Unit 2 Chapter 10 Answers

8/16/2019 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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Page 10 of 16


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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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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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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Unit 2 Chapter 10 Answers