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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,
 n  n i i
a  b    a b .
i 0  i 
n
n
The Binomial Theorem
Proof: Use induction on n.
 Base case: Let n = 0. Then


(a + b)0 = 1 and
 0  0 i i  0  0  0 0
 a b   a b  1.

i 0  i 
0
0

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
a  b k 1  a  b a  b k

 k  k i i
 a  b   a b
i 0  i 
k
 k  k i 1 i k  k  k i i 1
   a
b    a b
i 0  i 
i 0  i 
k
Proof, continued
 k  k i 1 i k 1  k  k i i 1 k 1
 a    a
b    a b  b
i 1  i 
i 0  i 
k
k
k


 k  k i 1 i
k 1
k i 1 i
a
 a    a
b   
b  b k 1
i 1  i 
i 1  i  1
k 1
a
k 1
k
  k   k   k i 1 i
 a
      
b  b k 1
i 1   i   i  1 
k
Proof, continued
 k  1 k i 1 i
k 1


 a  
a
b b

i 1  i 
k 1 k  1

 k i 1 i
a
  
b.
i 0  i 
k 1


k
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,
1  x 
n
 1  nx.
n(n  1) 2
1  x   1  nx 
x .
2
n(n  1) 2 n(n  1)( n  2) 3
n
1  x   1  nx 
x 
x.
2
6
n
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
n!
n1!n2 ! nk !
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?