1. The document contains 14 multi-part mathematics questions involving sequences and series. 2. Many questions involve finding the common ratio (r), first term (a), or sum (Sn) of sequences that satisfy given properties or equations relating to arithmetic and geometric progressions. 3. Proofs are provided that certain expressions are arithmetic or geometric progressions, or that certain series representations are valid.

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