Lecture 34 Section 6.7 Wed, Mar 28, 2007

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Lecture 34 Section 6.7 Wed, Mar 28, 2007 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

Proof, continued Thus, the statement is true for all n  0. 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?