9.5 The Binomial Theorem

advertisement
Digital Lesson
The Binomial Theorem
The binomial theorem provides a useful method for raising any
binomial to a nonnegative integral power.
Consider the patterns formed by expanding (x + y)n.
(x + y)0 = 1
1 term
(x + y)1 = x + y
2 terms
(x + y)2 = x2 + 2xy + y2
3 terms
(x + y)3 = x3 + 3x2y + 3xy2 + y3
4 terms
(x + y)4 = x4 + 4x3y + 6x2y2 + 4xy3 + y4
5 terms
6 terms
(x + y)5 = x5 + 5x4y + 10x3y2 + 10x2y3 + 5xy4 + y5
Notice that each expansion has n + 1 terms.
Example: (x + y)10 will have 10 + 1, or 11 terms.
Copyright © by Houghton Mifflin Company, Inc. All rights reserved.
2
Consider the patterns formed by expanding (x + y)n.
(x + y)0 = 1
(x + y)1 = x + y
(x + y)2 = x2 + 2xy + y2
(x + y)3 = x3 + 3x2y + 3xy2 + y3
(x + y)4 = x4 + 4x3y + 6x2y2 + 4xy3 + y4
(x + y)5 = x5 + 5x4y + 10x3y2 + 10x2y3 + 5xy4 + y5
1. The exponents on x decrease from n to 0.
The exponents on y increase from 0 to n.
2. Each term is of degree n.
Example: The 5th term of (x + y)10 is a term with x6y4.”
Copyright © by Houghton Mifflin Company, Inc. All rights reserved.
3
The coefficients of the binomial expansion are called binomial
coefficients. The coefficients have symmetry.
(x + y)5 = 1x5 + 5x4y + 10x3y2 + 10x2y3 + 5xy4 + 1y5
The first and last coefficients are 1.
The coefficients of the second and second to last terms
are equal to n.
Example: What are the last 2 terms of (x + y)10 ? Since n = 10,
the last two terms are 10xy9 + 1y10.
The coefficient of xn–ryr in the expansion of (x + y)n is written  n 
or nCr . So, the last two terms of (x + y)10 can be expressed  r 
as 10C9 xy9 + 10C10 y10 or as 10  xy 9 + 10  y10.
9 
 
Copyright © by Houghton Mifflin Company, Inc. All rights reserved.
10 
 
4
The triangular arrangement of numbers below is called Pascal’s
Triangle.
1
0th row
1
1
1+2=3
1
6 + 4 = 10
1
1
2
3
4
1st row
1
3
6
2nd row
1
4
3rd row
1
1 5 10 10 5 1
4th row
5th row
Each number in the interior of the triangle is the sum of the two
numbers immediately above it.
The numbers in the nth row of Pascal’s Triangle are the binomial
coefficients for (x + y)n .
Copyright © by Houghton Mifflin Company, Inc. All rights reserved.
5
Example: Use the fifth row of Pascal’s Triangle to generate the
6
6
sixth row and find the binomial coefficients  ,  , 6C4 and 6C2 .
1 5
5th row
1
5
10
10
5
1
1
6 15
20 15 6
6th row
1
6 6
  1
0  
6
 
 2
6
 
 3
6 6
  5
 4  
6
 
6
6C0
6C2
6C3
6C4
6C6
6C1
6
 =6=
1
6C5
6
  and 6C4 = 15 = 6C2.
5
There is symmetry between binomial coefficients.
nCr = nCn–r
Copyright © by Houghton Mifflin Company, Inc. All rights reserved.
6
Example: Use Pascal’s Triangle to expand (2a + b)4.
0th row
1
1
1
1
1
2
3
4
1st row
1
3
6
2nd row
1
3rd row
1
4
1
4th row
(2a + b)4 = 1(2a)4 + 4(2a)3b + 6(2a)2b2 + 4(2a)b3 + 1b4
= 1(16a4) + 4(8a3)b + 6(4a2b2) + 4(2a)b3 + b4
= 16a4 + 32a3b + 24a2b2 + 8ab3 + b4
Copyright © by Houghton Mifflin Company, Inc. All rights reserved.
7
The symbol n! (n factorial) denotes the product of the first n
positive integers. 0! is defined to be 1.
1! = 1
4! = 4 • 3 • 2 • 1 = 24
6! = 6 • 5 • 4 • 3 • 2 • 1 = 720
n! = n(n – 1)(n – 2)  3 • 2 • 1
Formula for Binomial Coefficients For all nonnegative
n!
integers n and r,
n
Cr 
(n  r )!r !
7!
7!
7
Example: 7 C3 


(7  3)! • 3! 4! • 3! 4! • 3!
(7 • 6 • 5 • 4) • (3 • 2 • 1) 7 • 6 • 5 • 4


 35
(4 • 3 • 2 • 1) • (3 • 2 • 1) 4 • 3 • 2 • 1
Copyright © by Houghton Mifflin Company, Inc. All rights reserved.
8
Example: Use the formula to calculate the binomial coefficients
 12  and  50  .
C
,
C
,
10 5 15 0
 
1
 
 48 
10!
10!
(10 • 9 • 8 • 7 • 6) • 5! 10 • 9 • 8 • 7 • 6



 252
10 C5 
(10  5)! • 5! 5! • 5!
5! • 5!
5 • 4 • 3• 2 •1
10!
10!
1! 1

  1
10 C0 
•
•
(10  0)! 0! 10! 0! 0! 1
50!
(50 • 49) • 48! 50 • 49
 50 
50!



 1225
  
•
•
•
2! 48!
2 1
 48  (50  48)! • 48! 2! 48!
12 
12! 12 • 11! 12
12!


  12
  
•
•
1
1  (12  1)! • 1! 1! 1! 11! 1!
Copyright © by Houghton Mifflin Company, Inc. All rights reserved.
9
Binomial Theorem
n 1
( x  y)  x  nx y 
n
n
 nCr x
n r
y 
r
 nxy
n 1
y
n
n!
with nCr 
(n  r )!r !
Example: Use the Binomial Theorem to expand (x4 + 2)3.
(x 4  2)3  3 C0(x 4 )3  3 C1( x 4 ) 2 (2)  3 C2(x 4 )(2) 2  3 C3(2)3
 1 (x 4 )3  3( x 4 ) 2 (2)  3(x 4 )(2) 2  1 (2)3
 x12  6 x8  12 x 4  8
Copyright © by Houghton Mifflin Company, Inc. All rights reserved.
10
Although the Binomial Theorem is stated for a binomial which
is a sum of terms, it can also be used to expand a difference of
terms.
Simply rewrite
(x + y) n as (x + (– y)) n
and apply the theorem to this sum.
Example: Use the Binomial Theorem to expand (3x – 4)4.
(3x  4) 4  (3x  (4)) 4
 1(3x) 4  4(3x)3 (4)  6(3x) 2 (4) 2  4(3x)(4)3  1(4) 4
 81x 4  4(27 x3 )(4)  6(9 x 2 )(16)  4(3x)(64)  256
 81x 4  432 x3  864 x 2  768x  256
Copyright © by Houghton Mifflin Company, Inc. All rights reserved.
11
Example: Use the Binomial Theorem to write the first three
terms in the expansion of (2a + b)12 .
12 
12 
12 
12
11
(2a  b)   (2a)   (2a) b   (2a)10 b 2  ...
0 
1 
2 
12
12!
12!
12!
12 12
11 11

(2 a ) 
(2 a )b 
(210 a10 )b 2  ...
(12  0)! • 0!
(12  1)! • 1!
(12  2)! • 2!
 (212 a12 )  12(211 a11 )b  (12 • 11)(210 a10 )b 2  ...
 4096 a12  24576 a11b  135168 a10b 2  ...
Copyright © by Houghton Mifflin Company, Inc. All rights reserved.
12
Example: Find the eighth term in the expansion of (x + y)13 .
Think of the first term of the expansion as x13y 0 . The power of
y is 1 less than the number of the term in the expansion.
The eighth term is 13C7 x 6 y7.
13! (13 • 12 • 11 • 10 • 9 • 8) • 7!

13 C7 
6! • 7!
6! • 7!
13 • 12 • 11 • 10 • 9 • 8

 1716
6 • 5 • 4 • 3 • 2 •1
Therefore, the eighth term of (x + y)13 is 1716 x 6 y7.
Copyright © by Houghton Mifflin Company, Inc. All rights reserved.
13
Download