1 / 8

Section 8.5 Binomial Theorem

Section 8.5 Binomial Theorem. Binomial Expansions. (a + b) 1 =. a + b. (a + b) 2 =. a 2 + 2ab + b 2. (a + b) 3 =. a 3 + 3a 2 b + 3ab 2 + b 3. (a + b) 4 =. a 4 + 4a 3 b + 6a 2 b 2 + 4ab 3 + b 4. (a + b) 5 =. a 5 + 5a 4 b + 10a 3 b 2 + 10a 2 b 3 + 5ab 4 + b 5.

jcutler
Download Presentation

Section 8.5 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. Section 8.5Binomial Theorem

  2. Binomial Expansions (a + b)1 = a + b (a + b)2 = a2 + 2ab + b2 (a + b)3 = a3 + 3a2b + 3ab2 + b3 (a + b)4 = a4 + 4a3b + 6a2b2 + 4ab3 + b4 (a + b)5 = a5 + 5a4b + 10a3b2 + 10a2b3 + 5ab4 + b5 Pasqual's Triangle 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1

  3. We can use Pascal’s triangle to expand binomials, but it becomes large and cumbersome when the powers of the binomial are large. Therefore, the coefficients in a binomial expansion are often given in terms of factorials. Binomial Coefficients n r For nonnegative integers n and r, with n ³r, the expression is called a binomial coefficient and is defined by n r n! = r! (n – r)! 6 2 6! 6 · 5 · 4! 30 = = = = 15 2! · 4! 2! · 4! 2 3! 3 0 = = 1 0! · 3!

  4. The Binomial Theorem For any positive integer, n n 0 n 1 n 2 n 3 n n (a + b)n = an + an-1b + an-2b2 + an-3b3 + . . . + bn Expand: (x + 2)5 5 4 5 5 5 0 5 1 5 2 5 3 x5 + x4(2)1 + x3(2)2 + x2(2)3+ x(2)4 + (2)5 = = (1)(x5) + 5(x4)(2) + 10(x3)(4) + 10(x2)(8) + 5(x)(16) + (1)(32) = x5 + 10x4 + 40x3 + 80x2 + 80x + 32

  5. Expand: (3x – 2y)4 4 0 4 1 4 2 4 3 4 4 (3x)4 + (3x)3(-2y)1 + (3x)2(-2y)2 + (3x)1(-2y)3+ (-2y)4 = = (1)(81x4) + 4(27x3)(-2y) + 6(9x2)(4y2) + 4(3x)(-8y3) + (1)(16y4) = 81x4 – 216x3y + 216x2y2 – 96xy3 + 16y4 Find the first three terms of the following expansion 20 0 20 1 20 2 (x2 – 1)20 (x2)20 + (x2)19(-1)1 + (x2)18(-1)2 + . . . = = (1)(x40) + 20(x38)(-1) + 190(x36)(1) + . . . = x40 – 20x38+ 190x36 – . . .

  6. Find the 8th term of the expansion of (2x – 3)12 12 7 8th term = (2x)5(-3)7 = 792(32x5)(-2187) = - 55,427,328 x5 In the expansion of (x + 2y)15 , find the term containing y8 15 8 the term containing y8 = (x)7(2y)8 = 6435(x7)(256y8) = 1,647,360 x7y8

  7. find and simplify f(x + h) – f(x) h f(x) = x5 (x + h)5 – x5 f(x + h) – f(x) h = h = (x5 + 5x4h + 10x3h2 + 10x2h3 + 5xh4 + h5) – x5 h = 5x4h + 10x3h2 + 10x2h3 + 5xh4 + h5 h = 5x4 + 10x3h + 10x2h2 + 5xh3 + h4

  8. Homework Pirates of Penzance

More Related