1 / 12

The Binomial Theorem

The Binomial Theorem. Pascal’s Triangle and the Binomial Theorem. ( x + y ) 0. = 1. ( x + y ) 1. = 1 x + 1 y. = 1 x 2 + 2 xy + 1 y 2. ( x + y ) 2. = 1 x 3 + 3 x 2 y + 3 xy 2 + 1 y 3. ( x + y ) 3. ( x + y ) 4.

leandra-knapp
Télécharger la présentation

The Binomial Theorem

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. The Binomial Theorem

  2. Pascal’s Triangle and the Binomial Theorem (x + y)0 = 1 (x + y)1 = 1x + 1y = 1x2 + 2xy + 1y2 (x + y)2 = 1x3 + 3x2y + 3xy2 +1y3 (x + y)3 (x + y)4 = 1x4 + 4x3y + 6x2y2 + 4xy3 + 1y4 (x + y)5 = 1x5 + 5x4y + 10x3y2 + 10x2y3 + 5xy4 + 1y5 (x + y)6 = 1x6 + 6x5y1 + 15x4y2 + 20x3y3 + 15x2y4 + 6xy5 + 1y6 7.5.2

  3. The Binomial Theorem The Binomial Theorem is a formula used for expanding powers of binomials. Each term of the answer is the product of three first-degree factors. For each term of the answer, an a and/or b is taken from each first-degree factor. (a + b)3 = (a + b)(a + b)(a + b) = a3 + 3a2b+ 3ab2 + b3 • The first term has no b. It is like choosing no b from three b’s. • The combination 3C0 is the coefficient of the first term. • The second term has one b. It is like choosing one b from three b’s. • The combination 3C1 is the coefficient of the second term. • The third term has two b’s. It is like choosing two b’s from three • b’s. The combination 3C2 is the coefficient of the first term. • The fourth term has three b’s. It is like choosing three b’s from • three b’s. The combination 3C3 is the coefficient of the third term. (a + b)3 = 3C0a3 + 3C1a2b+ 3C2ab2 + 3C3b3 7.5.3

  4. Pascal’s Triangle and the Binomial Theorem The numerical coefficients in a binomial expansion can be found in Pascal’s triangle. Pascal’s Triangle Pascal’s Triangle Using Combinatorics (a + b)0 n = 0 1st Row 0C0 1 (a + b)1 n = 1 1C1 2nd Row 1C0 1 1 (a + b)2 n = 2 2C1 2 1 2C0 2C2 3rd Row 1 (a + b)3 n = 3 3 3 1 3C0 3C1 3C2 3C3 4th Row 1 (a + b)4 4 n = 4 1 6 1 4C1 4C2 4C3 4C4 4 4C0 5th Row 1 5 1 5 10 10 (a + b)5 5C5 5C0 5C1 5C2 5C3 5C4 n = 5 6th Row 7.5.4

  5. Binomial Expansion - the General Term (a + b)3= a3 + 3a2b+ 3ab2 + b3 The degree of each term is 3. For the variable a, the degree descends from 3 to 0. For the variable b, the degree ascends from 0 to 3. (a + b)3= 3C0a3 - 0b0 + 3C1a3 - 1b1+ 3C2a3 - 2b2 + 3C3a3 - 3b3 (a + b)n= nC0an - 0b0 + nC1an - 1b1+ nC2an - 2b2 + … + nCkan - kbk The general term is the (k + 1)th term: tk + 1 = nCk an - kbk 7.5.5

  6. Binomial Expansion - Practice n = 4 a = 3x b = 2 Expand the following. a) (3x + 2)4 + 4C3(3x)1(2)3 + 4C2(3x)2(2)2 + 4C4(3x)0(2)4 + 4C1(3x)3(2)1 = 4C0(3x)4(2)0 + 1(16) = 1(81x4) + 4(27x3)(2) + 6(9x2)(4) + 4(3x)(8) = 81x4 + 216x3 + 216x2 + 96x +16 n = 4 a = 2x b = -3y b) (2x- 3y)4 + 4C3(2x)1(-3y)3 + 4C2(2x)2(-3y)2 = 4C0(2x)4(-3y)0 + 4C1(2x)3(-3y)1 + 4C4(2x)0(-3y)4 = 1(16x4) + 4(8x3)(-3y) + 6(4x2)(9y2) + 4(2x)(-27y3) + 81y4 = 16x4 - 96x3y + 216x2y2 - 216xy3 + 81y4 7.5.6

  7. Binomial Expansion - Practice c) (2x - 3x-1)5 = 5C0(2x)5(-3x-1)0 + 5C1(2x)4(-3x-1)1 + 5C2(2x)3(-3x-1)2 n = 5 a = 2x b = -3x-1 + 5C4(2x)1(-3x-1)4 + 5C3(2x)2(-3x-1)3 + 5C5(2x)0(-3x-1)5 = 1(32x5) + 5(16x4)(-3x-1) + 10(8x3)(9x-2) + 10(4x2)(-27x-3) + 1(-81x-5) + 5(2x)(81x-4) = 32x5 - 240x3 + 720x1 - 1080x-1 + 810x-3 - 81x-5 7.5.7

  8. Finding a Particular Term in a Binomial Expansion a) Find the eighth term in the expansion of (3x - 2)11. tk + 1 = nCkan - kbk n = 11 a = 3x b = -2 k = 7 t7 + 1 = 11C7 (3x)11 - 7(-2)7 t8 = 11C7 (3x)4(-2)7 = 330(81x4)(-128) = -3 421 440 x4 b) Find the middle term of (a2 - 3b3)8. n = 8, therefore, there are nine terms. The fifth term is the middle term. tk + 1 = nCkan - kbk n = 8 a = a2 b = -3b3 k = 4 t4 + 1 = 8C4 (a2) 8 - 4(-3b3)4 t5 = 8C4 (a2) 4(-3b3)4 = 70a8(81b12) = 5670a8b12 7.5.8

  9. Finding a Particular Term in a Binomial Expansion Find the constant term of the expansion of tk + 1 = nCkan - kbk = x0 x18 - 3k tk + 1 = 18Ck (2x)18 - k(-x-2)k n = 18 a = 2x b = -x-2 k = ? 18 - 3k = 0 -3k = -18 k = 6 tk + 1 = 18Ck 218 - kx18 - k(-1)kx-2k tk + 1 = 18Ck 218 - k (-1)kx18 - kx-2k tk + 1 = 18Ck 218 - k (-1)kx18 - 3k Substitute k = 6: t6 + 1 = 18C6 218 - 6 (-1)6x18 - 3(6) Therefore, the constant term is 76 038 144. t7= 18C6 212 (-1)6 t7= 76 038 144 7.5.9

  10. Finding a Particular Term in a Binomial Expansion Find the numerical coefficient of the x11 term of the expansion of tk + 1 = nCkan - kbk = x11 x20 - 3k tk + 1 = 10Ck (x2)10 - k(-x-1)k n = 10 a = x2 b = -x-1 k = ? 20 - 3k = 11 -3k = -9 k = 3 tk + 1 = 10Ckx20 - 2k(-1)kx-k tk + 1 = 10Ck(-1)kx20 - 2kx-k tk + 1 = 10Ck (-1)kx20 - 3k Substitute k = 3: t3 + 1 = 10C3 (-1)3x20 - 3(3) Therefore, the numerical coefficient of the x11 term is -120. t4= 10C3 (-1)3x11 t4= -120 x11 7.5.10

  11. Finding a Particular Term of the Binomial One term in the expansion of (2x - m)7 is -15 120x4y3. Find m. tk + 1 = nCkan - kbk n = 7 a = 2x b = -m k = 3 tk + 1 = 7C3 (2x)4(-m)3 tk + 1 = (35) (16x4)(-m)3 -15 120x4y3 = (560x4)(-m)3 Therefore, m is 3y. 3y = m 7.5.11

  12. THANK YOU

More Related