n → ∞ n
A series is said to be divergent if it is not convergent.
2. Sum of the first n positive integers is
n 1
∑ r = n(n + 1)
r = 1 2
Sum of the squares of the first n positive integers is
n
2
∑ r = 1 n(n + 1)(2n + 1)
r = 1 6
Sum of the cubes of the first n positive integers is
n 1
2
3
∑ r = n (n + 1) 2
r = 1 4
th
3. For an arithmetic progression with first term a and common difference d, the n term is u = a + (n – 1) d.
n
Sum of the first n terms is
S = n [2a + (n – 1)d]
2
n
131




02 STPM Math T T1.indd 131 3/28/18 4:21 PM

Mathematics Term 1 STPM Chapter 2 Sequences and Series
th
4. For a geometric progression with first term a and common ratio r, the n term is u = ar n – 1 .
n
Sum of the first n terms is n
S = a(r – 1) , for r > 1
r – 1
n
n
and S = a(1 – r ) , for r < 1
1 – r
n
lim
n
For |r| , 1 and large values of n such that x → ∞ r → 0, the sum (to infinity) is
a
S = 1 – r
∞
n
n!
5. 1 2 = (n – r)! r! , n, r  Z , 0  r  n.
+
r
2 = n(n – 1)(n – 2) … (n – r + 1)
r!
n
n
1 2 = 1 n – r 2
r
n
n
1 2 1 r + 1 2 1 n + 1 2
=
+
r + 1
r
6. The binomial theorem
n
n
n
n
2
r
n
(1 + x) = 1 + 1 2 x + 1 2 x + … + 1 2 x + … + 1 n – 1 2 x n – 1 + x , for n  Z .
+
n
r
1
2
(1 + x) = 1 + nx + n(n – 1) x + … + n(n – 1) … (n – r + 1) x + … , for n  Q.
r
n
2
2! r!
STPM PRACTICE 2
1. Write down the n term of the sequence 0, 2, 6, 12, 20, …. Find n, if the n term is 210.
th
th
th
2. Find the first three terms and the n term of the series with the sum to n terms, S , given by
n
1
(a) S = 4n – n (b) S = (n + 1)!
n
n
3. Find the value of
30 8 n n
(a) ∑ (r + 2) (b) ∑ 2 r (c) ∑ 2r (d) ∑ r(r!)
r = 1 r = 1 r = 1 r = 1
n
n
4. Show that ∑ (r + 1)2 r – 1 = n2 .
r = 1
n
1
2
5. (a) Prove that ∑ r = n(n + 1)(2n + 1).
r = 1 6
(b) If S = 1(n) + 2(n – 1) + 3(n – 2) + … + r(n + 1 – r) + … + n(1)
and T = 1(n – 1) + 2(n – 2) + 3(n – 3) + … + r(n – r) + … + (n – 1)(1),
where n is a positive integer, show that
S + T = 1 n(n + 1)(2n + 1)
6
4
1
6. If u = (2n + 1)(2n + 3) , show that u n – 1 – u = (2n – 1)(2n + 1)(2n + 3) .
n
n
Hence, find the sum to n terms of the series
1 + 1 + … + 1 + …
1 · 3 · 5 3 · 5 · 7 (2r – 1)(2r + 1)(2r + 3)
132
02 STPM Math T T1.indd 132 3/28/18 4:21 PM

Mathematics Term 1 STPM Chapter 2 Sequences and Series

7. The arithmetic mean of 1 and 1 is 1 .
a + b b + c a + c
2
2
Find, in terms of b, the arithmetic mean of a and c .
th
8. Find the smallest value of n such that the n term of the arithmetic series 12.0 + 10.7 + 9.4 + 8.1 + …
is negative. Find, also, the smallest value of n such that the sum to n terms is negative.

2
9. The sum of the first n terms of an arithmetic sequence u , u , u , … is given by S = 2n + n. Find an
3
n
1
2
explicit formula and recursive formula for u .
n
10. A sequence is u , u , u , … is defined by u = 4n – 1. The difference between successive terms of the
2
n
3
1
2
sequence forms a new sequence w , w , w , …. 2
2
1
3
(a) Find an explicit formula for w in terms of n.
n
(b) Show that w , w , w , … forms an arithmetic sequence, and state its first term and common difference.
3
2
1
(c) Find the sum of the first n terms of the sequence w , w , w , … in terms of u and w .
n
n
1
2
3
11. If S represents the sum of the first n terms of a geometric progression
n
1
1
1 + 1 + 1 2 2 + … + 1 2 n – 1 + …
2 2 2
and S represents the sum to infinity, find the smallest value of n such that S – S , 0.001.
n
3
12. A geometric sequence is defined by U r + 1 = 2 + U , and U = 2.
r
1
5
r – 1
3
(a) Write down each of the terms U , U and U in the form ∑ 2 1 2 m , and show that explicit formula
3
5
4
2
m = 0
3
r
3
for U is given by U = 5 1 – 1 24 .
r
r
5
(b) Determine the limit of U when r tends to infinity.
r
2
n
13. If S is the sum of the series 1 + 3x + 5x + … + (2n + 1)x , by considering (1 – x)S, show that
S = 1 + x – (2n + 3)x n + 1 + (2n + 1)x n + 2 , x ≠ 1.
(1 – x) 2
14. By bracketing the terms into pairs, show that
2
2
1 – 2 + 3 – 4 + … + (2n – 1) – (2n) = –n(2n + 1).
2
2
2
2
Deduce the sum of the series
2
2
2
2
2
2
2
(a) 1 – 2 + 3 – 4 + … + (2n – 1) – (2n) + (2n + 1) ,
2
2
2
2
2
2
(b) 25 – 26 + 27 – 28 + … + 49 – 50 .
15. The series of positive integers is grouped into four as follows:
(1, 2, 3, 4), (5, 6, 7, 8), (9, 10, 11, 12), …
Show that the sum of the integers in the k bracket is 2(8k – 3).
th
If the integers are similarly grouped with m integers in each bracket, find
(a) the last term in the n bracket,
th
(b) the first term in the n bracket.
th
th
If S represents the sum of the integers in the n bracket, find an expression for S in terms of m and n.
n
n
Hence, show that S , S , S are in arithmetic progression.
3n
2n
n
133
02 STPM Math T T1.indd 133 3/28/18 4:21 PM

Mathematics Term 1 STPM Chapter 2 Sequences and Series
16. The r term, u , of an infinite series is given by:
th
r
1
r 1 2
u = 1 4r – 1 + 1 2 4r + 1
2
2
k
(a) Express u in the form 2 4r + 1 ,where k is a constant .
r
(b) Find the sum of the first n terms of the series, and deduce the sum of the infinite series.
17. (a) Find the smallest value of n such that the sum of the first n terms of the geometric progression
3
2
1 + 0.99 + (0.99) + (0.99) + …
is more than 75% of the sum to infinity.
th
(b) Find the n term and the sum of the first n terms of the series.
2 1 + 11 + 111 + 1111 + 11111 + …
18. By expressing the recurring decimals as the sum of a constant and an infinite geometric series, obtain
each of the following decimals as a fraction in its lowest terms.
· · ·
· ·
· ·
(a) 0.2731 (b) 0.597 5 (c) 0.7015 0 1
19. The sum of the first n terms of a sequence u , u , u , … is given by S = 3 [1 – 2 ].
–n
3
2
n
1
(a) Show that u = 3(2 –(n + 1) ). 2
n
(b) Find u n + 1 in terms of u , and deduce that the sequence is a geometric sequence.
n
(c) Determine the sum of the series u , u , u , ….
3
2
1
20. Show that 1 – 2 = 5x .
1 – 2x 2 + x (1 – 2x)(2 + x)
3
Hence, expand 5x in ascending powers of x up to the term in x .
(1 – 2x)(2 + x)
4
4
21. (a) Expand fully the expansions of ( 5 – 3) and ( 5 + 3) to evaluate ( 5 – 3) + ( 5 + 3) .
4
4
(b) Hence, using the inequality of 2 , 5 , 3 and the result from part (a), show that
751 , ( 5 + 3) , 752.
4
p
22. Write down the first four terms in ascending powers of x for the expansion of (1 + ax) . Given that the
first three terms of this expansion is 1 + 2x + 11 x , show that 2pa(pa – a) = 11 and find the value of a
2
4
and p. State the range of values of x for which the expansion is valid.
1
1
1
2
23. Show that 25 1 – 25 22 — can be expressed in the form n 39 , where n is an integer to be found.
1
—
2
2
Expand (1 – x) as a series in ascending powers of x up to the term in x . By using the first two terms
— 1 1 p
2
of the expansion of (1 – x) with x = 25 2 , obtain an approximate value for 39 in the form q , where

p and q are integers.
24. Two positive integers, p and q, are connected by p = q + 1. Using the binomial expansion, show that
the expression p – 2nq – 1 can be divided exactly by q for all positive integers n. By choosing suitable
2n
2
15
16
values for p and n, show that 3 – 33 can be divided exactly by 4, and hence, show that 3 + 5 can be
divided exactly by 4.
25. Use the binomial theorem to expand 1 + x as a series in ascending powers of x up to the term in x ,
2
1 – x
with |x| , 1. By substituting x = 1 into your result, show that 11 is approximately 663 .
10 200
134
02 STPM Math T T1.indd 134 3/28/18 4:21 PM

Mathematics Term 1 STPM Chapter 2 Sequences and Series

1
n
1
26. Show that the first three non-zero terms in the expansion of 1 – 1 2 — 1 in ascending powers of 1 2 is
n
n
1
1
1 2
1 – 1 2 2 – 1 1 3 , and find the term in 1 2 4 .

2 n
n
n
1
—
10
By giving n a suitable value, use the first four non zero terms of the series to find the value of (0.9) ,
giving your answer correct to five decimal places.
n 1 2n
2
2
27. (a) Given that ∑ r = n(n + 1)(2n + 1), obtain an expression, in its simplest terms, for ∑ (2r – 1) .
r = 1 6 r = n + 1
∞ 1
(b) Find ∑ = , expressing your answer as a fraction in its lowest terms. Hence, express the recurring
r = 1 10 3r
· · ·
decimal 0.1 0 8 as a fraction in its simplest form.
2
28. (a) Find the sum of the arithmetic progression
1, 4, 7, 10, 13, 16, …, 1000.
Now, each third term of the progression, i.e. 7, 16, …, is removed. Find the sum of the remaining
terms.
th
(b) The r term, u , of a series is given by
r
1
r 1 2
u = 1 3r – 2 + 1 2 3r – 1
n B 3 3
1
Express ∑ u in the form A 1 – n2 , where A and B are constants. Find A, B and the sum of the
r = 1 r 27
series.
1
—
2
4
29. (a) Find the binomial expansion of (1 + x) , for small values of x, up to the term in x , with coefficients
in their simplest forms. By substituting x = 1 in your expression, show that 17 ≈ 8317 .
4
16 4096
5

(b) Express (x + 2) – (x – 2) as a polynomial in x, and hence, find the exact value of ( 5 + 2) –
5
5
5
( 5 – 2) .
By assuming that 0 , 5 – 2 , 1 , deduce that the difference between ( 5 + 2) and an integer is
5
4
less than 1 .
1024
2
30. Express f(x) = x + 5x in the form A + B + C , where A, B and C are constants.
(1 + x)(1 – x) 2 1 + x 1 – x (1 – x) 2
If the expansion of f(x) in ascending powers of x is
r
2
3
c + c x + c x + c x + … + c x + … ,
r
2
0
3
1
find c , c , c and show that c = 11.
0
1
2
3
Express c in terms of r.
r
4
n
31. (a) Show that for a fixed number x ≠ 1, 2x + 2x + 2x + … + 2x is a geometric series, and find its sum in
3
2
terms of x and n.
(b) The series U (x) is given by
n
n
3
2
U (x) = x + 3x + 5x + … + (2n – 1)x , for x ≠ 1.
n
By considering U (x) – xU (x) and using the result from (a), show that
n
n
2
U (x) = x + x – (2n + 1)x n + 1 + (2n – 1)x n + 2
n
(1 – x)
2
15 14
r
Hence, determine the value of ∑ (2r – 1)3 and deduce the value of ∑ (2r + 1)3 r + 2 .
r = 1 r = 1
135
02 STPM Math T T1.indd 135 3/28/18 4:21 PM

ANSWERS




Chapter 1 Functions Exercise 1.2
1. (b) c, d (c) {1, 2, 3, 4, 5}
Exercise 1.1 (d) {a, b, c, d} (e) No
1. (a) –9 (b) 7 (c) 0 (d) 0 2. (a) X Y
2. (a) (i) 3 (ii) 1 (iii) 0 –2 3
(b) (i) 24 (ii) 99 –1
3. (a) [–2, 6] or –2 < f(x) < 6 0 1
1
f(x ) 2 7
6 (b) {–2, –1, 0, 1, 2}
(c) {1, 3, 7}
(d) No
3. (a)
Y
x
–1 0 3
X –5
–3
–2
1 –1
(b) [–1, 3] or –1 < f(x) < 3 3 3
5 5
f(x )
7 7
3 9 11
15
(b) –1, 15
(c) Yes
x
–1 0 3 1
4. No. f(0) = f(1) = 0,
–1 2
a
1
9
9
(c) 3 – , 44 or – < f(x) < 4 5. 1 + a 2 ; {g : 0 , y < }
4 4 2
–1
f(x ) 6. (a) f : x ⟼ x + 2, x ∈ R
–1
(b) f : x ⟼ x – 1 , x ∈ R, x > 1
4
–1
(c) f : x ⟼ x + 1, x > 0
–1
(d) g : x ⟼ 3 + x + 1 , x > –1
(e) g : x ⟼ x + 9 , x . –9
–1
x
–2 0 3 2 + 3x
(f) h : x ⟼ x , x ≠ 0
–1
–1
– 9 _ (g) h : x ⟼ 2(x + 1) , x ≠ 1
4 x – 1
1 – x 1
–1
4. (a) No (b) Yes (c) No 7. f (x) = 2x – 1 , , x < 1
2
1
5. (a) 1, 4, , 64 (b) 4 (c) R +
2 9. (a) f(x ) (b) f(x )
6. x + 2 f f –1
7. 4 – x 2 f –1 f
8. (a) cos (1 + x) (b) cos (cos x) x 0 x
(c) 1 + cos x 0
9. (a) 1 – x , –1 , x , 1 (b) 1 – x
2
2
(c) 2x – x 4
2
10. (a) 4x – 6x + 1 (c) f(x ) f (d) f(x ) f
(b) 2x + 6x – 1 f –1 f –1
2
(c) 4x – 9
(d) x + 6x + 14x + 15x + 5 x x
4
2
3
0 0
11. (a) 6 (b) 25
(c) 2a – 4a + 9 (d) 4x – 12x + 14
2
2
2
(e) 2x + 4x + 9 (f) x + 10x + 30
4
2
268
Answers STPM Math T S1.indd 268 3/28/18 4:25 PM

Mathematics Term 1 STPM Answers
Exercise 1.3 4. (a) y (b) y
1. (a) y (b) y
–1
x
0 2
x
0
3 8 –1 2 –2
x 0 x (c) y (d) y
3 0 8
– – –
3
4
x
(c) y 0 2 x
0
– 2 2
8
-8
x (e) y
0
6
– –
5
2. (a) y (b) y
x
0
1 2
x
0
1 1
3 – – –
– 5. (a) y
2 x 3 3
0 –1
(c) y x
-1 0 1
57
–
16
2
(b) y
x
0 5
–
8
x
-1 0 1
3. (a) y (b) y
x
0
1
6. (a) y (b) y
–1 1
x
0 1
–– x
3 x _ 0
_ 1
3
-1 0
2 _ _ 1
2
(c) y
(c) y
x
5 –1 0
x
0 3
10


269





Answers STPM Math T S1.indd 269 3/28/18 4:25 PM