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The binomial theorem 1

1 1. 1 2 1. 1 3 3 1. The binomial theorem 1. Objectives:. Pascal’s triangle. Coefficient of (x + y) n when n is large. Notation:. n c r. Expansion of (x + y) n for n = 2, 3 and 4. (x + y) 2 =. x(x + y) + y(x + y) =. x 2 + 2xy + y 2.

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The binomial theorem 1

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  1. 1 1 1 2 1 1 3 3 1 The binomial theorem 1 Objectives: Pascal’s triangle Coefficient of(x + y)nwhen n is large Notation: ncr

  2. Expansion of (x + y)n for n = 2, 3 and 4 (x + y)2 = x(x + y) + y(x + y) = x2 + 2xy + y2 (x + y)3 = (x + y)(x + y)2 = (x + y)(x2 + 2xy + y2) = x(x2 + 2xy + y2) + y(x2 + 2xy + y2) = x3+ 2x2y + xy2 + x2y + 2xy2+ y3 = x3 + 3x2y + 3xy2 + y3 (x + y)4 = (x + y)(x + y)3 = x3 + 3x2y + 3xy2 + y3 = x(x3 + 3x2y + 3xy2 + y3) + y(x3 + 3x2y + 3xy2 + y3) = x4 + 3x3y + 3x2y2+ xy3 + x3y + 3x3y2 + 3x2y3+ y4 = x4 + 4x3y + 6x2y2 + 4xy3 + y4

  3. Expansion of (x + y)n (x + y)1 = 1x + 1y (x + y)2 = 1x2 + 2xy + 1y2 (x + y)3 = 1x3 + 3x2y + 3xy2 + 1y3 (x + y)4 = 1x4 + 4x3y + 6x2y2 + 4xy3 + 1y4 (x + y)5 = 1x5 + 5x4y + 10x3y2 + 10x2y3 + 5xy4 + 1y5 (1 + y)5 = 1(1)5 + 5(1)4y + 10(1)3y2 + 10(1)2y3 + 5(1)y4 + 1y5 = 1 + 5y + 10y2 + 10y3 + 5y4 + y5

  4. Examples: Write down the expansions: (x + 3y)2 = (x + y)2 = 1x2 + 2xy + 1y2 (x + 3y)2 = 1x2 + 2x(3y) + 1(3y)2 = x2 + 6xy + 9y2 (4 + y)3 = (x + y)3 = 1x3 + 3x2y + 3xy2 + 1y3 (4 + y)3 = 1(4)3 + 3(4)2y + 3(4)y2 + 1y3 = 64 + 48y + 12y2 + y3 (2a + 3b)3 = (x + y)3 = 1x3 + 3x2y + 3xy2 + 1y3 (2a + 3b)3 = 1(2a)3 + 3(2a)23b + 3(2a)(3b)2 + 1(3b)3 = 8a3 + 36a2b + 54ab2 + 27b3

  5. Examples: Write the coefficient of x3 in the expansion of (2x- 3)4. (x + y)4 = 1x4 + 4x3y + 6x2y2 + 4xy3 + 1y4 x3 is the 2nd term : 4(2x)3(-3) = 4(8x3)(-3) = -96x3 The required coefficient is - 96 In the expansion of (1 + bx)4, the coefficient of x3 is 1372. Find the constant b. (x + y)4 = 1x4 + 4x3y + 6x2y2 + 4xy3 + 1y4 y3 is the 4th term : 4(1)(bx)3 = 4b3x3 = 1372  4b3 = 1372  b3 = 343  b = 7

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