1 / 22

320 likes | 1.45k Vues

The Binomial Theorem. Lecture 34 Section 6.7 Wed, Mar 28, 2007. The Binomial Theorem. Theorem: Given any numbers a and b and any nonnegative integer n , . The Binomial Theorem. Proof: Use induction on n . Base case: Let n = 0. Then ( a + b ) 0 = 1 and

Télécharger la présentation
## The 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

**The Binomial Theorem**Lecture 34 Section 6.7 Wed, Mar 28, 2007**The Binomial Theorem**• Theorem: Given any numbers a and b and any nonnegative integer n,**The Binomial Theorem**• Proof: Use induction on n. • Base case: Let n = 0. Then • (a + b)0 = 1 and • Therefore, the statement is true when n = 0.**Proof, continued**• Inductive step • Suppose the statement is true when n = k for some k 0. • Then**Proof, continued**• Therefore, the statement is true when n = k + 1. • Thus, the statement is true for all n 0.**Example: Binomial Theorem**• Expand (a + b)8. • C(8, 0) = C(8, 8) = 1. • C(8, 1) = C(8, 7) = 8. • C(8, 2) = C(8, 6) = 28. • C(8, 3) = C(8, 5) = 56. • C(8, 4) = 70.**Example: Binomial Theorem**• Therefore, (a + b)8 = a8 + 8a7b + 28a6b2 + 56a5b3 + 70a4b4 + 56a3b5 + 28a2b6 + 8ab7 + b8.**Example: Calculating 1.018**• Compute 1.018 on a calculator. • What do you see?**Example: Calculating 1.018**• Compute 1.018 on a calculator. • What do you see? • 1.018 = 1.0828567056280801.**Example: Calculating 1.016**• 1.018 = (1 + 0.01)8 = 1 + 8(0.01) + 28(0.01)2 + 56(0.01)3 + 70(0.01)4 + 56(0.01)5 + + 28(0.01)6 + 8(0.01)7 + (0.01)8 = 1 + .08 + .0028 + .000056 + .00000070 + .0000000056 + .000000000028 + + .00000000000008 + .0000000000000001 = 1.0828567056280801.**Example: Approximating (1+x)n**• Theorem: For small values of x, and so on.**Example**• For example, (1 + x)8 1 + 8x + 28x2 when x is small. • Compute the value of (1 + x)8 and the approximation when x = .03. • Do it again for x = -.03.**Expanding Trinomials**• Expand (a + b + c)3.**Expanding Trinomials**• Expand (a + b + c)3. • (a + b + c)3 = ((a + b) + c)3 = (a + b)3 + 3(a + b)2c + 3(a + b)c2 + c3, = (a3 + 3a2b + 3ab2 + b3) + 3(a2 + 2ab + b2)c + 3(a + b)c2 + c3.**Expanding Trinomials**= a3 + 3a2b + 3ab2 + b3 + 3a2c + 6abc + 3b2c + 3ac2 + 3bc2 + c3. • What is the pattern?**Expanding Trinomials**• (a + b + c)3 = (a3 + b3 + c3) + 3(a2b + a2c + ab2 + b2c + ac2 + bc2) + 6abc.**The Multinomial Theorem**• Theorem: In the expansion of (a1 + … + ak)n, the coefficient of a1n1a2n2…aknk is**Example: The Multinomial Theorem**• Expand (a + b + c + d)3. • The terms are • a3, b3, c3, d3, with coefficient 3!/3! = 1. • a2b, a2c, a2d, ab2, b2c, b2d, ac2, bc2, c2d, ad2, bd2, cd2, with coefficient 3!/(1!2!) = 3. • abc, abd, acd, bcd, with coefficient 3!/(1!1!1!) = 6.**Example: The Multinomial Theorem**• Therefore, (a + b + c + d)3 = a3 + b3 + c3 + d3 + 3a2b + 3a2c + 3a2d + 3ab2 + 3b2c + 3b2d + 3ac2 + 3bc2 + 3c2d + 3ad2 + 3bd2 + 3cd2 + 6abc + 6abd + 6acd + 6bcd.**Example: The Multinomial Theorem**• Find (a + 2b + 1)4.**Actuary Exam Problem**• If we expand the expression (a + 2b + 3c)4, what will be the sum of the coefficients?

More Related