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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.

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The Binomial Theorem

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  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

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