Transcript Document

Copyright © 2011 Pearson Education, Inc.
Slide 11.5-1
Chapter 11: Further Topics in Algebra
11.1Sequences and Series
11.2Arithmetic Sequences and Series
11.3Geometric Sequences and Series
11.4Counting Theory
11.5The Binomial Theorem
11.6Mathematical Induction
11.7Probability
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Slide 11.5-2
11.5 The Binomial Theorem
The binomial expansions
( x  y )0  1
( x  y )1  x  y
( x  y ) 2  x 2  2 xy  y 2
( x  y )3  x 3  3 x 2 y  3xy 2  y 3
( x  y ) 4  x 4  4 x 3 y  6 x 2 y 2  4 xy 3  y 4
( x  y )5  x 5  5 x 4 y  10 x 3 y 2  10 x 2 y 3  5 xy 4  y 5
reveal a pattern.
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Slide 11.5-3
11.5 A Binomial Expansion Pattern
• The expansion of (x + y)n begins with x n and ends
with y n .
• The variables in the terms after x n follow the
pattern x n-1y , x n-2y2 , x n-3y3 and so on to y n .
With each term the exponent on x decreases by 1
and the exponent on y increases by 1.
• In each term, the sum of the exponents on x and y
is always n.
• The coefficients of the expansion follow Pascal’s
triangle.
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Slide 11.5-4
11.5 Pascal’s Triangle
Pascal’s Triangle
Row
0
1
1
1
1
1
1
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2
3
4
5
1
1
3
6
10
1
2
1
4
10
3
1
5
4
1
5
Slide 11.5-5
11.5 Pascal’s Triangle
• Each row of the triangle begins with a 1 and ends
with a 1.
• Each number in the triangle that is not a 1 is the
sum of the two numbers directly above it (one to
the right and one to the left.)
• Numbering the rows of the triangle 0, 1, 2, …
starting at the top, the numbers in row n are the
coefficients of x n, x n-1y , x n-2y2 , x n-3y3, … y n in
the expansion of (x + y)n.
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Slide 11.5-6
11.5 Binomial Coefficients
Binomial Coefficient
For nonnegative integers n and r, with r < n,
n
n!
 
 r  r !(n  r )!
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Slide 11.5-7
11.5 Binomial Coefficients
 n
• The symbols n Cr and   for the binomial
r
coefficients are read “n choose r”
 n
• The values of   are the values in the nth row
r
 3
of Pascal’s triangle. So   is the first number
 0
 3
in the third row and   is the third.
 2
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Slide 11.5-8
11.5 Evaluating Binomial Coefficients
 6
Example Evaluate (a)  
 2
8
(b)  
 0
Solution
 6
6!
6! 6  5  4  3  2 1


 15
(a)   
 2  2!(6  2)! 2!4! 2 1  4  3  2 1
 8
8!
8!
8!


1
(b)   
 0  0!(8  0)! 0!8! 1 8!
Copyright © 2011 Pearson Education, Inc.
Slide 11.5-9
11.5 The Binomial Theorem
Binomial Theorem
For any positive integer n,
 n  n 1  n  n  2 2  n  n 3 3
( x  y)  x    x y    x y    x y
1
 2
 3
 n  nr r
 n  n 1
n
 ...    x y  ...  
xy

y

r
 n  1
n
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n
Slide 11.5-10
11.5 Applying the Binomial Theorem
Example Write the binomial expansion of ( x  y )9 .
Solution Use the binomial theorem
9 8  9 7 2 9 6 3
( x  y)  x    x y    x y    x y
1
 2
 3
 9 5 4  9 4 5  9 3 6  9 2 7
 x y  x y  x y  x y
 4
 5
 6
 7
9 8
   xy  y 9
8
9
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9
Slide 11.5-11
11.5 Applying the Binomial Theorem
9! 8
9! 7 2 9! 6 3
( x  y)  x 
x y
x y 
x y
1!8!
2!7!
3!6!
9! 5 4 9! 4 5 9! 3 6
9! 2 7

x y 
x y 
x y 
x y
4!5!
5!4!
6!3!
7!2!
9! 8

xy  y 9
8!1!
 x9  9 x8 y  36 x 7 y 2  84 x 6 y 3  126 x 5 y 4  126 x 4 y 5
9
9
 84 x3 y 6  36 x 2 y 7  9 xy 8  y 9
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Slide 11.5-12
11.5 Applying the Binomial Theorem
5
b

Example Expand  a   .
2

Solution Use the binomial theorem with
b
xa , y
and n = 5,
2
2
5
5
  4 b    3 b
b

5
 a    a  1a      2 a   
2

   2    2
5
3
4
5
5
5
  2 b    b  b
 a     a      
 3  2   4  2   2 
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Slide 11.5-13
11.5 Applying the Binomial Theorem
Solution
5
b

5
4 b
3 b
 a    a  5a     10a   
2

 2
 2
3
4
2
5
 b
 b  b
10a     5a       
 2
 2  2
5 4 5 3 2 5 2 3 5 4 1 5
5
 a  a b  a b  a b  ab  b
2
2
4
16
32
2
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Slide 11.5-14
11.5 rth Term of a Binomial Expansion
rth Term of the Binomial Expansion
The rth term of the binomial expansion of (x + y)n,
where n > r – 1, is
 n  n( r 1) r 1
y .

x
 r  1
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Slide 11.5-15
11.5 Finding a Specific Term of a
Binomial Expansion
10
(
a

2
b
)
Example Find the fourth term of
.
Solution Using n = 10, r = 4, x = a, y = 2b in the
formula, we find the fourth term is
10  7
3
7
3
7 3
  a (2b)  120a 8b  960a b .
3
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Slide 11.5-16