5 binomial newton
DESCRIPTION
mmTRANSCRIPT
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