chapter 2 sequencess and series
TRANSCRIPT
FULLY WORKED SOLUTIONS
2CHAPTER
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 = 3 132
(a + u13
) = 234
a = 3 – 12d ... 1 132
(a + 3) = 234 ... 2
From 2 , a = 33
When a = 33, 33 = 3 – 12d ⇒ d = – 52
S25
= 252
3(2)(33) + (24)1– 5224
= 252
(66 – 60) = 75
2 (a) a = 1, r = 54
Sn =
a(rn – 1)r – 1
= 1542
n
– 1
154 – 12 = 431542
n
– 14 (b) S
n > 20
431542n
– 14 > 20
1542n
– 1 > 5
1542n
> 6
n lg 54
> lg 6
n > 8.03 The least number of terms is 9.
3 312 – 224 + 332 – 424 + … + 3(2n – 1)2 – (2n)24 = –3 – 7 – 11… a = –3, d = – 4
Sn = n
2 3–6 + (n – 1)(– 4)4
= n2
(–6 – 4n + 4) = n2
(– 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 u
n = a + (n – 1)(2) = a + 2n – 2
When n = 20, u20
= a + 2(20) – 2 = a + 38S
20 = 1120
202
(a + u20
) = 1120
10(a + a + 38) = 11202a = 74a = 37
u20
= a + 38 = 37 + 38 = 75
5 (a) a = 3, d = 4, Sn = 820
Sn = n
2[2a + (n – 1)d ]
820 = n2
(6 + 4n – 4)
1640 = n(2 + 4n)n(1 + 2n) = 820
n + 2n2 – 820 = 0(2n + 41)(n – 20) = 0
n = 20 since n = – 412
is not a solution.
T20
= 3 + 19(4) = 79
Chapter 2.indd 1 6/24/2015 5:36:59 PM
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, S
8 = 20, S
13 = 39
(i) S8 = 8p + 64q
8p = 20 – 64q
p = 52
– 8q … 1
S13
= 3913p + 169q = 39 … 2
Substitute 1 into 2 ,
13152 – 8q2 + 169q = 39
652
– 104q + 169q = 39
65q = 132
q = 110
p = 52
– 81 1102 = 17
10(ii) u
n = S
n – S
n – 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)
= 1710
+ 110
(2n – 1)
= 1710
+ n5
– 110
= 8 + n5
(iii) To show that it is an A.P.,
un = 1
5 (8 + n)
un – 1
= 15
[8 + (n – 1)]
= 15
(7 + n)
un – u
n – 1 = 1
5
⇒ Series is an A.P. with d = 15 = 0.2.
7 If J, K, M is a G.P., then
KJ
= MK
K 2 = MJ … 1
(ar k – 1)2 = (ar m – 1) (ar j – 1)
r 2k – 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) Sn
a ln
n
nn
n = + = + +[ ]
= +
2 22000 400 1600
2400 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
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
8a(1 – r10)
1 – r – a(1 – r5)
1 – r = – 11
8a(1 + r5)(1 – r5)
1 – r – a(1 – r5)
1 – r = – 11
8 … 2
a(1 – r5)1 – r
(1 + r5 – 1) = – 118
Substitute 1 into 2 ,44(r5) = – 11
8
r5 = – 132
r = – 12
[Shown]
a31 – 1– 122
541 – 1– 1
22 = 44
a11 + 1322 = 66
a = 64
Sn = a
1 – r
= 64
1 – 1– 122
= 4223
10 (a) r = 3 – 13 + 1
× 3 – 13 – 1
= 3 – 2 3 + 1
3 – 1 = 2 – 3
u3 = u
2r
= ( 3 – 1)(2 – 3)= 2 3 – 3 – 2 + 3= 3 3 – 5
u4 = u
3r
= (3 3 – 5)(2 – 3)= 6 3 – 3(3) – 10 + 5 3= 11 3 – 19
(b) r = 2 – 3 [< 1]
S∞ = a1 – r
= 3 + 1
1 – (2 – 3 )
= 3 + 13 – 1
× 3 + 13 + 1
= 3 + 2 3 + 12
= 2 + 3
11 (a) S
6
S3
= 78
, u2 = – 4
8S6 = 7S
3 … 1
ar = – 4
a = – 4r … 2
83a(1 – r6)1 – r 4 = 73a(1 – r3)
1 – r 48(1 – r6) = 7(1 – r3)
8 – 8r6 = 7 – 7r3
8r6 – 7r3 – 1 = 0(r3 – 1)(8r3 + 1) = 0
r3 = 1 or – 18
r = – 12
since r = 1 is
not a solution.a = – 4
1– 122
= 8
(b) r = 85
× 12
= 45
, a = 2
Sn > 9.9
a(1 – rn)1 – r
> 9.9
231 – 1452
n41152
> 9.9
1 – 1452n
> 0.99
1452n
< 0.01
n lg 45
< lg (0.01)
n > 20.64 The least number of terms is 21.
12 u3 = S
2
ar2 = a(1 – r2)
1 – rar2(1 – r) = a(1 + r)(1 – r)
ar2 = a(1 + r)r2 – r – 1 = 0
r = –(–1) ± (–1)2 – 4(1)(–1)
2
= 1 ± 52
a = 2
When r = 1 + 52
[|r| > 1], S∞ doesn’t exist
When r = 1 – 52
[|r| < 1],
Chapter 2.indd 3 6/24/2015 5:37:04 PM
4 ACE AHEAD Mathematics (T) First Term Second Edition© Oxford Fajar Sdn. Bhd. 2015
S∞ = a1 – r
= 2
1 – 1 – 5
2
= 2
12 – 1 + 52 2
= 41 + 5
× 1 – 51 – 5
= 4 – 4 5
1 – 5 = 5 – 1
13 a = 3, r = 0.4 (a) u
n < 0.02
ar n – 1 < 0.023(0.4)n – 1 < 0.02
(0.4)n – 1 < 1
150
(n – 1)lg 0.4 < lg 1
150
n – 1 > 5.47n > 6.47
The least number of terms is 7. (b) S∞ – S
n < 0.01
a1 – 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 ,S
n – rS
n = 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∞
=
a1 – r
– a(1 – r n)
1 – ra
1 – r
=
a1 – r
[1 – (1 – r n)]
a1 – r
= rn [Shown]
(a) u4 = 18, u
7 = 16
3
ar3 = 18 … 1 ar6 = 163
… 2
2
1, r3 = 116
3 2 × 118
r3 = 1 8272
r = 1232 a1232
3
= 18
a = 2434
S∞ = 1243
4 21 – 2
3
= 7294
(b) S∞ – S
n
S∞
< 0.001
rn < 0.001
1232n
< 0.001
n lg 23
< lg 0.001
n > 17.04 The least value of n is 18.
15 2r – 1
r(r – 1) –
2r + 1r(r + 1)
= (r + 1)(2r – 1) – (2r + 1)(r – 1)r(r – 1)(r + 1)
= 2r2 – r + 2r – 1 – 2r2 + 2r – r + 1r(r – 1)(r + 1)
= 2rr(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 + 1r(r + 1)4
Chapter 2.indd 4 6/24/2015 5:37:06 PM
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 – 1r(r – 1)4
= f (2) – f (n + 1)
= 32
– 2n + 1n(n + 1)
[Proven]
12
Σr = 2
∞ 2(r – 1)(r + 1)
= 12
lim 332
– 2n + 1n(n + 1)4n → ∞
= 12
lim 332
–
2n
+ 1n2
1 + 1n4n → ∞
= 12
132 – 02 = 34
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 1r(r + 1)(r + 2)
= 12
Σr = 1
n 2r(r + 1)(r + 2)
= 12
Σr = 1
n 3 1
r(r + 1) – 1
(r + 1)(r + 2)4 3f (r) =
1r(r + 1)4
= 12
Σr = 1
n [ f (r) – f (r + 1)]
= 12
[ f (1) – f (n + 1)]
= 12
312
– 1(n + 1)(n + 2)4
= 12
3n2 + 3n + 2 – 22(n + 1)(n + 2)4
= n2 + 3n4(n + 1)(n + 2)
17 (a) 1
(2r – 1)(2r + 1) ≡
A
2r – 1 +
B
2r – 11 ≡ A(2r + 1) + B(2r – 1)
Let r = – 12
, Let r = – 12
,
l = B(–2) 1 = A(2)
B = – 12
A = 12
1
(2r – 1)(2r + 1) =
12(2r – 1)
– 1
2(2r + 1)
= 121
12r – 1
– 1
2r + 12 (b) Σ
r = n
2n
1(2r – 1)(2r + 1)
= Σr = n
2n 3 1
2(2r – 1) –
12(2r + 1)4
= 12
Σr = n
2n [ f (r) – f (r + 1)] 3f (r) =
12r – 14,
= 12
[ f (n) – f (2n + 1)]
= 12
3 12n – 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 31n
+ 1n2
8 – 2n
– 1n24 = 0
n → ∞
18 Let 1(2r + 1)(2r + 3)
≡ A2r + 1
+ B2r + 3
l ≡ A(2r + 3) + B(2r + 1)
Let r = – 32
Let r = – 12
1 = B(–2) 1 = A(2)
B = – 12
A = 12
1(2r + 1)(2r + 3)
= 12(2r + 1)
– 12(2r + 3)
Σr = 1
n 1(2r + 1)(2r + 3)
= 12
Σr = 1
n 3 1
2r + 1 – 1
2r + 34 = 1
2 Σr = 1
n [ f (r) – f (r + 1)] 3f (r) =
12r + 14
= 12
[ f (1) – f (n + 1)]
= 12
313 – 12n + 34
= 16
– 12(2n + 3)
Chapter 2.indd 5 6/24/2015 5:37:07 PM
6 ACE AHEAD Mathematics (T) First Term Second Edition© Oxford Fajar Sdn. Bhd. 2015
To test for the convergence,
lim 316 – 12(2n + 3)4 = 1
6n → ∞
The series converges to 16
.
S` = 1
6
19 Σn = 25
N 1 1
2n – 1 –
12n + 12
= Σn = 25
N[ f (n) – f (n + 1)] 3f (n) =
12n – 14
= f (25) – f (N + 1)
= 1
50 – 1 –
12N + 1
= 17
– 1
2N + 1
Σn = 25
∞ u
n = lim 117 – 1
2n + 12 = 17n → ∞
20 1r!
– 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= Σ
r = 1
n [ f (r) – f (r + 1)],
3f (r) = 1r!4
= f (1) – f (n + 1)
= 1 – 1(n + 1)!
21 r + 1r + 2
– rr + 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= Σ
r = 1
n [ f (r) – f (r – 1)],
3f (r) = r + 1r + 24
= f (n) – f (0)
= n + 1n + 2
– 12
= 2(n + 1) – (n + 2)2(n + 2)
= 2n + 2 – n – 22(n + 2)
= n2(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)
= – 14
Σr = 1
n – 4(2r – 1)(2r + 1)(2r + 3)
= – 14
Σr = 1
n [ f (r) – f (r – 1)]
3f (r) = 1
(2r + 1)(2r + 3)4 = – 1
4[ f (n) – f (0)]
= – 143
1(2n + 1)(2n + 3)
– 134
= 112
– 14(2n + 1)(2n + 3)
= 4n2 + 6n + 2n + 3 – 312(2n + 1)(2n + 3)
= 4n2 + 8n12(2n + 1)(2n + 3)
= n2 + 2n3(2n + 1)(2n + 3)
23 Let 1(4n – 1)(4n + 3)
≡ A4n – 1
+ B4n + 3
l ≡ A(4n + 3) + B(4n – 1)
Let n = – 34
Let n = 14
1 = B(– 4) 1 = A(4)
B = – 14
A = 14
1(4n – 1)(4n + 3)
= 141
14n – 1
– 14n + 32
Chapter 2.indd 6 6/24/2015 5:37:09 PM
ACE AHEAD Mathematics (T) First Term Second Edition© Oxford Fajar Sdn. Bhd. 2015
7
Let f (r) = 14r + 3
14
Σr = 1
n 1 1
4r – 1 – 1
4r + 32 = 1
4 Σr = 1
n [ f (r – 1) – f (r)]
= 14
[ f (0) – f (n)]
= 14
313 – 14n + 34
= 14
34n + 3 – 33(4n + 3) 4 = n
3(4n + 3)
24 f (x) = x + x + 1
1f (x)
= 1
x + x + 1 × x – x + 1
x – x + 1
= x – x + 1x – (x + 1)
= x + 1 – x
Σx = 1
24 1f (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 2x2 + 6x + 0x2 + 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 – 4x + 2
+ 4x + 4
Σx = 1
n 31 – 4
x + 2 + 4
x + 44 = Σ
x = 1
n 1 – 4 Σ
x = 1
n 3 1
x + 2 – 1
x + 44 = n – 4313 – 1
5 + 1
4 – 1
6 + 1
5 – 1
7 + …
+ 1n + 1
– 1n + 3
+ 1n + 2
– 1n + 44
= n – 4313 + 14
– 1n + 3
– 1n + 44
= n – 43 712
– 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 – 843(n + 3)(n + 4)
= n – 7n2 + 25n3(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)
xx
x xx
x x
xx
xx x
x x
+ = + + +
+ +
= +
15
110
1
101
51 1
5
55 4 3
2
23 4 5
5 33
3 5
33 2
10 101
51 1
13
13
1
1
+ + +
+
− = + − + − +
−
xx
x x
xx
x xx
xx
x
3
33
5 35 3
3
3 31 1
1 15 10
101
51
= − + −
+ − = + + +
+
x xx x
xx
xx
x x x
x x
+
× − +
−
13
31 1
53
3
xx x
x x
Chapter 2.indd 7 6/24/2015 5:37:11 PM
8 ACE AHEAD Mathematics (T) First Term Second Edition© Oxford Fajar Sdn. Bhd. 2015
xx
x xx
x x
xx
xx x
x x
+ = + + +
+ +
= +
15
110
1
101
51 1
5
55 4 3
2
23 4 5
5 33
3 5
33 2
10 101
51 1
13
13
1
1
+ + +
+
− = + − + − +
−
xx
x x
xx
x xx
xx
x
3
33
5 35 3
3
3 31 1
1 15 10
101
51
= − + −
+ − = + + +
+
x xx x
xx
xx
x x x
x x
+
× − +
−
13
31 1
53
3
xx x
x x
Terms containing x4
= + −( ) + ( )= − += −
xx
x x x x
x x x
x
5 3 3
4 4 4
4
35 3 10
3 15 10
2
Coefficient of x4 = –2
(b) Tn
ra b
rx x
rx x
r
rn r r
r r
r r r
+−
− −
− −
=
= ( ) ( )
= ( )
=
1
2 6 1
12 2
62
62
62
(( ) −r rx12 3
The term independent of x is the term where 12 – 3r = 0
r = 4
28 (a) 11 – 32
x25
(2 + 3x)6
= 31 + 5C11– 3
2x2 + 5C
21– 32
x22
+ …4 [26 + 6C
1(2)5(3x) + 6C
2(2)4(3x)2 + … ]
= 11 – 152
x + 452
x22(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)–
12
= 31 + –11!
(x) + –1(–2)2!
(x)2 +
–1(–2)(–3)3!
(x)3 + …414–
122 31 + x2
4 4–
12
= 12
(1 – x + x2 – x3 + …)1 + – 1
21! 1
x2
4 2
+ – 1
21– 322
2! 1x2
4 22
+ …
= 12
(1 – x + x2 – x3 + …) 11 – x2
8 + …2
13
21
3
2
1 53
210
3
2
10
25
5
22
3
− − = − +
= − + + + −
x x x x
x x x x
x
3
25
3
2
115
25 10
9
43
1027
8
27
4
34
4
2 2 2
3
+ + +
= − − + + + −
+
x x x
x x x x x
x xx x x x
x x x x x
x
+ + + +
= − − + + + −
−
581
166
115
25
45
230 10
135
4
13
4 4
2 2 3 4
3
…
55
2
405
164 4x x+ +
=
…
115
2
35
5
15
4
515
162 3 4- + - -x x x x
Chapter 2.indd 8 6/24/2015 5:37:13 PM
ACE AHEAD Mathematics (T) First Term Second Edition© Oxford Fajar Sdn. Bhd. 2015
9
= 12
11 – x2
8 – x + x3
8 + x2 – x32
= 12
11 – x + 78
x2 – 78
x3 + …2where |x| < 1
30 Let f (x) ≡ A1 – x
+ Bx + C1 + 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 – 2BB = 0
Hence, f (x) = 31 – x
– 21 + x2
f (x) = 3(1 – x)–1 – 2(1 + x2)–1
= 331 + –11!
(–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)13
= 31 + 13
1! (–3x) +
131– 2
322!
(–3x)2
+
131– 2
321– 532
3!(–3x)3 + …4
= 1 – x – x2 – 53
x3 – … where |x| < 13
When x = 18
,
(1 – 3x)13 = 1582
13 =
3 52
≈ 1 – 18
– 11822
– 531
182
3
– …
3 5 ≈ 13151536
(2)
≈ 1315768
= 1.17 [2 decimal places]
32 ( ) ( )( )( )
!( )
(
1 1 22 2 1
2
1 2 2 1
1
2 2
2
- = + - +-
- +
= + + -( ) +
+
y
y
n n yn n
y
ny n n y
…
…
)) ( )( )( )
!( )- = + -( ) +
- - -+
= - + +( ) +
2 2
2
1 22 2 1
2
1 2 2 1
n n yn n
y
ny n n y
……
1
11 1
1 2 2 1
1 2 2 1
22 2
2
−+
= −( ) +( )= − + −( ) +( )
− + +(
−y
yy y
ny n n y
ny n n
nn 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
11 4 8
1
50
1
16
22 2−
+= − + +
= =
y
yny n y
y n
n …
Let ,
11
50
11
50
1 41
16
1
508
1
10
1
50
49
51
79 601
80 00
1
82 2
1
8
-
+» - +
»
00
1 1
Chapter 2.indd 9 6/24/2015 5:37:15 PM
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 + 10a2x2 + … (1 + bx2)4 = 1 + 4bx2 + … (1 – x)10 (1 + 2x2)3 – (1 + ax)5 (1 + bx2)4
= (1 – 10x + 45x2 + …)(1 + 6x2 + …)– (1 + 5ax + 10a2x2 + …)(1 + 4bx2 + …)
= 1 + 6x2 – 10x + 45x2 – 1 – 4bx2 – 5ax– 10a2x2 + …
–10 – 5a = 0 6 + 45 – 4b – 10a2 = 0a = –2 6 + 45 – 4b – 40 = 0
b = 114
34 (a) 1x + 1x2
5
1x – 1x2
3
= 3x5 + 5(x)411x2 + 10x31 1x2 2 + 10x21 1
x3 2+ 5x1 1
x4 2 + 1x5 4 3x3 + 3(x)21–1
x 2+ 3x1– 1
x22
+ 1– 1x2
3
4 = 1x5 + 5x3 + 10x + 10
x + 5
x3 + 1
x5 21x3 – 3x + 3
x – 1
x32 To obtain x4 term,
= … + x513x2 + 5x3(–3x) + 10x(x3) + …
= 3x4 – 15x4 + 10x4 = –2x4
The coefficient of x4 term is –2.
(b) (1 + x)15 – 5 + 3x
5 + 2x
= 31 +
151!
(x) +
151– 4
522!
(x)2
+
151– 4
521– 952
3!(x)3 + …4
– (5 + 3x)(5 + 2x)–1
= 11 + 15
x – 225
x2 + 6125
x32– (5 + 3x)(5)–111 + 2
5x2
–1
= 11 + 15
x – 225
x2 + 6125
x32– (5 + 3x)115231 + –1
1!125
x2+ –1(–2)
2! 125x2
2
+ –1(–2)(–3)3! 125x2
3
4
= 1 + 15
x – 225
x2 + 6125
x3 – 15
(5 + 3x)
11 – 25
x + 425
x2 – 8125
x32= 1 + 1
5x – 2
25x2 + 6
125x3 – 1
5 15 – 2x
+ 45
x2 – 825
x3 + 3x – 65
x2 + 1225
x32= 1 + 1
5 x – 2
25 x2 + 6
125 x3 – 1 – 1
5 x + 2
25 x2
– 4125
x3 = 2125
x3 where |x| < 25 [Shown]
Hence, p = 2125
By using x = 0.02,
(1.02)15 –
125350 2
125250 2
= 2125
(0.02)3
= 1.28 × 10–7 [Shown] 35 (1 + ax + bx2)7
= [(1 + ax) + (bx2)]7
= 7Cr (1 + ax)7 – r (bx2)r
x term: = 7C
0(1 + ax)7
= 7C0[7C
1 (ax)]
= 7ax x2 term: = 7C
0(1 + ax)7 + 7C
1(1 + ax)6 (bx2)
= 7C0[7C
2(ax)2] + 7C
1[6C
0(ax)0 (bx2)]
= 21a2x2 + 7bx2
= (21a2 + 7b)x2
7a = 1 21a2 + 7b = 0
a = 17
2111722
+ 7b = 0
7b = – 37
b = – 349
Let (1 + ax + bx2)7 = 1 + x
(1 + x)17 = 1 + 1
7x – 3
49x2 + … 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 + –11!
(–2x)
+ –1(–2)2!
(–2x)2 + –1(–2)(–3)3!
(–2x)3 + …4
Chapter 2.indd 10 6/24/2015 5:37:17 PM
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)12
= 1 +
121!
(x) +
121– 1
222!
(x)2
+
121– 1
221– 322
3! (x)3
+
121– 1
221– 3221– 5
224!
(x)4 + …
= 1 + x2
– x2
8 + x3
16 – 5
128 x4 + … … 1
14
(6 + x) – (2 + x)–1
= 6 + x4
– 2–131 + x24
– 1
= 6 + x4
– 1231 + –1
1! 1x
22 + –1(–2)2!
1x22
2
+ –1(–2)(–3)3!
1x22
3
+ –1(–2)(–3)(– 4)4!
1x22
4
4 = 6 + x
4 – 1
211 – x2
+ x2
4 – x3
8 + x4
16 – …2
= 6 + x4
– 12
+ x4
– x2
8 + x3
16 – x4
32 + …
= 1 + x2
– x2
8 + x3
16 – x4
32 … 2
To obtain the error,
1 – 2 , 11 + x2
– x2
8 + x3
16 – 5
128x42
– 11 + x2
– x2
8 + x3
16 – x4
32 2
= – 5128
x4 + x4
32
= – x4
128 The error is approximately x4
128. [Shown]
38 ( )a b ab
aa
b
a
ab
aa
b
+ = +
= +
+ = +
1
2
1
2 1
2
1
2
1
2
1
2 1
2
1 1
1 11
2
aa
b
a
b
a
a
+−
+− −
+
=
1
2
1
21
2
1
2
1
21
1
22
3
2
3
1
2
!
!
…
112 8
16
1
12 8
2
2
3
3
1
2
1
2
1
2
22
2
+ − +
+
−( ) = −
= − − −
b
a
b
a
b
a
a b ab
a
ab
a
b
a
… … bb
a
a b a b ab
a
b
a
ab
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 41
4
1
8 4
129
256
1
23
- = +( )
æ
èçç
ö
ø÷÷
=
5 3 5 3 5 3
2
129
2565 3 2
5 3 2256
129512
129
-( ) +( ) = -
=
+( ) =+( ) = ´
=
1 – 2
2
1
Chapter 2.indd 11 6/24/2015 5:37:19 PM
12 ACE AHEAD Mathematics (T) First Term Second Edition© Oxford Fajar Sdn. Bhd. 2015
39 y = 1
1 + 3x + 1 + x ×
1 + 3x – 1 + x1 + 3x – 1 + x
= 1 + 3x – 1 + x
1 + 3x – (1 + x)
= 12x
1 1 + 3x – 1 + x2 [Shown]
12x3(1 + 3x)
12 – (1 + x)
124
= 12x531 +
121!
(3x) +
121– 1
222!
(3x)2
+
121– 1
221– 322
3!(3x)3 + …4 – 31 +
121!
(x)
+
12
1– 122
2!(x)2 +
121– 1
221– 322
3!(x)3 + …46
= 12x311 + 3
2 x – 9
8 x2 + 27
16 x3+ …2 – 11 + x
2
– x2
8 + x3
16 + …24
= 12x1x – x2 + 13
8 x3 + …2
= 12
– x2
+ 1316
x2
Using x = 1100
,
y = 1
1 + 3x + 1 + x
= 1
103100
+ 101100
= 10
103 + 101 ≈ 1
2 –
1 110022
+ 13161
11002
2
≈ 79 213160 000
[Shown]
40 (a) 3 – 5x + 3x2
≡ A(1 + x2) + (B + Cx)(1 – 2x)
Let x = 12
,
54
= 54
(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 = 1C = –1
(b) (1 – 2x)–1 = 31 + –11!
(–2x) + –1(–2)2!
(–2x)2
+ –1(–2)(–3)3!
(–2x)34= 1 + 2x + 4x2 + 8x3
(1 + x2)–1 = 31 + –11!
(x2) + …4= 1 – x2
(c) (3 – 5x + 3x2)(1 – 2x)–1(1 + x2)–1
= 11 – 2x
+ 2 – x1 + 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
39
1
3
1
31 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
Sa r
r
S
n
n
n
n
=−( )−
=
−
−
= −
=∞
1
1
10
71
1
9
11
9
5
121
1
9
55
12
Chapter 2.indd 12 6/24/2015 5:37:22 PM