unit 2 chapter 10 answers
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Unit 2 Answers: Chapter 10 © Macmillan Publishers Limited 2013
Chapter 10 Binomial Theorem
Try these 10.1
(a)! !
( 3)!
n n
n=
+ ( 3)( 2)( 1) !n n n n+ + + ×
1
( 1)( 2)( 3)n n n=
+ + +
(b) n! + (n + 1)! + (n - 2)!= n(n – 1) (n – 2)! + (n + 1)n (n – 1) (n – 2)! + (n – 2)!= (n – 2)! [n(n – 1) + (n + 1) n(n – 1) + 1]= (n – 2)! [n
2 – n + n
3 – n + 1]
= (n – 2)! (n3 + n
2 – 2n + 1)
Exercise 10A
1 (a) (1 + x)4 = 1 + 4C 1 x +4C 2 x
2 + 4C 3 x3 + x4
= 1 + 4 x + 6 x2 + 4 x
3 + x
4
(b) (1 − x)5 = 1 + 5C 1(− x) +5C 2(− x)
2 + 5C 3(− x)3 + 5C 4(− x)
4 + (− x)5
= 1 − 5 x + 10 x2 − 10 x3 + 5 x4 − x5 (c) (1 + 2 x)
5 = 1 +
5C 1(2 x) +
5C 2(2 x)
2 +
5C 3(2 x)
3 +
5C 4(2 x)
4 + (2 x)
5
= 1 + 10 x + 40 x2 + 80 x
3 + 80 x
4 + 32 x
5
(d)
4 2 3 44 4 4
1 2 32 2 2 2 2
1 1
3 3 3 3 3
x C x C x C x x
− = + − + − + − + −
2 3 48 8 32 1613 3 27 81
x x x x= − + − +
(e) 6 6 6 5 6 4 2 6 3 31 2 3(3 ) 3 (3) ( ) (3) ( ) (3) ( ) x C x C x C x+ = + + + 6 2 4 6 5 6
4 5(3) ( ) (3)( ) ( )C x C x x+ + +
2 3 4 5 6729 1458 1215 540 135 18 x x x x x x= + + + + + +
2 5 5 5 4 5 3 2 5 2 3 5 4 51 2 3 4(3 2 ) 3 (3) ( 2 ) (3) ( 2 ) (3) ( 2 ) (3) ( 2 ) ( 2 ) x C x C x C x C x x− = + − + − + − + − + − 2 3 4 5243 810 360 720 240 32 x x x x x= − + − + −
3 9 9 9 2 9 31 2 3(1 2 ) 1 ( 2 ) ( 2 ) ( 2 ) x C x C x C x− = + − + − + − 2 31 18 144 672 (the first four terms) x x x= − + −
4 (a) 10 10 10 9 10 8 21 2(4 ) 4 (4) ( ) (4) ( ) x C x C x− = + − + −
= 1 048 576 – 2 621 440 x + 2 949 120 x2
(b) 15 15 15 15 21 2(1 ) 1 ( ) ( ) x C x C x− = + − + − 21 15 105 x x= − +
(c)
10 210 10 9 10 8
1 21 1 1
2 2 (2) (2)3 3 3
x C x C x
+ = + +
251201024 12803
x x= + +
5 (a) (1 − 3 x)10
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Term in 6 10 4 66 (1) ( 3 ) x C x= −
= 210 × 729 x6
= 15 3090 x6
Coefficienticient of x6 = 15 3090
(b) (2 + 3 x)12 Term in x
6 =
12C 6(2)6 (3 x)
6
= 924 × 64 × 729 x6 = 43 110 144 x
6
Coefficienticient of x6 = 43 110 144
(c) (1 − 2 x)9
Term in 6 9 3 66 (1) ( 2 ) x C x= −
684(64) x=
65376 x=
(d) (3 + x)15
Term in 6 15 9 66 (3) ( ) x C x=
= 5005 × 19 683 x6
= 98 513 415 x6
6 (1 − 3 x)8
(a) There are 9 terms
(b) Term in x5 =
8C 5(1)
3 (−3 x)
5 = 56 × (−243) x
5
= −13 608 x5 (c) The fifth term is the term in x
4:
Fifth term 8 4 44 (1) ( 3 )C x= −
= 70 × 81 x4
= 5670 x
4
7 (a) (1 + 4 x)7
Fifth term 7 3 4 4 44 (1) (4 ) 35 256 8960 .C x x x= = × =
(b) (2 − x)11
Sixth term 11 6 5 5 55(2) ( ) 462 64( ) 29 568C x x x= − = × − = −
(c)
122
33
x
−
Seventh term
612 6 6 6
62
(3) 924 64 59 1363
C x x x
= − = × =
8 Expanding (1−
2 x)
6
up to x
3
:6 6 6 2 6 31 2 3(1 2 ) 1 ( 2 ) ( 2 ) ( 2 ) x C x C x C x− = + − + − + − + …
= 1 − 12 x + 60 x2 − 160 x
3 + …
6 2 3(1 ) (1 2 ) (1 ) (1 12 60 160 ...) x x x x x x− − = − − + − +
Coefficienticient of x3 = −160 − 60 = −220
9 (2 − x) (3 + x)5
= (2 − x) (35 + 5C 1(3)4 ( x) + 5C 2(3)
3 ( x)2+ …)
= (2 − x) (243 + 405 x + 270 x2+ …)
= 486 + 810 x + 540 x2 − 243 x − 405 x
2 + … (up to x
2)
= 486 + 567 x + 135 x2 + …Since 486 + 567 x + 135 x
2 = a + bx + cx
2
⇒ a = 486, b = 567, c = 13510 (2 − x)5
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Third term 5 3 2 22 (2) ( ) 80 .C x x= − =
(1 + ax)6
Fourth term 6 3 3 3 33(1) ( ) 20C ax a x= =
Coefficienticient of third term = 12
Coefficienticient of fourth term
∴ 80 =
320
2
a
a3 = 8⟹ a = 2
11
82
x x
−
General term of the expansion is8
8 2 ( )
r r
r C x x
−
−
8 8 82 ( )r r r r r C x x− − += −
8 8 8 2( ) 2r r r r C x− − += −
For the term independent of x: –8 + 2r = 0, r = 4
∴ The term independent of x is4 8 8 4
4( ) 2C −− = 70 × 16 = 1120
12
103
2
35 x
x
−
General term of the expansion is10
10 323 ( 5 )
r
r r C x x
−
− 10 10 2 10 3(3) ( ) ( 5)r r r r r C x x− − −= −
10 10 20 2 3(3) ( 5)r r r r r C x− − + += −
10 10 20 5(3) ( 5)r r r r C x− − += −
For the term independent of x
−20 + 5r = 05r = 20r = 4
∴ The term independent of x is10 6 4
4 (3) ( 5)C − = 210 × 729 × 625
= 95 681 250
13
152
3
24 x
x
+
General term of the expansion is:15
15 2
3
2(4 )
r r
r C x x
−
15 15 45 3 2(2) 4r r r r r C x x− − +=
15 15 45 5(2) (4)r r r r C x− − +=
For the term independent of x: −45 + 5r = 0r = 9
∴ The term independent of x is:
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15 6 99 (2) (4)C = 5005 × 64 × 262 144
= 83 969 966 080
14
64
2
13
2
x
x
−
General term of the expansion is:6
6 4
2
1( 3 )
2
r r
r C x x
−
−
6 1 6 2 6 4(2 ) ( ) ( 3) ( )r r r r r C x x− − − −= −
6 6 12 6(2) ( 3)r r r r C x− + − += −
For the term independent of x: −12 + 6r = 0r = 2
∴ The term independent of x: 6 4 22 (2) ( 3)C − −
= 15 × 116
× 9
=135
16
15 (a) (1 + 2 x)5 = 1 + 5C 1(2 x) +
5C 2(2 x)2 + …
= 1 + 10 x + 40 x2 + …
(b) (1 − 3 x)5 = 1 +
5C 1(−3 x) +5C 2(−3 x)
2 + …
= 1 − 15 x + 90 x2 + …
(1 + 2 x)5 (1 − 3 x)
5 = [(1 + 2 x) (1 − 3 x)]
5
= (1 − x − 6 x2)
5
= (1 + 10 x + 40 x2+ …) (1 − 15 x + 90 x
2+ …)
Coefficienticient of x2 = 90 − 150 + 40 = −2016 5 5 5 4 5 3 21 2(2 ) 2 (2) ( ) (2) ( ) x C x C x+ = + + + …
= 32 + 80 x + 80 x2 + …
(1 + px + qx2) (2 + x
5)
= (1 + px + qx2) (32 + 80 x + 80 x
2 + …)
= 32 + 80 x + 80 x2 + 32 px + 80 px2 + 32qx2 + …= 32 + x (80 + 32 p) + x
2(80 + 80 p + 32q) + …
Coefficienticient of x = 16⇒ 80 + 32 p = 16
p = −280 + 80 p + 32q = 16
80 − 160 + 32q = 16
96 332
q = =
17 (1 − 4 x)7 = 1 + 7C 1(−4 x) +
7C 2(−4 x)2
= 1 − 28 x + 336 x2 + …
(1 + 2 x − 3 x2) (1 − 4 x)
7 = (1 + 2 x − 3 x
2) (1 − 28 x + 336 x
2 + …)
Coefficienticient of x2 = 336 + (2) (−28) −3
= 336 − 56 − 3= 277
18 (a)
6 26 6
1 21 1 1
1 14 4 4
x C x C x
− = + − + − + …
26 1514 16
x x= − + + …
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23 1512 16
x x= − + + …
(b) (2 + x)6 = 26 + 6C 1 (2)5 ( x) + 6C 2 (2)
4 ( x)2 + …= 64 + 192 + 240 x
2 + …
6661 11 (2 ) 1 (2 )
4 4 x x x x
− + = − +
621 12
2 4 x x x
= + − −
621 12
2 4 x x
= + −
2 23 151 (64 192 240 + )2
16
x x x x
= − + + +
+ … …
Coefficienticient of x2 3 15240 (192) (64)2 16 = − +
240 288 60 12= − + =
19 (a)
61
23
y
−
Term in y4
46 2
41
23
C y
= −
4115 481
y= × ×
460
81 y=
Coefficienticient of y4
60 20
81 27= =
(b)
104
y y
−
General term of the expansion is
10 10 4 r
r r C y
y
− −
10 10 ( 4) ( )r r r r C y y− −= −
10 10 2( 4)r r r C y −= −
10 − 2r = 4
⇒ 2r = 6r = 3
Coefficienticient of y4 10 33( 4) 7680C = − = −
20 (1 − 2 x)2 (1 + px)
8
2 8 8 21 2(1 4 4 ) (1 ( ) ( ) ) x x C px C px= − + + + +
2 2 2(1 4 4 ) (1 8 28 ) x x px p x= − + + + + 2 2 2 21 8 28 4 32 4 px p x x px x= + + − − + + …
= 1 + x(8 p – 4) + x2(28 p2 – 32 p + 4) + …Since 2 8 2(1 2 ) (1 ) 1 20 x px x qx− + ≡ + + +
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2 2 21 (8 4) (28 32 4) 1 20 x p x p p x qx⇒ + − + − + = + +
8 4 20 3 p p⇒ − = ⇒ = 228 32 4 p p q− + =
228(3) 32(3) 4q∴ = − + = 160
Exercise 10B
1 1/ 3 2 3
1 2 1 2 5
1 3 3 3 3 3(1 ) 1
3 2! 3! x x x x
− − −
+ = + + + + …
2 31 1 51 , for 1 13 9 81
x x x x= + − + + − <
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6 13 2 33
1 2 1 2 5
1 3 3 3 3 31 3 (1 3 ) 1 ( 3 ) ( 3 ) ( 3 )
3 2! 3! x x x x x
− − −
− = − = + − + − + − +…
2 35 1 1
1 , for 3 3 3 x x x x= − + − − < <
7
11 11 (2 ) 2 1
2 2
x x
x
−− − = − = − −
2 31 ( 1) ( 2) ( 1) ( 2) ( 3)
1 ( 1)2 2 2! 2 3! 2
x x x − − − − − = + − − + − + − +
2 31 1 1 112 2 4 8
x x x
= + + + +
2 31 1 1 1 , for 2 2
2 4 8 16
x x x x= + + + + − <
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2 31 ( 1) ( 2) ( 1)( 2)( 3)
(2 ) 1 ( 1)2 2 2! 2 3! 2
x x x x
− − − − − = + + − − + − + − +
2 31 1 1 1
(2 ) 1 ...2 2 4 8 x x x x
= + + + + +
2 3 2 3 41 1 1 1 1 12 ...2 2 4 2 4 8
x x x x x x x
= + + + + + + + +
2 31 12 2 ...2 2
x x x
= + + + +
2 31 112 4
x x x= + + + + …
(c)2
2 1(1 ) (1 ) (2 )2
x x x
x
−+ = + ++
12 1 1(1 2 ) (2) 12
x x x
−− = + + +
2 321 1 ( 1) ( 2) 1 ( 1) ( 2) ( 3) 1(1 2 ) 1 ( 1)
2 2 2! 2 3 2
! x x x x x
− − − − − = + + + − + +
…
+
2 2 31 1 1 1(1 2 ) 12 4
2 8
x x x x x
= + + − + − …
+
2 3 2 3 2 31 1 1 1 1 11 22 2 4 8 2 2
x x x x x x x x
= − + − + − + + …+ −
2 31 3 1 1
12 2 4 8 x x x
= + + − …
+
2 31 3 1 1
2 4 8 16 x x x= + + − +
11 2 12
22 (1 3 )
1 3 x
x
−= −−
2 2 2 2 3( 1) ( 2) ( 1) ( 2) ( 3)2 1 ( 1) ( 3 ) ( 3 ) ( 3 )2! 3!
x x x− − − − −
= + − − + − + −
2 4( 1) ( 2) ( 3) ( 4) ( 3 )4!
x− − − −
+ − +
2 4 6 8
2 [1 3 9 27 81 ] x x x x= + + + + …+ 2 4 6 82 6 18 54 162 ... x x x x= + + + + +
12 1/ 2 2 3
1 1 1 1 3
1 2 2 2 2 2Using (1 ) 1
2 2! 3! x x x x
− − −
+ = + + + +…
2 31 1 112 8 16
x x x= + − + + …
1/ 2 2 31 1 10.02 (1 0.02) 1 (0. 02) (0.02) (0.02)2 8 16
x = ⇒ + = + − + + …
1.02 1 0.01 0.00005 0.0000005⇒ = + − + +…
1.0099505= = 1.0100 (4 dp)
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13 4
1
(0.97)
4
4
1(1 )
(1 )
x
x
−= −
−
22 3 4( 4)( 5) ( 4)( 5)( 6) ( 4)( 5)( 6)( 7)1 ( 4)( ) ( ) ( ) ( ) ...
2! 3! 4! x x x x
− − − − − − − − −= + − − + − + − + − +
2 3 41 4 10 20 35 ... x x x x= + + + + + 4 2 3 40.03 (1 0.03) 1 4(0.03) 10(0.03) 20(0.03) 35(0.03) x −= ⇒ − = + + + +
4(0.97) 1 0.12 0.009 0.00054 0.00003− ≈ + + + +
= 1.129757
≡ 1.1296 (4 dp)
14 2 2( 2) ( 3)
(1 2 ) 1 ( 2) (2 ) (2 ) ...2!
x x x− − −
+ = + − + +
= 1 − 4 x + 12 x2 +…2
2 21 (1 ) (1 2 )1 2
x x x
x
− − = − + +
2 2(1 2 ) (1 4 12 ) x x x x= − + − + + … 2 2 21 4 12 2 8 x x x x x= − + − + + +…21 6 21 x x≈ − +
152 2 3
( )(1 ) (2 ) (1 )
x x f x
x x x
+ +=
+ + −
2
2 3(1 ) (2 ) (1 ) (1 ) (2 ) (1 )
x x A B C x x x x x x+ + ≡ + ++ + − + + −
∴ 2 2 3 (2 )(1 ) (1 )(1 ) (1 )(2 ) x x A x x B x x C x x+ + ≡ + − + + − + + +
When x = –1, 2 = 2 A ⟹ A = 1
When x = –2, –3 = 3 B ⟹ B = –1
When x = 1, 6 = 6C ⟹ A = 1
1 1 1( )
1 2 1 f x
x x x= − +
+ + −
1 1 1(1 ) (2 ) (1 ) x x x− − −= + − + + − 1
1 11
(1 ) 1 (1 )2 2
x
x x
−− −
= + − + + −
22
2
( 1)( 2) 1 1 ( 1)( 2) 11 ( 1)( ) 1 ( 1)
2! 2 2 2! 2
( 1)( 2)1 ( 1)( ) ( )
2!
x x x x
x x
− − − − ≈ + − + − + − +
− − + + − − + −
2 2 21 1 11 12 4 8
x x x x x x= − + − + − + + +
23 1 15
2 4 8 x x= + +
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16 3/ 2 2 3
3 1 3 1 1
3 2 2 2 2 2(1 ) 1 ( ) ( ) ( ) ...
2 2! 3! x x x x
−
− = + − + − + − +
2 33 3 11 ...2 8 16
x x x= − + + +
3 1,
2 16a b∴ = − =
17 2 2
6 4
1- 2(1 2 ) (1 3 ) 1 3
x A Bx C
x x x x
+ +≡ +
− + +
26 4 (1 3 ) ( ) (1 2 ) x A x Bx C x⇒ + ≡ + + + −
1 77 4
2 4 x A A= ⇒ = ⇒ =
0 4 0 x A C C = ⇒ = + ⇒ =
2Coefficient of 0 3 2 , 2 12, 6 x A B B B⇒ = − = =
2 2
6 4 4 6
1 2(1 2 ) (1 3 ) 1 3
x x
x x x x
+∴ ≡ +
−− + +
1 2 3( 1) ( 2) ( 1) ( 2) ( 3)4(1 2 ) 4 1 ( 1) ( 2 ) ( 2 ) ( 2 )2! 3!
x x x x− − − − − −
− = + − − + − + − +
2 34 (1 2 4 8 ...) x x x= + + + + 2 34 8 16 32 x x x= + + +
2 1 2 2 2( 1) ( 2)6 (1 3 ) 6 1 ( 1) (3 ) (3 )2!
x x x x x− − −
+ = + − +
2 46 (1 3 9 ) x x x= − + 36 18 x x= − +
2 3 3
2
6 44 8 16 32 6 18
(1 2 ) (1 3 )
x x x x x x
x x
+∴ = + + + + − +
− +
2 34 14 16 14 x x x= + + +
18 15 2 3
1 4 1 4 9
1 5 5 5 5 5(1 ) 1 ( ) ( ) ( )
5 2! 3! x x x x
− − −
+ = + + + +
2 31 2 615 25 125
x x x= + − + −
1/5
1/55 5 1
31 32 1 (32 1) 32 132
= − = − = −
1/ 51/ 5 132 1
32
= −
2 31 1 2 1 6 1
2 15 32 25 32 125 32
≈ + − − − + −
2[1 0.00625 0.000078 0.00000 46484]= − − −
1.9873(4 dp)=
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Review exercise 10
1 (a + x)5 + (1 – 2 x)
4
= a5 +
5C 1 a4 x +
5C 2 a3 x
2 +…+ 1 + (4)(–2 x) +
4C 2(–2 x)2.
Coefficient of x2
=5
C 2 a3
+4
C 2 (–2)2
= 10a3 + 2410a
3 + 24 = 664
10a3 = 640
a3 = 64
a = 3 64 = 4
2 ( ) ( ) ( ) ( )7 2 37 7 6 7 5 7 4
1 2 32 2 2 2 p x p C p x C p x C p x+ = + + + + …
Coefficient of x2 = 7C 2 p5 (4) = 84 p5
Coefficient of x3 =
7C 3 p
4(8) = 280 p
4
Since the coefficient of x2 = coefficient of x3
⇒ 84 p5 = 280 p
4
p =280 10
84 3=
3 (4 – 3 x)1
14
n
x
−
21 ( 1) 1
(4 3 ) 1 ( ) ...4 2! 4
n n x n x x
− = − + − + − +
21 1(4 3 ) 1 ( 1) ...4 32
x nx n n x
= − − + − +
= 4 – nx +
1
8 n (n – 1) x
2
– 3 x +
23
...4 nx +
2 1 34 ( 3) ( 1)8 4
x n x n n n
= + − − + − + +…
∴ 4 – 13 x + px2 =
2 1 34 ( 3) ( 1)8 4
x n x n n n
+ − − + − +
Equating coefficients of x : – n –3 = –13
n + 3 = 13 ⇒ n = 10
1 3 1 3( 1) (10) (9) (10)
8 4 8 4 p n n n= − + = +
= 18.754 (3 – 2 x)
30
Term in x3 =
30C 3 (3)
27 (–2 x)
3 ⇒ coefficient of x
3 =
30C 3 (3)
27 (–2)
3
Term in x4 =
20C 4 (3)
26 (–2 x)
4 ⇒ coefficient of x
4 =
20C 4 (3)
26 (–2)
4
30 27 33
30 26 44
(3) ( 2)
(3) ( 2)
p C
q C
−=
−
30!(3)
27!3!30!
( 2)26!4!
×=
× −
2
9= −
5 (5 + px) (3 – x)6
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= (5 + px) (3 + 6C 1 (– x)) +…= (5 + px) (3 – 6 x) up to the term in x Coefficient of x = –30 + 3 p = 03 p = 30
p = 10
6 4 1
(1 )(1 2 ) 1 1 2
x A B
x x x x
−≡ +
− + − +
4 x – 1≡ A (1 + 2 x) + B (1 – x)
x = 1⇒ 3 = 3 A ⇒ A = 1
1
2 x = − ⇒ –3 =
3
2 B ⇒ B = –2
4 1 1 2
(1 )(1 2 ) 1 1 2
x
x x x x
−∴ ≡ −
− + − +
1 2 31 ( 1)( 2) ( 1)( 2)( 3)(1 ) 1 ( 1)( ) ( ) ( )
1 2! 3!
x x x x
x
− − − − − −= − = + − − + − + −−
+…
= 1 + x + x2 + x
3+…, for –1
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1
1 12 (3) 13
x
−
− = −
22 1 ( 1)( 2) 1
1 ( 1) ...3 3 2! 3
x x − −
= + − − + − +
22 1 11 ...3 3 9
x x
= + + +
22 2 2 ..., for3 9 27
x x= + + + –3
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= 1.12616 (5 dp)
11 2 2 3( 2)( 3) ( 2)( 3)( 4)
(1 ) 1 ( 2)( ) ( ) ( ) ...2! 3!
x x x x− − − − − −
− = + − − + − + − +
2 31 2 3 4 ... x x x= + + + +
When x = 0.002, (1 – 0.002) –2 = 1 + 2(0.002) + 3(0.002)2 + 4 (0.002)3 +…(0.998)
–2 = 1 + 0.004 + 0.000012 + 0.000000032
= 1.004012032
2
11.00401
0.998∴ = (5 dp)
12 2/3 2 3
2 5 2 5 8
2 3 3 3 3 3(1 3 ) 1 (2 ) (3 ) (3 )
3 2! 3! x x x x
−
− − − − −
+ = + − + +
4
2 5 8 11
3 3 3 3(3 ) ...
4! x
− − − −
+ +
4
2
3Coefficient of x∴ =
5
3×
8×
3
11
3× 43×
4 3 2× ×
110
3=
13 (4 + x)1/2
= 41/2
1/ 21
14
x
+
2 3
1 1 1 1 3
1 1 1 12 2 2 2 22 1 ...2 4 2! 4 3! 4
x x x
− − −
= + + + +
2 31 1 12 1 ...8 128 1024
x x x
= + − + +
2 31 1 12 ...4 64 512
x x x= + − + +
When x = 0.004
1/2 2 31 1 1(4 0.004) 2 (0.004) (0.004) (0.004) ...4 16 512
+ = + − + +
≈
2 + 0.001 – 0.000001 + 0.000000000125= 2.000999= 2.0010 (4 dp)
14
12
2 2 x x
+
General term is
12C r ( x
2)
12– r 2 r
x
12 24-2 -2r r r r C x x= 12 24 3 2r r r C x
−=
For the term independent of x: 24 – 3r = 0r = 8
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Unit 2 Answers: Chapter 10 © Macmillan Publishers Limited 2013
The term independent of x is12
C 8 28 = 495 × 256 = 126 720
15
9
2
43 x
x
−
General term is9 9
2
4(3 )
r
r r C x
x
− −
9 9 9 23 ( 4) ( )r r r r r C x x− − −= −
9 9 9 3( 4) 3r r r r C x− −= −
Term independent of x: 9 – 3r = 0⇒ r = 3
∴ 9C3 (–4)
3 (3)
6 = 84 × (−64) × 729 = –3 919 104
16 ( ) ( ) ( )8 2
8 7 68 8
1 2
2 2 2... x x C x C x
x x x
− = + − + − +
4 8 3 8 2 2
1 2
( 2) ( 2) ... x C x C x= + − + − +
Coefficient of x2 =
8C 2 (–2)
2 = 28 × 4 = 112
17 (2 + px)5 = 2
5 +
5C 1 (2)
4 ( px) +
5C 2 (2)
3 ( px)
2 + …
= 32 + 80 px + 80 p2 x
2
80 p = 262/3
p =1
3
q = 80 p2 = 80
21 80
3 9
=
18 1/ 2 1/ 21
(1 ) (1 )1
x x x
x
−+ = + −−
− − −
≈ + + + − − + −
2 2
1 1 1 3
1 12 2 2 21 1 ( ) ( )
2 2! 2 2! x x x x
2 21 1 1 31 12 8 2 8
x x x x
≡ + − + +
≈ + + + + −2 2 21 3 1 1 1
12 8 2 4 8
x x x x x
211
2
x x= + +
Let x =1
10,
11
101
110
+
−
≈ 1 +2
1 1 1
10 2 10
+
11/10 1 11
9 /10 10 200≈ + +
11 200 20 1
2009
+ +≈
22111 3
200≈ ×
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Unit 2 Answers: Chapter 10 © Macmillan Publishers Limited 2013
66311
200≈
19 1/ 2 2
1 1
1 2 21 2 (1 2 ) 1 (2 ) (2 )2 2!t t t t
−
+ = + = + + +
= + − +21
12
t t
11 (1 ) (1 )1
pt pt qt
qt
−+ = + ++
2( 1)( 2)(1 ) 1 ( 1) ( ) ( ) ...2!
pt qt qt − −
= + + − + +
2 2(1 ) (1 ) pt qt q t ≈ + − + 2 2 21 qt q t pt pqt ≈ − + + −
≈ 1 + t ( p – q) + t 2
(q2
– pq)Equating coefficients of t : p – q = 1
Equating coefficient of t 2: q
2 – pq =
1 1 1 3( ) ,
2 2 2 2q p q q p− ⇒ − − = − ⇒ = =
20 3 (1 3 ) x− 1
3(1 3 ) x= −
( )2 3
1 2 1 2 5
1 3 3 3 3 31 ( 3 ) 3 ( 3 )
3 2! 3! x x x
− − −
= + − + − + − +…
2 3
5 1 11 ..., for3 3 3 x x x x= − − − + − < <
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