Bab 9 . TEOREMA BINOMIALHalaman 130, Exercise 9A1.. Write down the expansion of each the followinga.. (2x +y)2 = (2x + y) (2x + y) = 4x2 + 4xy + y2b.. (5x + 3y)2 = (5x + 3y)(5x + 3y) = 25x2 + 30xy + 9y2c.. (4 + 7p)2 = 16 + 56p + 49p2d.. (1 8t)2 = 1 16t + 64t2e.. (1 - 5x2)2 = 1 10x2 + 25x4f.. (2 +x3)2g.. (x2 + y3)3 = x6 + 3x4y3 + 3x2y6 + y9h.. (3x2 + 2y3)3 = 27x6 + 54x4y3 + 36x2y6 + 8y9

2.. Write down the expansion of each the followinga.. (x+2)3b.. (2p+3q)3 c.. (1-4x)3 = 1 12x + 48x2 64x3d.. (1-x3)3 = 1 3x3 + 3x6 x9

3.. Find the coefficient of x in the expansion ofa.. (3x + 7)2 2. 3x.7 = 42x koef 42 b.. (2x+ 5)3 3.2x.52 = 150x koef 150

4.. Find the coefficient of x2 in the expansion ofa.. (4x+5)3 3.(4x)2.5 = 240x2 koef240 b..(1 -3x)4 6.12.(-3x)2 = 54x2 koef 54

5.. Expand each of the following expressions a.. (1 + 2x)5 = 1 + 5.2x + 10.(2x)2 + 10. (2x)3+5.(2x)4+ (2x)5 = 1 + 10x + 40x2 + 80x3 + 80x4 + 32x5b.. (p+2q)6 = 1.p6 + 6.p5.2q + 15.p4.(2q)2 + 20.p3.(2q)3 + 15.p2.(2q)4 + 6.p.(2q)5 + 1.(2q)6= p6 + 12p5q + 60.p4q2 + 160.p3q3 + 240.p2q4 + 192.pq5 + 64q6 c.. (2m 3n)4 = 1.(2m)4 + 4.(2m)3.(-3n) + 6.(2m)2.(-3n)2 + 4.2m.(-3n)3 + 1. (-3n)4= 16m4 96m3 n + 216m2 n2 216mn3 + 81n4 d.. (1 + 1/2x)4 = 1 + 4 (1/2x) + 6 (1/2x)2 + 4(1/2x)3 + 1 (1/2x)4 = 1 + 2x + 6/8x + 1/2x2 + 1/16x4

6.. Find the coefficient of x3 in the expansion ofa.. (1+3x)5 10.12.(3x)3 = 270x3 koef 270b.. (2 - 5x)4 4.2.(-5x)3 = 1000x3 koef 1000

7.. Expand ( 1 + x + 2x2 )2. Check your answer with a number substitution( 1 + x + 2x2 )2 = ( 1 + (x + 2x2) )2 = 1 + 2. (x + 2x2) + 1. (x + 2x2)2 = 1 + ( 2x + 4x2 ) + ( x2 + 4x3 + 4x4 )= 1 + 2x + 5x2 + 4x3 + 4x4Misal x = 1 maka( 1 + x + 2x2 )2 = ( 1 + 1 + 2.12 )2 = 42 = 161 + 2x + 5x2 + 4x3 + 4x4 = 1 + 2.1 + 5.12 + 4.13 + 4.14 = 1 + 2 + 5 + 4 + 4 = 16

8.. Write down the expansion (x+4)3 and hence expand (x+1) (x+4)3(x+4)3 = 1.x3 + 3.x2.4 + 3.x.42 + 1.43 = x3 + 12x2 + 48x + 64(x+1) (x+4)3 = (x+1).( x3 + 12x2 + 48x + 64) = x4 + 12x3 + 48x2 + 64x + x3 + 12x2 + 48x + 64= x4 + 13x3 + 60x2 + 112x + 64

9.. Expand (3x + 2)2 ( 2x + 3 )3(3x + 2)2 = 9x2 + 12x + 4( 2x + 3 )3 = 8x3 + 36x2 + 54x + 27(3x + 2)2 ( 2x + 3 )3 = ( 9x2 + 12x + 4 ) ( 8x3 + 36x2 + 54x + 27 ) = .

10.. In the expansion of (1+ax)4 , the coefficient of x3 is 1372. Find the constant a4.1.(ax)3 = 4a3 x3 koef 4a3 = 1372 a3 = 1372 : 4 a = = 7

11.. Expand (x + y)11 = ..12.. Find the coefficient of x6 y6 in the expansion of (2x+ y)12 = .

Halaman 134 Exercise 9B1.. a.. = 35

2.. Find the coefficient x3 in the expansion of each of the followinga.. (1 + x)5 12 x3 = 10 x3 koef = 10b.. (1 x)8 15 ( -x)5 koef = - 56 c.. (1+x)11 d.. (1-x)16

3.. Find the coefficient x5 in the expansion of each of the followinga..(2+x)7 = 22 x5 b.. (3-x)8 = 33(-x)5 c.. (1+2x)9 = 14 (2x)5 d.. (1-1/2x)12 = 17(-1/2x)5

4.. Find the coefficient of x6 y8 in the expansion of each of the followinga.. (x+y)14 = b.. (2x+y)14 = 26 c.. (3x-2y)14 = 36 (-2)8 d..(4x +1/2y)14 = 46(1/2)8

5.. Find the first four terms in the expansion in ascending powers ofx of followinga.. (1+x)13 = + x + x2 + x3 b.. (1-x)15 = - +x2 +x3c.. (1+3x)10 = + (3x) +(3x)2+ (3x)3 d..(2-5x)7 = 27 + 26 (-5x) + 25(-5x)2+ 24(-5x)3

6.. Find the first three terms in the expansion in ascending powers x of the followinga.. (1+x)22b.. (1-x)30c.. (1-4x)18d.. (1+6x)19

7..Find the first three terms in the expansion, in ascending powers of x, of (1+2x)8. By substituting x=0.01, find an approximation to 1.028(1+2x)8 = 1 + + 4x2 = 1 + 16x + 112x2(1+2. 0,01)8 = 1 + 16(0,01) + 112 (0,01)2 = 1 + 0,16 + 0,0112 = 1,1712

8.. Find the first three terms in the expansion, in ascending powers of x, of (2+5x)12. By substituting a suitable value for x, find an approximation to 2.00512 to 2 decimal place(2+5x)12 = 212 + 211.5x+ 210 . (5x)2 = 4096 + 122880 x + 1689600 (x2 subts x = 0.001= 4096 + 122,88 + 1,6896 = 4218,88

9..Expand (1+2x)16 up to and including the terms in x3. Deduce the coefficient of x3 in the expantion of (1+3x)(1+2x)16 (1+2x)16 = 1 + 2x + 4x2 + 8x3 = 1 + 32x + 480x2 + 4480x3(1+3x)(1+2x)16 koef x3 = 4480.1 + 480.3 = 5920

10.. Expand (1-3x)10 up to and including the terms in x2. Deduce the coefficient of x2 in the expantion of (1+3x)2(1-2x)10 = (1 + 12x + 9x2) (1 20x + 180x2 - ) koef x2 = 1.180 + 12.-20 + 9.1 = 180-240+9= -51

11.. Given that the coefficient of x in the expansion of ( 1+ax) (1+5x)40 adalah 207, determain the value of ( 1+ax) (1+5x)40 = (1+ax)( 1+5x+ ) koef x = 1.5 + a.1 = 5 +a = 207 a = 202

13. Given that the expansion of (1+ax)n begin 1 + 36x + 576x2, find the value a and n(1+ax)n = 1 + ax + a2x2 = 1 + an x + a2 x2 an = 36, a2 = 576 n=9, a=4

MISCELLANEOUS EXERCISE 9 halaman 1353.. b. Find the coefficient of x3y5 in the expansion of (5a + 1/2b)8 = 53 (1/2)57.. = x6 + 3.x4. + 3.x2. + = x6 + 3.x3 + 3 + 10.. koef x2 = 3. (x4)1.(4/x)211.. koef x = 014.. a.. 1/1000 b.. 1 /200015.. x = 1020.. (x3)3(-1/x)2122.. (27x3 + 135x2 + 225x + 125 ) (27x3 - 135x2 + 225x - 125) = 730270x2 + 250 = 730 x = 4/3 25.. c. = 26.. = = = =27.. b.. 56a = 28b = 21c8a = 4b = 3cA= 12, b = 24, c= 3227..c.. (n r ) x = (n-r) = (r+1) x = ( r+1) = N x = n x = 28.. + = = = = = = =29.. ( 1 + 3x)18 subts x = 0,000130.. (2)4 = (2)4 + 4(2)3 () + 6(2)2()2 + 4(2)3 +()4=64 + 64 + 144 + 24 = 217 + 88

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