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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

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the binomial theorem

The Binomial Theorem

Lecture 34

Section 6.7

Wed, Mar 28, 2007

the binomial theorem1
The Binomial Theorem
  • Theorem: Given any numbers a and b and any nonnegative integer n,
the binomial theorem2
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
Proof, continued
  • Inductive step
    • Suppose the statement is true when n = k for some k 0.
    • Then
proof continued2
Proof, continued
    • Therefore, the statement is true when n = k + 1.
  • Thus, the statement is true for all n 0.
example binomial theorem
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 theorem1
Example: Binomial Theorem
  • Therefore,

(a + b)8 = a8 + 8a7b + 28a6b2 + 56a5b3

+ 70a4b4 + 56a3b5 + 28a2b6 + 8ab7 + b8.

example calculating 1 01 8
Example: Calculating 1.018
  • Compute 1.018 on a calculator.
  • What do you see?
example calculating 1 01 81
Example: Calculating 1.018
  • Compute 1.018 on a calculator.
  • What do you see?
  • 1.018 = 1.0828567056280801.
example calculating 1 01 6
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
Example: Approximating (1+x)n
  • Theorem: For small values of x,

and so on.

example
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
Expanding Trinomials
  • Expand (a + b + c)3.
expanding trinomials1
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 trinomials2
Expanding Trinomials

= a3 + 3a2b + 3ab2 + b3 + 3a2c + 6abc + 3b2c + 3ac2 + 3bc2 + c3.

  • What is the pattern?
expanding trinomials3
Expanding Trinomials
  • (a + b + c)3 = (a3 + b3 + c3)

+ 3(a2b + a2c + ab2 + b2c + ac2 + bc2)

+ 6abc.

the multinomial theorem
The Multinomial Theorem
  • Theorem: In the expansion of

(a1 + … + ak)n,

the coefficient of a1n1a2n2…aknk is

example the multinomial theorem
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 theorem1
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.

actuary exam problem
Actuary Exam Problem
  • If we expand the expression

(a + 2b + 3c)4,

what will be the sum of the coefficients?