The Binomial Theorem

Pascal’s Triangle and the Binomial Theorem (x + y)0 = 1 (x + y)1 = 1x + 1y (x + y)2 = 1x2 + 2xy + 1y2 (x + y)3 = 1x3 + 3x2y + 3xy2 +1 y3 (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

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 = 3C0 a3 + 3C1 a2b + 3C2 ab2 + 3C3 b3 7.5.3

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 n = 0 (a + b)0 1st Row 0C0 1 2nd Row n = 1 (a + b)1 1 1C0 1C1 1 3rd Row n = 2 (a + b)2 1 2 1 2C0 2C1 2C2 n = 3 (a + b)3 4th Row 1 3 3 1 3C0 3C1 3C2 3C3 n = 4 (a + b)4 5th Row 1 4 6 4 1 4C0 4C1 4C2 4C3 4C4 n = 5 (a + b)5 1 5 10 10 5 1 5C0 5C1 5C2 5C3 5C4 5C5 6th Row 7.5.4

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 = 3C0 a3 - 0b0 + 3C1 a3 - 1b1 + 3C2 a3 - 2b2 + 3C3 a3 - 3b3 (a + b)n = nC0 an - 0b0 + nC1 an - 1b1 + nC2 an - 2b2 + … + nCk an - kbk The general term is the (k + 1)th term: tk + 1 = nCk an - k bk 7.5.5

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

Binomial Expansion - Practice (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 + 5C3(2x)2(-3x-1)3 + 5C4(2x)1(-3x-1)4 + 5C5(2x)0(-3x-1)5 = 1(32x5) + 5(16x4)(-3x-1) + 10(8x3)(9x-2) + 10(4x2)(-27x-3) + 5(2x)(81x-4) + 1(-81x-5) = 32x5 - 240x3 + 720x1 - 1080x-1 + 810x-3 - 81x-5 7.5.7

Finding a Particular Term in a Binomial Expansion a) Find the eighth term in the expansion of (3x - 2)11. tk + 1 = nCk an - k bk 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 = nCk an - k bk 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

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

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

Finding a Particular Term of the Binomial One term in the expansion of (2x - m)7 is -15 120x4y3. Find m. tk + 1 = nCk an - k bk 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

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