Review ……

(a + b)2 =

(a + b)3= (a + b)(a + b)(a + b)(a2 + 2ab + b2)(a + b)

=

(a + b)(a + b)= a2 + 2ab + b2

= a3 + 3a2b + 3ab2 + b3

(a + b)4= a4 + 4a3b + 6a2b2 + 4ab3 + b4

(a + b)5= a5 + 5a4b + 10a3b2 + 10a2b3 + 5ab4 + b5

Penjabaran Binomial (Binomial Newton)

1

(a + b)2 (a + b)3

(a + b)4

(a + b)5

1 1(a + b)1

(a + b)0

1 2 11 13 3

61 14 4

110

1 55 10

SEGITIGA PASCAL(Pascal Triangle)

0C0

(a + b)2 (a + b)3

(a + b)4

(a + b)5

1C0 1C1(a + b)1

(a + b)0

2C0

SEGITIGA PASCAL

2C1 2C2

3C0 3C1 3C2 3C3

4C0 4C1 4C2 4C3 4C4

5C0 5C1 5C2 5C3 5C4 5C5

(Pascal Triangle)

= 0C0

(a + b)2

(a + b)3

(a + b)4

+ 1C1(a + b)1

(a + b)0

a0b0

= 1C0

a1b0 a0b1

+ 2C1= 2C0

a2b0 a1b1 + 2C2 a0b2

+3C1= 3C0

a3b0 a2b1+3C2 a1b2+3C3 a0b3

(a + b)n

=

3

0r3Cr a3-rbr

= 4Cr a4-rbr

4

0r

= nCr an-rbr

n

0r

Bentuk umumBinomial

Newton(Newton’s Binomial)

(x + 3)4

Example :

+4C1= 4C0

x430 x331+4C2 x232

+4C3 x133

= 1.

x4.1

Describe the from of :

+4C4 x034

+4.x3.3+6.x2.9 +4.x1.27+1.x0.81

= x4 +12x3 +54 x2 +108x1 +81

coefficient : x4 1

x3 12x 108

(x - 2)6

Example :

Calculate the coefficient x4 in a advancement of (x – 2)6

=

Coefficient : x4

15

means r = 2

60

6Cr x6-r(-2)r

6

0r

Answers :

6C2 x6-2(-2)2

.4x4

a). (2x - 1)5

Exercise :1). Describe the following forms :

b). (3x + 2)7

c). (p + 2q)4

d).

2). Determine the coefficient for :

a). x4 from (x – 2)6

c). x3 from (3 – 2x)5

d). p3q4 from (2p + q)7

b). x5 from (2x + 1)7 e). x4 from

5x2)(x

10x1)(2x

1a). (2x - 1)5

Answers :

+5C1= 5C0

(2x)5(-1)0

=1.

32x5.1 +5. 16x4.(-1)

+ 8x3.1

= 32x5 - 80x4 +80 x3 - 40 x2 +10

(2x)4(-1)1

+5C4 5C3 (2x)2(-1)3 (2x)1(-1)4

+5C2 (2x)3(-1)2

+ +5C

5

(2x)0(-1)5

101

04x2.(-1)

+5. 2x.1

+ 1.(-1)

1+x - 1

1b). (3x + 2)7

Answers :

+7C

1

= 7C0

(3x)7 .20

=1. 2187x7.1

+7. 729x6.2 + 243x5.4

=2187x7

10206x3

+ 22680

x4

6048 x2 +1344

(3x)6.21

+7C

4

7C3 (3x)4.23

(3x)3.24

+7C2 (3x)5.22

+ +7C5 (3x)2.25

21.

35.81x4.8

+35.27x3.16

+ 9x2.3221.

+

x + 128

+7C

7

7C6 (3x)1.26

(3x)0.27+

7.3x.64 +1.

1.128

+

+ x6 20412

x5 ++15120

+

1c). (p + 2q)4

Answers :

+4C1= 4C0

p4(2q)0 p3(2q)1

+4C2 p2(2q)2

+4C3 p1(2q)3

= 1.

p4.1

+4C4 p0(2q)4

+4. p3.2q +6. p2.4q2

+4. p1.8q3

+1.p0.16q4

= p4 +8 p3q +24 p2q2

+3

2p1q3

+16q4

1d).

Answers :

+5C1= 5C0

x5

=1.

x5.1+ 5. -2x3 + 4x1

= x5 - 10

x3 +40 x - +

x4

+5C

4

5C3 x2 x1

+5C2 x3

+ +5C5 x0

10

10

+5. +1

+

5x2)(x

0x2)( 1

x2)( 2

x2)(

3x2)( 4

x2)( 5

x2)(

)( x8

)(3x

16 )5x

32(

x80 )(

3x

80 - 5x

32

(2x + 1)7

2b). Coefficient x5 from (2x + 1)7

=

Coefficient : x5

21.

means r = 2

672

7Cr (2x)7-r.1r

7

0r

Answers :

7C2 (2x)7-2.12

.1(2x)5

(3 - 2x)5

2c). Coefficient x3 from (3 - 2x)5

=

Coefficient : x3

10.

means r = 3

-720

5Cr 25-r.(-2x)r

5

0r

Answers :

5C3 .35-3.(-2x)3

.-8x332

(2p + q)7

2d). Coefficient p3q4 from (2p + q)7

=

Coefficient :

p3q4

35.

means r = 4

280

7Cr (2p)7-r. qr

7

0r

Answers :

7C4 .(2p)7-4. q4

. q48p3

2e). Coefficient x4

from

=

Coefficient : x4

120

means r = 3

-15360

10Cr (2x)10-r

10

0r

Answers :

10C3 (2x)10-3

128x7

.

10x1)(2x

10x1)(2x r

x1)(

3x1)(

3x

1

PROBLEMS

1. By using Newton’s Binomial rule, describe the following forms :

a). (x + 3)8

b). (2x – 3y)5

2. Determine the coefficient for :

a). 4th part from (x – 3y)6

b). 6th part from (x + 8)7

3. Determine the coefficient for :

a). x10 from the advancement of (x3 – 2x)6

b). x2y9 from the advancement of (2x - y)5