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The Binomial Theorem (9.5)

The Binomial Theorem (9.5). I. Pascal's Triangle. (x + y) 0. 1. (x + y) 1. 1 1 . x. +. y. (x + y) 2. 1 2 1 . x 2. +. xy. +. y 2. (x + y) 3. 1 3 3 1 . x 3. +. x 2 y. +. xy 2. +. y 3. (x + y) 4. +. xy 3.

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The Binomial Theorem (9.5)

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  1. The Binomial Theorem (9.5)

  2. I. Pascal's Triangle (x + y)0 1 (x + y)1 1 1 x + y (x + y)2 1 2 1 x2 + xy + y2 (x + y)3 1 3 3 1 x3 + x2y + xy2 + y3 (x + y)4 + xy3 1 4 6 4 1 x4 + x3y + x2y2 + y4 • There are n + 1 terms in the expansion of (x + y)n • The degree of each term in the expansion of (x + y)n is n • In each successive term of the expansion of (x + y)n, the exponent of x decreases by 1 and the exponent of y increases by 1. • The first term is always xn and the last term is always yn

  3. A. Using Pascal's Triangle Example: a) Write row 7 of Pascal’s Triangle 1 7 21 35 35 21 7 1 b) Expand (x + y)7 NOTE: multiplying this out would work but would be time- consuming = x7 + 7x6y + 21x5y2 + 35x4y3 + 35x3y4 + 21x2y5 + 7xy6 + y7 Think of x as the first term Think of y as the second term

  4. A. Using Pascal's Triangle Example: c) Use Pascal’s Triangle to expand (2a  3b)4 4th row of Pascal’s Triangle: 1 4 6 4 1 x4 + x3y + x2y2 + xy3 + y4 (2a – 3b)4 = [(2a) + ( -3b)]4 let x = 2a let y = -3b the first term the second term So (2a + 3b)4 = 1(2a)4+ 4(2a)3(-3b) + 6(2a)2(-3b)2 + 4(2a)(-3b)3 + 1(-3b)4 = 16a4 96a3b + 216a2b2 216ab3 + 81b4

  5. II. Binomial Theorem A. Analyze RECALL: (x + y)4 = x4 + 4x3y + 6x2y2 + 4xy3 + y4 Look at x3y: There are four numbers being multiplied, but only 1 y. How many combinations? RECALL: nCr = C(n, r) = n! (n – r)!r!

  6. EXAMPLE: Expand (x + y)4 Out of 4 spots, choose 0 y’s 4! = 1  1x4 1st term: x4 4C0 = (4 – 0)! 0! 4!  4x3y = 4! = 4 2nd term: x3y 4C1 = 3! (4 – 1)! 1! Out of 4 spots, choose 1 y 4! 4C2 = = 4! = 6  6x2y2 3rd term: x2y2 (4 – 2)! 2! 2!2! 4! = 4! = 4  4xy3 4C3 = 4th term: xy3 1!3! (4 – 3)! 3! 4! 4C4 = = 1  1y4 5th term: y4 (4 – 4)! 4! Use TI Calculator! Answer: (x + y)4 = 1x4 + 4x3y + 6x2y2 + 4xy3 + 1y4

  7. B. Binomial Theorem Def: For any (+) integer, n, the expansion of (x + y)n is nC0 xn + nC1 xn– 1y + nC2 xn – 2y2 + … + nCn yn SUMMARIZE:

  8. EXAMPLE: Expand (x – 2y)16 16C0 (x)16 + 16C1 (x)15(-2y) + 16C2(x)14(-2y)2 + 16C3(x)13(-2y)3 + 16C4(x)12 (-2y)4+ 16C5 (x)11 (-2y)5 + 16C6(x)10(-2y)6 + 16C7(x)9(-2y)7 + 16C8(x)8(-2y)8 + 16C9(x)7(-2y)9 + 16C10(x)6(-2y)10 + 16C11(x)5(-2y)11 + 16C12(x)4(-2y)12 + 16C13(x)3(-2y)13 + 16C14(x)2(-2y)14 + 16C15(x)(-2y)15 + 16C16(-2y)16 = x16 – 32x15y + 480x14y2 – 4480x13y3 + 29120x12y4 – 139776x11y5 + 512512x10y6 – 1464320x9y7 + 3294720x8y8 – 5857280x7y9 + 8200192x6y10 – 8945664x5y11 + 7454720x4y12 – 4587520x3y13 + 1966080x2y14 – 524288xy15 + 65536y16

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