The Binomial Theorem Lecture 29 Section 6.7 Mon, Apr 3, 2006

The Binomial Theorem Theorem: Given any numbers a and b and any nonnegative integer n, (a + b) n =  k = 0..n C(n, k)a n – k b k. Proof: Use induction on n. Basic step: Let n = 0. Then (a + b) 0 = 1.  k=0..0 C(0, k)a 0 – k b k = C(0, 0)a 0 b 0 = 1. Therefore, the statement is true when n = 0.

Proof, continued Inductive step Suppose the statement is true for some n  0. Then

Proof, continued

Therefore, the statement is true for n + 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 = a 8 + 8a 7 b + 28a 6 b a 5 b a 4 b a 3 b a 2 b 6 + 8ab 7 + b 8.

Example: Calculating Compute on a calculator. What do you see?

Example: Calculating = ( ) 8 = 1 + 8(0.01) + 28(0.01) (0.01) (0.01) (0.01) (0.01) 6 + 8(0.01) 7 + (0.01) 8 = =

Example: Approximating (1+x) n Theorem: For small values of x, and so on.

Example For example, (1 + x) 8  1 + 8x + 28x 2 when x is small. Compute the value of (1 + x) 8 and the approximation when x =.03.

Newton’s Generalization of the Binomial Theorem Theorem: Given any numbers a, b, and n, (a + b) n =  k=0..  C(n, k)a n – k b k. where C(n, k) = [n(n – 1)…(n – k + 1)]/k! Note that n need not be an integer nor positive.

Newton’s Generalization Expand (a + b) -1 in a series, showing the first 5 terms. Compute the first 5 coefficients C(-1, 0) = 1. C(-1, 1) = -1. C(-1, 2) = (-1)(-2)/2! = 1. C(-1, 3) = (-1)(-2)(-3)/3! = -1. C(-1, 4) = (-1)(-2)(-3)(-4)/4! = 1.

Newton’s Generalization Therefore,

Newton’s Generalization Expand (a + b) 5/2 in a series, showing the first 5 terms.

Newton’s Generalization Therefore,

Newton’s Generalization If 0 < b < a, then this series will converge rapidly.

Newton’s Generalization Approximate (1.2) 5/2. Let a = 1 and b = 0.2 = 1/5. Then b/a = 1/5.

The Multinomial Theorem Theorem: In the expansion of (a 1 + … + a k ) n, the coefficient of a 1 n 1 a 2 n 2 …a k n k is n!/(n 1 !n 2 !…n k !) where n = n 1 + n 2 + … + n k.

Example: The Multinomial Theorem Expand (a + b + c + d) 3. The terms are a 3, b 3, c 3, d 3, with coefficient 3!/3! = 1. a 2 b, a 2 c, a 2 d, ab 2, b 2 c, b 2 d, ac 2, bc 2, c 2 d, ad 2, bd 2, cd 2, 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 = a 3 + b 3 + c 3 + d 3 + 3a 2 b + 3a 2 c + 3a 2 d + 3ab 2 + 3b 2 c + 3b 2 d + 3ac 2 + 3bc 2 + 3c 2 d + 3ad 2 + 3bd 2 + 3cd 2 + 6abc + 6abd + 6acd + 6bcd. Find (a + b + c) 4.

Actuary Exam Problem If we expand the expression (a + 2b + 3c) 4, what will be the sum of the coefficients?