Download BINOMIAL EXPANSIONS

702
(13-26)
Chapter 13
Sequences and Series
13.5 B I N O M I A L E X P A N S I O N S
In this
section
●
Some Examples
●
Obtaining the Coefficients
●
The Binomial Theorem
In Chapter 5 you learned how to square a binomial. In this section you will study
higher powers of binomials.
Some Examples
We know that (x y)2 x 2 2xy y 2. To find (x y)3, we multiply (x y)2 by
x y:
(x y)3 (x 2 2xy y 2)(x y)
(x 2 2xy y 2)x (x 2 2xy y 2)y
x 3 2x 2y xy 2 x 2y 2xy 2 y 3
x 3 3x 2y 3xy 2 y 3
The sum x 3 3x 2y 3xy 2 y 3 is called the binomial expansion of (x y)3. If we
again multiply by x y, we will get the binomial expansion of (x y)4. This method
is rather tedious. However, if we examine these expansions, we can find a pattern and
learn how to find binomial expansions without multiplying.
Consider the following binomial expansions:
(x (x (x (x (x (x y)0
y)1
y)2
y)3
y)4
y)5
1
xy
x 2 2xy y 2
x 3 3x 2y 3xy 2 y 3
x 4 4x 3y 6x 2y 2 4xy 3 y 4
x 5 5x 4y 10x 3y 2 10x 2y 3 5xy 4 y 5
Observe that the exponents on the variable x are decreasing, whereas the exponents
on the variable y are increasing, as we read from left to right. Also notice that the
sum of the exponents in each term is the same for that entire line. For instance, in
the fourth expansion the terms x 4, x 3y, x 2y 2, xy 3, and y 4 all have exponents with a
sum of 4. If we continue the pattern, the expansion of (x y)6 will have seven terms
containing x 6, x 5y, x 4y 2, x 3y 3, x 2y 4, xy 5, and y 6. Now we must find the pattern for the
coefficients of these terms.
Obtaining the Coefficients
If we write out only the coefficients of the expansions that we already have, we can
easily see a pattern. This triangular array of coefficients for the binomial expansions
is called Pascal’s triangle.
(x y)0 1
1
1
1
1
1
1
2
3
4
5
(x y)2 1x 2 2xy 1y 2
1
3
6
10
(x y)1 1x 1y
1
4
10
(x y)3 1x 3 3x 2y 3xy 2 1y 3
1
(x y)4 1x 4 4x 3y 6x 2y 2 4xy 3 1y 4
1
5
1
Coefficients in (x y)5
Notice that each line starts and ends with a 1 and that each entry of a line is the
sum of the two entries above it in the previous line. For instance, 4 3 1,
13.5
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Binomial Expansions
703
and 10 6 4. Following this pattern, the sixth and seventh lines of coefficients
are
1
6
15
20
15
6
1
1
7
21
35
35
21
7
1
Pascal’s triangle gives us an easy way to get the coefficients for the binomial
expansion with small powers, but it is impractical for larger powers. For larger
powers we use a formula involving factorial notation.
n! (n factorial)
If n is a positive integer, n! (read “n factorial”) is defined to be the product of
all of the positive integers from 1 through n.
calculator
close-up
You can evaluate the coefficients using either the factorial notation or nCr. The
factorial symbol and nCr are
found in the MATH menu
under PRB.
For example, 3! 3 2 1 6, and 5! 5 4 3 2 1 120. We also
define 0! to be 1.
Before we state a general formula, consider how the coefficients for (x y)4 are
found by using factorials:
4!
4! 0!
4!
3! 1!
4!
2! 2!
4!
1! 3!
4!
0! 4!
4321
1
43211
4321
4
3211
4321
6
2 1 2 1
Coefficient of x 4 (or x 4y 0 )
Coefficient of 4x 3y
Coefficient of 6x 2y 2
4321
4
1321
4321
1
14321
Coefficient of 4xy 3
Coefficient of y 4 (or x 0y 4)
Note that each expression has 4! in the numerator, with factorials in the denominator corresponding to the exponents on x and y.
The Binomial Theorem
We now summarize these ideas in the binomial theorem.
The Binomial Theorem
In the expansion of (x y)n for a positive integer n, there are n 1 terms,
given by the following formula:
n!
n!
n!
n!
(x y)n x n x n1y x n2y 2 . . . y n
n! 0!
(n 1)! 1!
(n 2)! 2!
0! n!
n
The notation is often used in place of
r
Using this notation, we write the expansion as
n!
(n r)!r!
in the binomial expansion.
(x y)n n x n n x n1y n x n2 y2 . . . n y n.
0
1
2
n
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Chapter 13
Sequences and Series
Another notation for n! is nCr . Using this notation, we have
(n r)!r!
(x y)n nC0 x n nC1x n1y nC2 x n2y 2 . . . nCn y n.
E X A M P L E
1
Using the binomial theorem
Write out the first three terms of (x y)9.
Solution
9!
9!
9!
(x y)9 x 9 x 8y x 7y 2 . . .
9! 0!
8! 1!
7! 2!
x 9 9x 8y 36x 7y 2 . . .
E X A M P L E
2
Using the binomial theorem
Write the binomial expansion for (x 2 2a)5.
Solution
We expand a difference by writing it as a sum and using the binomial theorem:
(x 2 2a)5 (x 2 (2a))5
5!
5!
5!
5!
(x 2)5 (x 2)4(2a)1 (x 2)3(2a)2 (x 2)2(2a)3
5!0!
4!1!
3!2!
2! 3!
5!
5!
(x 2)(2a)4 (2a)5
1! 4!
0! 5!
x 10 10x 8a 40x 6a 2 80x 4a 3 80x 2a 4 32a 5
E X A M P L E
3
calculator
Finding a specific term
Find the fourth term of the expansion of (a b)12.
Solution
The variables in the first term are a12b0, those in the second term are a11b1, those in
the third term are a10b2, and those in the fourth term are a9b3. So
close-up
n!
Because nCr , we
(n r)! r !
have
12!
12!
12C9 and 12C3 .
3! 9!
9! 3!
So there is more than one way
to compute 12!(9! 3!):
12!
a9b3 220a9b3.
9! 3!
The fourth term is 220a9b3.
Using the ideas of Example 3, we can write a formula for any term of a binomial
expansion.
Formula for the kth Term of (x y)n
For k ranging from 1 to n 1, the kth term of the expansion of (x y)n is
given by the formula
n!
x nk1y k1.
(n k 1)!(k 1)!
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705
7!
(a 2)2(2b)5 21a 4(32b5) 672a 4b 5
(7 6 1)!(6 1)!
13.5
E X A M P L E
4
Binomial Expansions
Finding a specific term
Find the sixth term of the expansion of (a 2 2b)7.
Solution
Use the formula for the kth term with k 6 and n 7:
We can think of the binomial expansion as a finite series. Using summation
notation, we can write the binomial theorem as follows.
The Binomial Theorem (Using Summation Notation)
For any positive integer n,
n
n!
(x y)n x ni y i
i0 (n i)!i!
E X A M P L E
5
or
(x y)n n
i x ni y i.
n
i0
Using summation notation
Write (a b)5 using summation notation.
Solution
Use n 5 in the binomial theorem:
(a b)5 5
5!
a 5i bi
(5 i)! i!
i0
WARM-UPS
True or false? Explain your answer.
1.
2.
3.
4.
5.
There are 12 terms in the expansion of (a b)12. False
The seventh term of (a b)12 is a multiple of a 5b7. False
For all values of x, (x 2)5 x 5 32. False
In the expansion of (x 5)8 the signs of the terms alternate. True
The eighth line of Pascal’s triangle is
1 8 28 56 70 56 28 8 1.
True
6. The sum of the coefficients in the expansion of (a b)4 is 24.
3
3!
7. (a b)3 a 3ibi True
i0 (3 i)! i!
8. The sum of the coefficients in the expansion of (a b)n is 2n.
9. 0! 1! True
7!
10. 21 True
5!2!
True
True
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Chapter 13
Sequences and Series
EXERCISES
Reading and Writing After reading this section, write out the
answers to these questions. Use complete sentences.
1. What is a binomial expansion?
The sum obtained for a power of a binomial is called a
binomial expansion.
2. What is Pascal’s triangle and how do you make it?
Pascal’s triangle gives the coefficients for (a b)n for
n 1, 2, 3, and so on. Each row starts and ends with a 1.
The other terms are obtained by adding the closest two
terms in the preceding row.
3. What does n! mean?
The expression n! is the product of the positive integers
from 1 through n.
4. What is the binomial theorem?
The binomial theorem gives the expansion of (a b)n.
Evaluate each expression.
5!
5. 10
2! 3!
8!
7. 56
5! 3!
6!
6. 5! 1!
9!
8. 2! 7!
6
36
Use the binomial theorem to expand each binomial. See Examples 1 and 2.
9. (r t)5 r 5 5r 4t 10r 3t 2 10r 2t 3 5rt 4 t 5
10. (r t)6 r 6 6r 5t 15r 4t 2 20r 3t 3 15r 2t 4 6rt 5 t 6
11. (m n)3 m3 3m2n 3mn2 n3
12. (m n)4 m4 4m3n 6m2n2 4mn3 n4
13. (x 2a)3 x 3 6ax 2 12a 2x 8a 3
14. (a 3b)4 a 4 12a 3b 54a 2b2 108ab3 81b4
15. (x 2 2)4 x 8 8x 6 24x 4 32x 2 16
16. (x 2 a 2)5 x10 5a2x 8 10a4x 6 10a6x 4 5a8x 2 a10
17. (x 1)7 x7 7x 6 21x 5 35x 4 35x 3 21x 2 7x 1
18. (x 1)6 x6 6x 5 15x 4 20x 3 15x 2 6x 1
Write out the first four terms in the expansion of each binomial.
See Examples 1 and 2.
19. (a 3b)12 a12 36a11b 594a10b2 5940a9b3
20. (x 2y)10 x 10 20x 9y 180x 8y 2 960x7y 3
21.
22.
23.
24.
(x 2 5)9 x 18 45x 16 900x 14 10,500x 12
(x 2 1)20 x 40 20x 38 190x 36 1140x 34
(x 1)22 x 22 22x 21 231x 20 1540x 19
(2x 1)8 256x 8 1024x7 1792x 6 1792x 5
x y
25. 2 3
a b
26. 2 5
10
8
5x9y 5x 8y 2 5x7y3
x 10
1024
768
256
144
a8
a7b 7a6b2 7a5b3
80
400
500
256
Find the indicated term of the binomial expansion. See Examples 3 and 4.
27. (a w)13, 6th term
28. (m n)12, 7th term
8 5
1287a w
924m6n6
16
29. (m n) , 8th term
30. (a b)14, 6th term
9 7
11,440m n
2002a9b5
8
31. (x 2y) , 4th term
32. (3a b)7, 4th term
5 3
448x y
2835a4b3
2
20
33. (2a b) , 7th term
34. (a 2 w 2)12, 5th term
635,043,840a28b6
495a16w8
Write each expansion using summation notation. See Example 5.
35. (a m)8
36. (z w)13
8
8!
a8imi
(8 i)! i!
i0
37. (a 2x)5
5
5!(2)i
a5ix i
(5 i)! i!
i0
13
13!
z13iwi
(13 i)! i!
i0
38. (w 3m)7
7
7!(3)i
w7imi
(7 i)!i!
i0
GET TING MORE INVOLVED
39. Discussion. Find the trinomial expansion for (a b c)3
by using x a and y b c in the binomial theorem.
a3 b3 c3 3a2b 3a2c 3ab2 3ac2 3b2c 3bc2 6abc
40. Discussion. What problem do you encounter when trying
to find the fourth term in the binomial expansion for
(x y)120 ? How can you overcome this problem? Find the
fifth term in the binomial expansion for (x 2y)100.
280,840x 117y 3, 62,739,600x 96y4
COLLABORATIVE ACTIVITIES
Lotteries Are Series(ous)
Roberto and his brother-in-law Horatio each have a child who
will be graduating from high school in 5 years. Each would like
to buy his child a car for a graduation present. Horatio decides to
buy two lottery tickets each week for the next 5 years, hoping to
win and buy a new car for his child. The lottery tickets are $2.00
each at the local convenience store. Roberto, who doesn’t
Grouping: Two to four students per group
Topic: Sequences and series
believe in lotteries, decides to set aside $4.00 each week and to
deposit this money in a savings account each quarter for the next
5 years to buy a used car. He finds a bank that will pay 5% yearly
interest compounded quarterly.