SHCEME FOR CHAPTER 2 SEQUENCES AND SERIES OXFORD FAJAR ACE BOOK

1. FULLY WORKED SOLUTIONS
2
CHAPTER
Focus STPM 2
ACE AHEAD Mathematics (T) First Term Second Edition
© Oxford Fajar Sdn. Bhd. 2015
1
SEQUENCES
AND SERIES
1 u13 = 3, S13 = 234
a + 12d = 3 13
2
(a + u13
) = 234
a = 3 – 12d ... 1 13
2
(a + 3) = 234 ... 2
From 2 , a = 33
When a = 33, 33 = 3 – 12d ⇒ d = – 5
2
S25 =
25
2 3(2)(33) + (24)1– 5
224
=
25
2
(66 – 60) = 75
2 (a) a = 1, r =
5
4
Sn =
a(rn
– 1)
r – 1
=
15
42
n
– 1
15
4
– 12
= 4315
42
n
– 14
(b) Sn
> 20
4315
42
n
– 14 > 20
15
42
n
– 1 > 5
15
42
n
> 6
n lg 5
4
> lg 6
n > 8.03
The least number of terms is 9.
3 312
– 22
4 + 332
– 42
4 + … + 3(2n – 1)2
– (2n)2
4
= –3 – 7 – 11…
a = –3, d = – 4
Sn =
n
2
3–6 + (n – 1)(– 4)4
=
n
2
(–6 – 4n + 4) =
n
2
(– 4n – 2)
= –n(2n + 1) [Shown]
(a) 12
– 22
+ 32
– 42
+ … + (2n – 1)2
– (2n)2
+ (2n + 1)2
= –n(2n + 1) + (2n + 1)2
= (2n + 1)(–n + 2n + 1)
= (2n + 1)(n + 1)
(b) 212
– 222
+ 232
– 242
+ … + 392
– 402
= (12
– 22
+ 32
– 42
+ … + 392
– 402
)
– (12
– 22
+ 32
– 42
+ … + 192
– 202
)
= S20
– S10
= –20(2(20) + 1) – [–10(2(10) + 1)]
= –820 – (–210) = –610
4 d = 2
un = a + (n – 1)(2) = a + 2n – 2
When n = 20, u20 = a + 2(20) – 2 = a + 38
S20 = 1120
20
2
(a + u20
) = 1120
10(a + a + 38) = 1120
2a = 74
a = 37
u20 = a + 38 = 37 + 38 = 75
5 (a) a = 3, d = 4, Sn = 820
Sn =
n
2
[2a + (n – 1)d ]
820 =
n
2
(6 + 4n – 4)
1640 = n(2 + 4n)
n(1 + 2n) = 820
n + 2n2
– 820 = 0
(2n + 41)(n – 20) = 0
n = 20 since n = – 41
2
is not a solution.
T20 = 3 + 19(4) = 79
Chapter 2.indd 1 6/24/2015 5:36:59 PM

2. 2 ACE AHEAD Mathematics (T) First Term Second Edition
© Oxford Fajar Sdn. Bhd. 2015
(b) Since they form an A.P.,
(p2
+ q2
)2
– (p2
– 2pq – q2
)2
= (p2
+ 2pq – q2
)2
– (p2
+ q2
)2
(p2
+ q2
– p2
– 2pq – q2
)(p2
+ q2
– p2
+ 2pq + q2
)
= (p2
+ 2pq + q2
+ p2
+ q2
)(p2
+ 2pq
– q2
– p2
– q2
)
(2p2
– 2pq)(2q2
+ 2pq) = (2p2
+ 2pq)
(2pq – 2q2
)
4pq(p – q)(q + p) = 4pq(p + q)(p – q)
LHS = RHS [Shown]
6 (a) Sn = pn + qn2
, S8 = 20, S13 = 39
(i) S8 = 8p + 64q
8p = 20 – 64q
p =
5
2
– 8q … 1
S13 = 39
13p + 169q = 39 … 2
Substitute 1 into 2 ,
1315
2
– 8q2 + 169q = 39
65
2
– 104q + 169q = 39
65q =
13
2
q =
1
10
p =
5
2
– 811
102 =
17
10
(ii) un = Sn
– Sn – 1
= pn + qn2
– [p(n – 1) + q(n – 1)2
]
= p(n – n + 1) + q[n2
– (n – 1)2
]
= p + q[(n + n – 1)(n – n + 1)]
= p + q(2n – 1)(1)
=
17
10
+
1
10
(2n – 1)
=
17
10
+
n
5
– 1
10
=
8 + n
5
(iii) To show that it is an A.P.,
un =
1
5
(8 + n)
un – 1 =
1
5
[8 + (n – 1)]
=
1
5
(7 + n)
un
– un – 1 =
1
5
⇒ Series is an A.P. with d =
1
5
= 0.2.
7 If J, K, M is a G.P., then
K
J
=
M
K
K 2
= MJ … 1
(ar k – 1
)2
= (ar m – 1
) (ar j – 1
)
r2k – 2
= rm – 1 + j – 1
2k – 2 = m + j – 2
k + k = m + j
k – m = j – k … 2
and
2k = m + j … 3
(k – m)lg J + (m – j)lg K + ( j – k)lg M
= lg J k – m
+ lg Km – j
+ lg M j – k
= lg J j – k
+ lg Km – j
+ lg M j – k
[from 2 ]
= ( j – k)(lg J + lg M) + lg K m – j
= ( j – k)lg JM + lg K m – j
= ( j – k)lg K2
+ (m – j)lg K
= 2( j – k)lg K + (m – j)lg K
= (2j – 2k + m – j)lg K
= ( j + m – 2k)lg K
= (2k – 2k)lg K [from 3 ]
= 0 [Proven]
8 (a) A.P., a = 200, d = 400
T a n d n
n
n = + −( ) = + −
= +
1 2000 1 400
400 1600
( )( )
= +400( 4)n
(b) S
n
a l
n
n
n
n
n = + = + +[ ]
= +
2 2
2000 400 1600
2
400 3600
( )
( )
= +200 ( 9)n n
(c) Sn
 200 000
200n(n + 9)  200 000
n2
+ 9n – 1000  0
n  –36.44135, n  27.44135
∴ n = 28
Chapter 2.indd 2 6/24/2015 5:37:02 PM

3. ACE AHEAD Mathematics (T) First Term Second Edition
© Oxford Fajar Sdn. Bhd. 2015
3
9 S5 = 44,
a(1 – r5
)
1 – r
= 44 … 1
S10
– S5 = – 11
8
a(1 – r10
)
1 – r
– a(1 – r5
)
1 – r
= – 11
8
a(1 + r5
)(1 – r5
)
1 – r
– a(1 – r5
)
1 – r
= – 11
8
… 2
a(1 – r5
)
1 – r
(1 + r5
– 1) = – 11
8
Substitute 1 into 2 ,
44(r5
) = – 11
8
r5
= – 1
32
r = – 1
2
[Shown]
a31 – 1– 1
22
5
4
1 – 1– 1
22
= 44
a11 +
1
322 = 66
a = 64
Sn =
a
1 – r
=
64
1 – 1– 1
22
= 422
3
10 (a) r =
3 – 1
3 + 1
×
3 – 1
3 – 1
=
3 – 2 3 + 1
3 – 1
= 2 – 3
u3 = u2
r
= ( 3 – 1)(2 – 3)
= 2 3 – 3 – 2 + 3
= 3 3 – 5
u4 = u3
r
= (3 3 – 5)(2 – 3)
= 6 3 – 3(3) – 10 + 5 3
= 11 3 – 19
(b) r = 2 – 3 [ 1]
S∞ =
a
1 – r
=
3 + 1
1 – (2 – 3 )
=
3 + 1
3 – 1
×
3 + 1
3 + 1
=
3 + 2 3 + 1
2
= 2 + 3
11 (a)
S6
S3
=
7
8
, u2 = – 4
8S6 = 7S3
… 1
ar = – 4
a = – 4
r
… 2
83a(1 – r6
)
1 – r 4 = 73a(1 – r3
)
1 – r 4
8(1 – r6
) = 7(1 – r3
)
8 – 8r6
= 7 – 7r3
8r6
– 7r3
– 1 = 0
(r3
– 1)(8r3
+ 1) = 0
r3
= 1 or – 1
8
r = – 1
2
since r = 1 is
not a solution.
a = – 4
1– 1
22
= 8
(b) r =
8
5
×
1
2
=
4
5
, a = 2
Sn
9.9
a(1 – rn
)
1 – r
9.9
231 – 14
52
n
4
11
52
9.9
1 – 14
52
n
0.99
14
52
n
0.01
n lg 4
5
lg (0.01)
n 20.64
The least number of terms is 21.
12 u3 = S2
ar2
=
a(1 – r2
)
1 – r
ar2
(1 – r) = a(1 + r)(1 – r)
ar2
= a(1 + r)
r2
– r – 1 = 0
r =
–(–1) ± (–1)2
– 4(1)(–1)
2
=
1 ± 5
2
a = 2
When r =
1 + 5
2
[|r| 1], S∞
doesn’t exist
When r =
1 – 5
2
[|r| 1],
Chapter 2.indd 3 6/24/2015 5:37:04 PM

4. 4 ACE AHEAD Mathematics (T) First Term Second Edition
© Oxford Fajar Sdn. Bhd. 2015
S∞ =
a
1 – r
=
2
1 –
1 – 5
2
=
2
12 – 1 + 5
2 2
=
4
1 + 5
×
1 – 5
1 – 5
=
4 – 4 5
1 – 5
= 5 – 1
13 a = 3, r = 0.4
(a) un 0.02
ar n – 1
0.02
3(0.4)n – 1
0.02
(0.4)n – 1
1
150
(n – 1)lg 0.4 lg
1
150
n – 1 5.47
n 6.47
The least number of terms is 7.
(b) S∞
– Sn
0.01
a
1 – r
–
a(1 – rn
)
1 – r
0.01
a
1 – r
(1 – 1 + r n
) 0.01
3
0.6
(r n
) 0.01
0.4n
2 × 10–3
n lg 0.4 lg (2 × 10–3
)
n 6.78
The least number of terms is 7.
14 Sn = a + ar + ar 2
+ … + ar n – 2
+ ar n – 1
… 1
rSn = ar + ar 2
+ … + ar n – 2
+
ar n – 1
+ ar n
… 2
1 – 2 ,
Sn
– rSn = a – ar n
Sn
(1 – r) = a(1 – r n
)
Sn =
a(1 – r n
)
1 – r
[Shown]
For S∞
to exist, |r| 1,
lim Sn
= S∞
=
a(1 – r ∞)
1 – r
=
a
1 – rn → ∞
S∞
– Sn
S∞
=
a
1 – r
–
a(1 – rn
)
1 – r
a
1 – r
=
a
1 – r
[1 – (1 – rn
)]
a
1 – r
= rn
[Shown]
(a) u4 = 18, u7 =
16
3
ar3
= 18 … 1 ar6
=
16
3
… 2
2
1
, r3
= 116
3 2 ×
1
18
r3
= 18
272
r = 12
32
a12
32
3
= 18
a =
243
4
S∞
=
1243
4 2
1 – 2
3
=
729
4
(b)
S∞
– Sn
S∞
0.001
rn
0.001
12
32
n
0.001
n lg
2
3
lg 0.001
n 17.04
The least value of n is 18.
15
2r – 1
r(r – 1)
–
2r + 1
r(r + 1)
=
(r + 1)(2r – 1) – (2r + 1)(r – 1)
r(r – 1)(r + 1)
=
2r2
– r + 2r – 1 – 2r2
+ 2r – r + 1
r(r – 1)(r + 1)
=
2r
r(r – 1)(r + 1)
=
2
(r – 1)(r + 1)
[Verified]
Σr = 2
n 2
(r – 1)(r + 1)
= Σr = 2
n
3 2r – 1
r(r – 1)
–
2r + 1
r(r + 1)4
Chapter 2.indd 4 6/24/2015 5:37:06 PM

5. ACE AHEAD Mathematics (T) First Term Second Edition
© Oxford Fajar Sdn. Bhd. 2015
5
= Σr = 2
n
[ f (r) – f (r + 1)] 3f (r) =
2r – 1
r(r – 1)4
= f (2) – f (n + 1)
=
3
2
– 2n + 1
n(n + 1)
[Proven]
1
2
Σr = 2
∞ 2
(r – 1)(r + 1)
=
1
2
lim 33
2
– 2n + 1
n(n + 1)4n → ∞
=
1
2
lim
33
2
–
2
n
+
1
n2
1 +
1
n
4n → ∞
=
1
2
13
2
– 02 =
3
4
16
1
r(r + 1)
–
1
(r + 1)(r + 2)
=
r + 2 – r
r(r + 1)(r + 2)
=
2
r(r + 1)(r + 2)
[Shown]
Σr = 1
n 1
r(r + 1)(r + 2)
=
1
2
Σr = 1
n 2
r(r + 1)(r + 2)
=
1
2
Σr = 1
n
3 1
r(r + 1)
– 1
(r + 1)(r + 2)4
3f (r) =
1
r(r + 1)4
=
1
2
Σr = 1
n
[ f (r) – f (r + 1)]
=
1
2
[ f (1) – f (n + 1)]
=
1
2 3
1
2
–
1
(n + 1)(n + 2)4
=
1
2 3
n2
+ 3n + 2 – 2
2(n + 1)(n + 2)4
=
n2
+ 3n
4(n + 1)(n + 2)
17 (a)
1
(2r – 1)(2r + 1)
≡
A
2r – 1
+
B
2r – 1
1 ≡ A(2r + 1) + B(2r – 1)
Let r = –
1
2
, Let r = –
1
2
,
l = B(–2) 1 = A(2)
B = –
1
2
A =
1
2
1
(2r – 1)(2r + 1)
=
1
2(2r – 1)
–
1
2(2r + 1)
=
1
21
1
2r – 1
–
1
2r + 12
(b) Σr = n
2n 1
(2r – 1)(2r + 1)
= Σr = n
2n
3
1
2(2r – 1)
–
1
2(2r + 1)4
=
1
2
Σr = n
2n
[ f (r) – f (r + 1)] 3f (r) =
1
2r – 14,
=
1
2
[ f (n) – f (2n + 1)]
=
1
2 3
1
2n – 1
–
1
4n + 14
=
1
2 1
4n + 1 – 2n + 1
(2n – 1)(4n + 1)2
=
n + 1
(2n – 1)(4n + 1)
=
an + b
(2n – 1)(4n + 1)
[Shown]
a = 1, b = 1
(c) lim 3
n + 1
(2n – 1)(4n + 1)4n → ∞
= lim
3
1
n
+
1
n2
8 – 2
n
– 1
n2
4= 0n → ∞
18 Let
1
(2r + 1)(2r + 3)
≡
A
2r + 1
+
B
2r + 3
l ≡ A(2r + 3) + B(2r + 1)
Let r = –
3
2
Let r = –
1
2
1 = B(–2) 1 = A(2)
B = –
1
2
A =
1
2
1
(2r + 1)(2r + 3)
=
1
2(2r + 1)
–
1
2(2r + 3)
Σr = 1
n 1
(2r + 1)(2r + 3)
=
1
2
Σr = 1
n
3 1
2r + 1
–
1
2r + 34
=
1
2
Σr = 1
n
[ f (r) – f (r + 1)] 3f (r) =
1
2r + 14
=
1
2
[ f (1) – f (n + 1)]
=
1
2 31
3
–
1
2n + 34
=
1
6
–
1
2(2n + 3)
Chapter 2.indd 5 6/24/2015 5:37:07 PM

6. 6 ACE AHEAD Mathematics (T) First Term Second Edition
© Oxford Fajar Sdn. Bhd. 2015
To test for the convergence,
lim 31
6
– 1
2(2n + 3)4 =
1
6n → ∞
The series converges to 1
6
.
S` =
1
6
19 Σn = 25
N
1 1
2n – 1
–
1
2n + 1
2
= Σn = 25
N
[ f (n) – f (n + 1)] 3f (n) =
1
2n – 1
4
= f (25) – f (N + 1)
=
1
50 – 1
–
1
2N + 1
=
1
7
–
1
2N + 1
Σn = 25
∞
un
= lim 11
7
– 1
2n + 1
2 =
1
7n → ∞
20 1
r!
– 1
(r + 1)!
=
(r + 1)! – r!
r!(r + 1)!
=
(r + 1)r! – r!
r!(r + 1)!
=
r
(r + 1)!
[Shown]
Σr = 1
n r
(r + 1)!
= Σr = 1
n
31
r!
– 1
(r + 1)!4
= Σr = 1
n
[ f (r) – f (r + 1)],
3f (r) =
1
r!4
= f (1) – f (n + 1)
= 1 – 1
(n + 1)!
21
r + 1
r + 2
–
r
r + 1
=
(r + 1)2
– r(r + 2)
(r + 1)(r + 2)
=
r2
+ 2r + 1 – r2
– 2r
(r + 1)(r + 2)
=
1
(r + 1)(r + 2)
[Shown]
Σr = 1
n 1
(r + 1)(r + 2)
= Σr = 1
n
3r + 1
r + 2
–
r
r + 14
= Σr = 1
n
[ f (r) – f (r – 1)],
3f (r) =
r + 1
r + 24
= f (n) – f (0)
=
n + 1
n + 2
– 1
2
=
2(n + 1) – (n + 2)
2(n + 2)
=
2n + 2 – n – 2
2(n + 2)
=
n
2(n + 2)
22 f (r) – f (r – 1)
=
1
(2r + 1)(2r + 3)
–
1
(2r – 1)(2r + 1)
=
2r – 1 – 2r – 3
(2r – 1)(2r + 1)(2r + 3)
=
– 4
(2r – 1)(2r + 1)(2r + 3)
[Shown]
Σr = 1
n 1
(2r – 1)(2r + 1)(2r + 3)
= – 1
4 Σr = 1
n – 4
(2r – 1)(2r + 1)(2r + 3)
= – 1
4 Σr = 1
n
[ f (r) – f (r – 1)]
3f (r) =
1
(2r + 1)(2r + 3)4
= – 1
4
[ f (n) – f (0)]
= – 1
43 1
(2n + 1)(2n + 3)
– 1
34
=
1
12
–
1
4(2n + 1)(2n + 3)
=
4n2
+ 6n + 2n + 3 – 3
12(2n + 1)(2n + 3)
=
4n2
+ 8n
12(2n + 1)(2n + 3)
=
n2
+ 2n
3(2n + 1)(2n + 3)
23 Let
1
(4n – 1)(4n + 3)
≡
A
4n – 1
+
B
4n + 3
l ≡ A(4n + 3) + B(4n – 1)
Let n = –
3
4
Let n =
1
4
1 = B(– 4) 1 = A(4)
B = –
1
4
A =
1
4
1
(4n – 1)(4n + 3)
=
1
41 1
4n – 1
–
1
4n + 32
Chapter 2.indd 6 6/24/2015 5:37:09 PM

7. ACE AHEAD Mathematics (T) First Term Second Edition
© Oxford Fajar Sdn. Bhd. 2015
7
Let f (r) =
1
4r + 3
1
4 Σr = 1
n
1 1
4r – 1
–
1
4r + 32
=
1
4 Σr = 1
n
[ f (r – 1) – f (r)]
=
1
4
[ f (0) – f (n)]
=
1
4 31
3
– 1
4n + 34
=
1
4 34n + 3 – 3
3(4n + 3)4 =
n
3(4n + 3)
24 f (x) = x + x + 1
1
f (x)
=
1
x + x + 1
×
x – x + 1
x – x + 1
=
x – x + 1
x – (x + 1)
= x + 1 – x
Σx = 1
24 1
f (x)
= – Σx = 1
24
1 x – x + 12
= – ( 1 – 2 + 2 – 3 + … +
24 – 25)
= – (1 – 5) = 4
25 Using long division,
1
x2
+ 6x + 8 2 x2
+ 6x + 0
x2
+ 6x + 8
–8
x(x + 6)
(x + 2)(x + 4)
≡ 1 –
8
(x + 2)(x + 4)
Let
8
(x + 2)(x + 4)
≡
A
(x + 2)
+
B
(x + 4)
,
8 ≡ A(x + 4) + B(x + 2)
Let x = – 4, Let x = –2,
8 = B(–2) 8 = A(2)
B = – 4 A = 4
x(x + 6)
(x + 2)(x + 4)
= 1 –
4
x + 2
+
4
x + 4
Σx = 1
n
31 –
4
x + 2
+
4
x + 44
= Σx = 1
n
1 – 4 Σx = 1
n
3 1
x + 2
–
1
x + 44
= n – 431
3
–
1
5
+
1
4
–
1
6
+
1
5
–
1
7
+ …
+
1
n + 1
–
1
n + 3
+
1
n + 2
–
1
n + 44
= n – 431
3
+
1
4
–
1
n + 3
–
1
n + 44
= n – 437
12
–
2n + 7
(n + 3)(n + 4)4
= n – 437(n2
+ 7n + 12) – 12(2n + 7)
12(n + 3)(n + 4) 4
= n –
7n2
+ 49n + 84 – 24n – 84
3(n + 3)(n + 4)
= n –
7n2
+ 25n
3(n + 3)(n + 4)
=
3n(n + 3)(n + 4) – n(7n + 25)
3(n + 3)(n + 4)
=
n[3n2
+ 7n + 12) – 7n – 25]
3(n + 3)(n + 4)
=
n[3n2
+ 21n + 36 – 7n – 25]
3(n + 3)(n + 4)
=
n(3n2
+ 14n + 11)
3(n + 3)(n + 4)
=
n(n + 1)(3n + 11)
3(n + 3)(n + 4)
[Shown]
26 ( ) ( )( ) ( )( )
( )( ) ( )
2 3 2 4 2 3 6 2 3
4 2 3 3
16 96 216
4 4 3 2 2
3 4
+ = + + +
+
= + +
x x x
x x
x xx x
x
x x x
2 3
4
4 4 3 2 2
216
81
2 3 2 4 2 3 6 2 3
4 2 3
+ +
+ -[ ] = + - + -
+ -
( ) ( )( ) ( )( )
( )( xx x
x x x
x
x x x
) ( )
( ) ( )
3 4
2 3
4
4 4
3
16 96 216 216
81
2 3 2 3 192 43
+ -
= - + - +
+ - + = + 22
2
2 3 2 2 3 2
192 2 432 2
1056 2
3
4 4
3
x
xLet =
+ - +
= +
=
,
( ) ( )
( )
27 (a)      
     
x
x
x x
x
x
x
x
x
x
x x
x x
+ = + + +
+ +
= +
1
5
1
10
1
10
1
5
1 1
5
5
5 4 3
2
2
3 4 5
5 33
3 5
3
3 2
10 10
1
5
1 1
1
3
1
3
1
1
+ + +
+
− = + − + − +
−
x
x
x x
x
x
x x
x
x
x
x
 
   
     
 3
Chapter 2.indd 7 6/24/2015 5:37:11 PM

8. 8 ACE AHEAD Mathematics (T) First Term Second Edition
© Oxford Fajar Sdn. Bhd. 2015
x x= + 55 33
3 5
3
3 2
10 10
1
5
1 1
1
3
1
3
1
1
+ + +
+
− = + − + − +
−
x
x
x x
x
x
x x
x
x
x
x
 
   
     
 
 
   
  
3
3
3
5 3
5 3
3
3 3
1 1
1 1
5 10
10
1
5
1
= − + −
+ − = + + +
+
x x
x x
x
x
x
x
x x x
x x 
 
  
+
× − +
−
1
3
3
1 1
5
3
3
x
x x
x x
Terms containing x4
= + −( )+ ( )
= − +
= −
x
x
x x x x
x x x
x
5 3 3
4 4 4
4
3
5 3 10
3 15 10
2
 
Coefficient of x4
= –2
(b) T
n
r
a b
r
x x
r
x x
r
r
n r r
r r
r r r
+
−
−
−
− −
=
= ( ) ( )
= ( )
=
1
2
6
1
12 2
6
2
6
2
6
2
 
 
 
 (( ) −r r
x12 3
The term independent of x is the term
where 12 – 3r = 0
r = 4
28 (a) 11 –
3
2
x2
5
(2 + 3x)6
= 31 + 5
C11–
3
2
x2 + 5
C21–
3
2
x2
2
+ …4
[26
+ 6
C1
(2)5
(3x) + 6
C2
(2)4
(3x)2
+ … ]
= 11 –
15
2
x +
45
2
x2
2(64 + 576x + 2160x2
)
= 64 + 576x + 2160x2
– 480x – 4320x2
+ 1440x2
= –720x2
+ 96x + 64 [Shown]
(b)
29 (a) 1 + 10(3x + 2x2
) +
10(9)
2
(3x + 2x2
)2
+
10(9)(8)
3!
(3x + 2x2
)3
= 1 + 30x + 20x2
+ 45(9x2
+ 12x3
)
+ 120(27x3
)
= 1 + 30x + 425x2
+ 3780x3
Coefficient of x3
= 3780
(b) (1 + x)–1
(4 + x2
)
–
1
2
= 31 +
–1
1!
(x) +
–1(–2)
2!
(x)2
+
–1(–2)(–3)
3!
(x)3
+ …4 14
–
1
2
2 31 +
x2
4 4
–
1
2
=
1
2
(1 – x + x2
– x3
+ …)1 +
– 1
2
1! 1x2
4 2
+
– 1
21– 3
22
2! 1x2
4 2
2
+ …
=
1
2
(1 – x + x2
– x3
+ …) 11 –
x2
8
+ …2
   
   
1
3
2
1
3
2
1 5
3
2
10
3
2
10
2
5
5
2
2
3
− − = − +








= − + + + −
x x x x
x x x x
x    
 

3
2
5
3
2
1
15
2
5 10
9
4
3
10
27
8
27
4
3
4
4
2 2 2
3
+ + +
= − − + + + −
+
x x x
x x x x x
x xx x x x
x x x x x
x
  + + + +
= − − + + + −
−
5
81
16
6
1
15
2
5
45
2
30 10
135
4
13
4 4
2 2 3 4
3
…
55
2
405
16
4 4
x x+ +
=
…
1
15
2
35
5
15
4
515
16
2 3 4
- + - -x x x x
Chapter 2.indd 8 6/24/2015 5:37:13 PM

9. ACE AHEAD Mathematics (T) First Term Second Edition
© Oxford Fajar Sdn. Bhd. 2015
9
=
1
2
11 –
x2
8
– x +
x3
8
+ x2
– x3
2
=
1
2
11 – x +
7
8
x2
–
7
8
x3
+ …2
where |x| 1
30 Let f (x) ≡
A
1 – x
+
Bx + C
1 + x2
1 + 2x + 3x2
≡ A(1 + x2
) + (Bx + C)(1 – x)
Let x = 1,
1 + 2 + 3 = A(2)
A = 3
Let x = 0,
1 = 3(1) + (0 + C)(1)
C = –2
Let x = –1,
1 – 2 + 3 = 3(2) + [–B + (–2)](2)
2 = 6 – 4 – 2B
B = 0
Hence, f (x) =
3
1 – x
– 2
1 + x2
f (x) = 3(1 – x)–1
– 2(1 + x2
)–1
= 331 +
–1
1!
(–x) +
–1(–2)
2!
(–x)2
+
–1(–2)(–3)
3!
(–x)3
+ …4
– 231 +
–1
1!
(x)2
+ …4
= 3(1 + x + x2
+ x3
+ …) – 2(1 – x2
+ …)
= 3 + 3x + 3x2
+ 3x3
– 2 + 2x2
= 1 + 3x + 5x2
+ 3x3
where |x| 1
[Shown]
31 (a)
( 5 + 2)6
– ( 5 – 2)6
8 5
=
[( 5 + 2)3
+ ( 5 – 2)3
][( 5 + 2)3
– ( 5 – 2)3
]
8 5
=
( 5 + 2 + 5 – 2)[( 5 + 2)2
– ( 5 + 2)( 5 – 2) + ( 5 – 2)2
]
( 5 + 2 – 5 + 2)[( 5 + 2)2
+ ( 5 + 2)( 5 – 2) + ( 5 – 2)2
]
8 5
=
[2 5(5 + 4 5 + 4 – 1 + 5
– 4 5 + 4)] [(4)(5 + 4 5
+ 4 + 1 + 5 – 4 5 + 4)]
8 5
= (17)(19)
= 323
(b) (1 – 3x)
1
3
= 31 +
1
3
1!
(–3x) +
1
31– 2
32
2!
(–3x)2
+
1
31– 2
321– 5
32
3!
(–3x)3
+ …4
= 1 – x – x2
– 5
3
x3
– … where |x|
1
3
When x =
1
8
,
(1 – 3x)
1
3 = 15
82
1
3
=
3
5
2
≈ 1 – 1
8
– 11
82
2
– 5
311
82
3
– …
3
5 ≈ 1315
1536
(2)
≈ 1315
768
= 1.17
[2 decimal places]
32 ( ) ( )
( )( )
!
( )
(
1 1 2
2 2 1
2
1 2 2 1
1
2 2
2
- = + - +
-
- +
= + + -( ) +
+
y
y
n
n y
n n
y
ny n n y
…
…
)) ( )
( )( )
!
( )-
= + -( ) +
- - -
+
= - + +( ) +
2 2
2
1 2
2 2 1
2
1 2 2 1
n
n y
n n
y
ny n n y
…
…
 1
1
1 1
1 2 2 1
1 2 2 1
2
2 2
2
−
+
= −( ) +( )
= − + −( ) +( )
− + +(
−y
y
y y
ny n n y
ny n n
n
n n
…
)) +( )
= − + + − +
+ −( )
= − +
∴
y
ny n n y ny
n y n n y
ny n y
2
2
2 2 2
2 2
1 2 2 1 2
4 2 1
1 4 8
…
( )
11
1
1 4 8
1
50
1
16
2
2 2−
+
= − + +
= =
y
y
ny n y
y n
n
 …
Let ,
1
1
50
1
1
50
1 4
1
16
1
50
8
1
10
1
50
49
51
79 601
80 00
1
8
2 2
1
8
-
+
» - +
»
     
  00
1 1
Chapter 2.indd 9 6/24/2015 5:37:15 PM

10. 10 ACE AHEAD Mathematics (T) First Term Second Edition
© Oxford Fajar Sdn. Bhd. 2015
33 (1 – x)10
= 1 – 10x + 45x2
+ …
(1 + 2x2
)3
= 1 + 6x2
+ …
(1 + ax)5
= 1 + 5ax + 10a2
x2
+ …
(1 + bx2
)4
= 1 + 4bx2
+ …
(1 – x)10
(1 + 2x2
)3
– (1 + ax)5
(1 + bx2
)4
= (1 – 10x + 45x2
+ …)(1 + 6x2
+ …)
– (1 + 5ax + 10a2
x2
+ …)(1 + 4bx2
+ …)
= 1 + 6x2
– 10x + 45x2
– 1 – 4bx2
– 5ax
– 10a2
x2
+ …
–10 – 5a = 0 6 + 45 – 4b – 10a2
= 0
a = –2 6 + 45 – 4b – 40 = 0
b =
11
4
34 (a) 1x +
1
x2
5
1x –
1
x2
3
= 3x5
+ 5(x)4
11
x2+ 10x3
11
x2 2+ 10x2
11
x3 2
+ 5x11
x4 2+
1
x5 43x3
+ 3(x)2
1–1
x 2
+ 3x1–
1
x2
2
+ 1–
1
x2
3
4
= 1x5
+ 5x3
+ 10x +
10
x
+
5
x3 +
1
x5 2
1x3
– 3x +
3
x
–
1
x32
To obtain x4
term,
= … + x5
13
x2+ 5x3
(–3x) + 10x(x3
) + …
= 3x4
– 15x4
+ 10x4
= –2x4
The coefficient of x4
term is –2.
(b) (1 + x)
1
5 – 5 + 3x
5 + 2x
= 31 +
1
5
1!
(x) +
1
51– 4
52
2!
(x)2
+
1
51– 4
521– 9
52
3!
(x)3
+ …4
– (5 + 3x)(5 + 2x)–1
= 11 +
1
5
x –
2
25
x2
+
6
125
x3
2
– (5 + 3x)(5)–1
11 +
2
5
x2
–1
= 11 +
1
5
x –
2
25
x2
+
6
125
x3
2
– (5 + 3x)11
5231 +
–1
1!12
5
x2
+
–1(–2)
2!
12
5
x2
2
+
–1(–2)(–3)
3! 12
5
x2
3
4
= 1 +
1
5
x –
2
25
x2
+
6
125
x3
–
1
5
(5 + 3x)
11 –
2
5
x +
4
25
x2
–
8
125
x3
2
= 1 +
1
5
x –
2
25
x2
+
6
125
x3
–
1
5 15 – 2x
+
4
5
x2
–
8
25
x3
+ 3x –
6
5
x2
+
12
25
x3
2
= 1 +
1
5
x –
2
25
x2
+
6
125
x3
– 1 –
1
5
x +
2
25
x2
–
4
125
x3
=
2
125
x3
where |x|
2
5
[Shown]
Hence, p =
2
125
By using x = 0.02,
(1.02)
1
5
–
1253
50 2
1252
50 2
=
2
125
(0.02)3
= 1.28 × 10–7
[Shown]
35 (1 + ax + bx2
)7
= [(1 + ax) + (bx2
)]7
= 7
Cr
(1 + ax)7 – r
(bx2
)r
x term:
= 7
C0
(1 + ax)7
= 7
C0
[7
C1
(ax)]
= 7ax
x2
term:
= 7
C0
(1 + ax)7
+ 7
C1
(1 + ax)6
(bx2
)
= 7
C0
[7
C2
(ax)2
] + 7
C1
[6
C0
(ax)0
(bx2
)]
= 21a2
x2
+ 7bx2
= (21a2
+ 7b)x2
7a = 1 21a2
+ 7b = 0
a =
1
7
2111
72
2
+ 7b = 0
7b = –
3
7
b = –
3
49
Let (1 + ax + bx2
)7
= 1 + x
(1 + x)
1
7
= 1 +
1
7
x –
3
49
x2
+ … where |x| 1
By using x = 0.014,
7 1.014 ≈ 1 +
1
7
(0.014) –
3
49
(0.014)2
≈ 1.001988 [6 decimal places]
36 (1 + x)7
1 – 2x
= (1 + x)7
(1 – 2x)–1
= [1 + 7x + 21x2
+ 35x3
+ …] 31 +
–1
1!
(–2x)
+
–1(–2)
2!
(–2x)2
+
–1(–2)(–3)
3!
(–2x)3
+ …4
Chapter 2.indd 10 6/24/2015 5:37:17 PM

11. ACE AHEAD Mathematics (T) First Term Second Edition
© Oxford Fajar Sdn. Bhd. 2015
11
= (1 + 7x + 21x2
+ 35x3
+ …) (1 + 2x + 4x2
+ 8x3
+ …)
= 1 + 2x + 4x2
+ 8x3
+ 7x + 14x2
+ 28x3
+ 21x2
+ 42x3
+ 35x3
+ …
= 1 + 9x + 39x2
+ 113x3
+ … where |x|
1
2
[Shown]
Using x = –0.01,
(0.99)7
(1.02)
≈ 1 + 9(–0.01) + 39(–0.01)2
+ 113(–0.01)3
+ …
≈ 0.914 [3 decimal places]
37 1 + x
= (1 + x)
1
2
= 1 +
1
2
1!
(x) +
1
21– 1
22
2!
(x)2
+
1
21– 1
221– 3
22
3!
(x)3
+
1
21– 1
221– 3
221– 5
22
4!
(x)4
+ …
= 1 +
x
2
–
x2
8
+
x3
16
–
5
128
x4
+ … … 1
1
4
(6 + x) – (2 + x)–1
=
6 + x
4
– 2–1
31 +
x
24
– 1
=
6 + x
4
–
1
231 +
–1
1!
1x
22 +
–1(–2)
2!
1x
22
2
+
–1(–2)(–3)
3!
1x
22
3
+
–1(–2)(–3)(– 4)
4!
1x
22
4
4
=
6 + x
4
–
1
211 –
x
2
+
x2
4
–
x3
8
+
x4
16
– …2
=
6 + x
4
–
1
2
+
x
4
–
x2
8
+
x3
16
–
x4
32
+ …
= 1 +
x
2
–
x2
8
+
x3
16
–
x4
32
… 2
To obtain the error,
1 – 2 , 11 +
x
2
–
x2
8
+
x3
16
–
5
128
x4
2
– 11 +
x
2
–
x2
8
+
x3
16
–
x4
32 2
= –
5
128
x4
+
x4
32
= –
x4
128
The error is approximately
x4
128
.[Shown]
38 ( )a b a
b
a
a
b
a
a
b
a
a
b
+ = +








= +
+ = +
1
2
1
2 1
2
1
2
1
2
1
2 1
2
1 1
1 1
1
2
   
  aa
b
a
b
a
a
  
    
 

+
−






+
− −
+




=
1
2
1
2
1
2
1
2
1
2
1
1
2
2
3
2
3
1
2
!
!
…
11
2 8
16
1
1
2 8
2
2
3
3
1
2
1
2
1
2
2
2
2
+ − +
+
−( ) = −
= − − −
b
a
b
a
b
a
a b a
b
a
a
b
a
b
a
… …
 
 bb
a
a b a b a
b
a
b
a
a
b
a
b
a
3
3
1
2
1
2
1
2
3
3
1
2
3
3
16
2
2
2
16
8
+
+( ) − −( ) = +
= +
… …
 
 
Let a = 4, b = 1
5 3 4
1
4
1
8 4
129
256
1
2
3
- = +
( )
æ
è
ç
ç
ö
ø
÷
÷
=
5 3 5 3 5 3
2
129
256
5 3 2
5 3 2
256
129
512
129
-( ) +( )= -
=
+( )=
+( )= ´
=
1 – 2
2
1
Chapter 2.indd 11 6/24/2015 5:37:19 PM

12. 12 ACE AHEAD Mathematics (T) First Term Second Edition
© Oxford Fajar Sdn. Bhd. 2015
39 y =
1
1 + 3x + 1 + x
×
1 + 3x – 1 + x
1 + 3x – 1 + x
=
1 + 3x – 1 + x
1 + 3x – (1 + x)
=
1
2x
1 1 + 3x – 1 + x[Shown]
1
2x3(1 + 3x)
1
2
– (1 + x)
1
2
4
=
1
2x531 +
1
2
1!
(3x) +
1
21– 1
22
2!
(3x)2
+
1
21– 1
22 1– 3
22
3!
(3x)3
+ …4– 31 +
1
2
1!
(x)
+
1
2
1– 1
22
2!
(x)2
+
1
21– 1
22 1– 3
22
3!
(x)3
+ …46
=
1
2x311 +
3
2
x –
9
8
x2
+
27
16
x3
+ …2 – 11 +
x
2
–
x2
8
+
x3
16
+ …24
=
1
2x1x – x2
+
13
8
x3
+ …2
=
1
2
–
x
2
+
13
16
x2
Using x =
1
100
,
y =
1
1 + 3x + 1 + x
=
1
103
100
+
101
100
=
10
103 + 101
≈
1
2
–
1 1
1002
2
+
13
161 1
1002
2
≈
79 213
160 000
[Shown]
40 (a) 3 – 5x + 3x2
≡ A(1 + x2
) + (B + Cx)(1 – 2x)
Let x =
1
2
,
5
4
=
5
4
(A)
A = 1
Let x = 0,
3 = 1(1 + 0) + (B + 0)(1)
B = 2
Let x = 1,
1 = 1(2) + (2 + C )(–1)
C + 2 = 1
C = –1
(b) (1 – 2x)–1
= 31 +
–1
1!
(–2x) +
–1(–2)
2!
(–2x)2
+
–1(–2)(–3)
3!
(–2x)3
4
= 1 + 2x + 4x2
+ 8x3
(1 + x2
)–1
= 31 +
–1
1!
(x2
) + …4
= 1 – x2
(c) (3 – 5x + 3x2
)(1 – 2x)–1
(1 + x2
)–1
=
1
1 – 2x
+
2 – x
1 + x2
= 1(1 – 2x)–1
+ (2 – x)(1 + x2
)–1
= 1(1 + 2x + 4x2
+ 8x3
) + (2 – x)(1 – x2
)
= 1 + 2x + 4x2
+ 8x3
+ 2 – 2x2
– x + x3
= 3 + x + 2x2
+ 9x3
⇒ a = 1, b = 2, c = 9
41 (a) ur
r r
r r
r
= +
= +
= +( )
=
+ + −
+ +
+
   
   
 
1
3
1
3
1
3
9
1
3
1
3
1 9
10
2 1 2 1 2
2 1 2 1
2 1
332 1r +
(b) It is a G.P. with a =
10
27
and r =
1
9
S
a r
r
S
n
n
n
n
=
−( )
−
=
−








−
= −








=∞
1
1
10
7
1
1
9
1
1
9
5
12
1
1
9
 
 
55
12
Chapter 2.indd 12 6/24/2015 5:37:22 PM