Download SECTION 8-4 Binomial Formula

8-4 Binomial Formula
581
point of view it appears that he will never finish the race.
This famous paradox is attributed to the Greek philosopher
Zeno, 495–435 B.C. If we assume the runner runs at 440
yards per minute, the times between tape breakings form an
infinite geometric progression. What is the sum of this
progression?
★
★
95. Geometry. If the midpoints of the sides of an equilateral
triangle are joined by straight lines, the new figure will be
an equilateral triangle with a perimeter equal to half the
original. If we start with an equilateral triangle with
perimeter 1 and form a sequence of “nested” equilateral triangles proceeding as described, what will be the total
perimeter of all the triangles that can be formed in this
way?
92. Puzzle. If a sheet of very thin paper 0.001 inch thick is torn
in half, and each half is again torn in half, and this process
is repeated for a total of 32 times, how high will the stack of
paper be if the pieces are placed one on top of the other?
Give the answer to the nearest mile.
96. Photography. The shutter speeds and f-stops on a camera
are given as follows:
93. Atmospheric Pressure. If atmospheric pressure decreases
roughly by a factor of 10 for each 10-mile increase in altitude up to 60 miles, and if the pressure is 15 pounds per
square inch at sea level, what will the pressure be 40 miles
up?
94. Zeno’s Paradox. Visualize a hypothetical 440-yard oval
racetrack that has tapes stretched across the track at the
halfway point and at each point that marks the halfway
point of each remaining distance thereafter. A runner running around the track has to break the first tape before the
second, the second before the third, and so on. From this
SECTION
8-4
Shutter speeds:
f-stops:
1 1 1
1
1
1
1, 12, 14, 18, 15
, 30, 60, 125
, 250
, 500
1.4, 2, 2.8, 4, 5.6, 8, 11, 16, 22
These are very close to being geometric progressions. Estimate their common ratios.
★★
97. Geometry. We know that the sum of the interior angles of
a triangle is 180°. Show that the sums of the interior angles
of polygons with 3, 4, 5, 6, . . . sides form an arithmetic sequence. Find the sum of the interior angles for a 21-sided
polygon.
Binomial Formula
• Factorial
• Binomial Formula
The binomial form
(a b)n
where n is a natural number, appears more frequently than you might expect. The
coefficients in the expansion play an important role in probability studies. The binomial formula, which we derive below, enables us to expand (a b)n directly for n
any natural number. Since the formula involves factorials, we digress for a moment
to introduce this important concept.
• Factorial
For n a natural number, n factorial—denoted by n!—is the product of the first n
natural numbers. Zero factorial is defined to be 1.
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8 Sequences and Series
DEFINITION 1
n Factorial
For n a natural number
n! n(n 1) . . . 2 1
1! 1
0! 1
It is also useful to note that:
Theorem 1
Recursion Formula for n Factorial
n! n (n 1)!
EXAMPLE 1
Evaluating Factorials
(A) 4! 4 3! 4 3 2! 4 3 2 1! 4 3 2 1 24
(B) 5! 5 4 3 2 1 120
7!
7 6!
(C)
7
6!
6!
8!
8 7 6 5!
(D)
336
5!
5!
Matched Problem 1
CAUTION
Find: (A) 6!
(B)
6!
5!
(C)
9!
6!
When reducing fractions involving factorials, don’t confuse the single integer n
with the symbol n!, which represents the product of n consecutive integers.
6!
2!
3!
6! 6 5 4 3!
6 5 4 120
3!
3!
Factorials are used in the definition of the important symbol
nr . This symbol
is frequently used in probability studies. It is called the combinatorial symbol and
is defined for nonnegative r and n, as follows:
8-4 Binomial Formula
DEFINITION 2
583
Combinatorial Symbol
For nonnegative integers r and n, 0 r n.
nr r!(nn! r)!
n(n 1)(n 2) . . . (n r 1)
r(r 1) . . . 2 1
The combinatorial symbol
nr also can be denoted by C
n,r
, nCr, or C(n, r) and
read as “n choose r.” Many calculators use nCr to denote the function that evaluates
the combinatorial symbol.
EXAMPLE 2
Evaluating the Combinatorial Symbol
(A)
(B)
Matched Problem 2
EXPLORE-DISCUSS 1
8!
8 7 6 5!
56
83 3!(88! 3)! 3!5!
3 2 1 5!
70 0!(77! 0)! 7!7! 1
Remember, 0! 1.
Find: (A)
92
(B)
5
5
(A) Compute the terms of the finite sequence
80, 81, 82, . . . , 88 (the
sequence is graphed in Fig. 1). Is the sequence arithmetic? Geometric? Which
term is the largest? The smallest? Find the sum of the corresponding series.
FIGURE 1
75
0
8
0
(B) Answer the same questions for the finite sequence
90, 91, 92, . . . , 99.
584
8 Sequences and Series
• Binomial Formula
We are now ready to try to discover a formula for the expansion of (a b)n using
ordinary induction; that is, we will look at a few special cases and postulate a general
formula from them. We will then try to prove that the formula holds for all natural
numbers, using mathematical induction. To start, we calculate directly the first five natural number powers of (a b)n, arranging the terms in decreasing powers of a:
(a b)1 a b
(a b)2 a2 2ab b2
(a b)3 a3 3a2b 3ab2 b3
(a b)4 a4 4a3b 6a2b2 4ab3 b4
(a b)5 a5 5a4b 10a3b2 10a2b3 5ab4 b5
Observations
1. The expansion of (a b)n has n 1 terms.
2. The power of a decreases by 1 for each term as we move from left to right.
3. The power of b increases by 1 for each term as we move from left to right.
4. In each term, the sum of the powers of a and b always adds up to n.
5. Starting with a given term, we can get the coefficient of the next term by
multiplying the coefficient of the given term by the exponent of a and dividing by the number that represents the position of the term in the series of
terms. For example, in the expansion of (a b)4, the coefficient of the third
term is found from the second term by multiplying 4 and 3 and then dividing by 2. Thus, the coefficient of the third term is (4 3)/2 6.
We now postulate the properties for the general case:
(a b)n a n n n 1
n(n 1) n 2 2
a b
a b
1
12
n!
n!
n!
an a n 1b a n 2b2
0!(n 0)!
1!(n 1)!
2!(n 2)!
n(n 1)(n 2) n 3 3 . . .
a b bn
123
0a 1a
n
n
n
b
n 1
2a
n
n!
n!
a n 3b3 . . . bn
3!(n 3)!
n!(n n)!
b n 2 2
3a
n
b ...
n 3 3
nb
n
n
Thus, we have arrived at the binomial formula using ordinary induction:
8-4 Binomial Formula
Theorem 2
585
Binomial Formula
For n a positive integer
k a
n
(a b)n n
nk k
b
k0
We now proceed to prove that the binomial formula holds for all natural numbers n using mathematical induction.
Proof
State the conjecture.
j a
n
Pn: (a b)n n
nj
bj
j0
Part 1
Show that P1 is true.
j a
1
1
1j
10 a 11 b a b (a b)
bj 1
j0
Thus, P1 is true.
Part 2
Show that if Pk is true, then Pk1 is true.
j a
k
Pk : (a b)k k
kj
bj
Assume Pk is true.
j0
k1
Pk1: (a b)k1 j0
k 1 k1 j j
a
b
j
Show P k1 is true.
We begin by multiplying both sides of Pk by (a b):
kja
k
(a b) k(a b) b j (a b)
kj
j0
The left side of this equation is the left side of P k1. Now we multiply out the right
side of the equation and try to obtain the right side of P k1:
2k a b . . . kkb (a b)
k
k
k
k
a
a b a b . . . ab 0
1
2
k
k
k
k
k
a b a b . . . ab b 0
1
k1
k
k
k
k
k
k
a
a b a b . . .
0
0
1
1
2
k
k
k
ab b
k 1 k k
(a b)k1 0k a 1k a
k
k1
k1
b
k2 2
k
k1 2
k
k1
k
k
k1 2
k
k
k1
k1 2
k
k1
586
8 Sequences and Series
We now use the following facts (the proofs are left as exercises; see Problems 49–51,
Exercise 11-4)
r k 1 kr k r 1 0k k 0 1 kk kk 11
to rewrite the right side as
k 0 1a
k1
k 1 1a b k 2 1a b . . .
k1
k1
ab b
k k 1
k
k1 2
k
k1
k1
j0
k 1 k1 j j
a
b
j
Since the right side of the last equation is the right side of Pk1, we have shown that
Pk1 follows from Pk .
Conclusion
EXAMPLE 3
Pn is true. That is, the binomial formula holds for all positive integers n.
Using the Binomial Formula
Use the binomial formula to expand (x y)6.
k x
6
Solution
(x y)6 6
6 k
yk
k0
60x 61x y 62x y 63x y 64x y 65xy 66y
6
5
4 2
3 3
2 4
5
6
x6 6x 5y 15x 4 y 2 20x 3 y 3 15x 2 y 4 6xy 5 y6
Matched Problem 3
EXAMPLE 4
Use the binomial formula to expand (x 1)5.
Using the Binomial Formula
Use the binomial formula to expand (3p 2q)4.
Solution
(3p 2q)4 [(3p) (2q)]4
k (3p)
4
4
a 3p, b 2q
4 k
(2q)k
k0
40(3p) 41(3p) (2q) 42(3p) (2q)
4
4
(3p)(2q) (2q)
3
4
4
3
2
2
3
81p4 216p3q 216p2q 2 96pq 3 16q 4
4
8-4 Binomial Formula
Matched Problem 4
EXPLORE-DISCUSS 2
587
Use the binomial formula to expand (2m 5n)3.
(A) Compute each term and also the sum of the alternating series
6
6
6
6
...
.
0
1
2
6
(B) What result about an alternating series can be deduced by letting a 1 and
b 1 in the binomial formula?
EXAMPLE 5
Using the Binomial Formula
Use the binomial formula to find the fourth and sixteenth terms in the expansion of
(x 2)20.
Solution
In the expansion of (a b)n, the exponent of b in the rth term is r 1 and the exponent of a is n (r 1). Thus,
Fourth term:
20 17
x (2)3
3
20 19 18 17
x (8)
321
9,120x17
Sixteenth term:
20 5
x (2)15
15
20 19 18 17 16 5
x (32,768)
54321
508,035,072x5
Matched Problem 5
Use the binomial formula to find the fifth and twelfth terms in the expansion of
(u 1)18.
Answers to Matched Problems
1. (A) 720
(B) 6
(C) 504
2. (A) 36
(B) 1
3. x5 5x 4 10x3 10x2 5x 1
4. 8m3 60m2n 150mn2 125n3
5. 3,060u14; 31,824u7
EXERCISE
8-4
A
4.
20!
17!
5. 4! 5!
7.
8!
3!5!
8.
6. (4 5)!
Evaluate each expression in Problems 1–12.
1. 7!
2. 5!
3.
15!
13!
10!
2!8!
9.
7!
0!(7 0)!
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10.
8 Sequences and Series
12!
12!(12 12)!
11.
10!
7!
12.
10!
3!
Write each expression in Problems 13–16 as the quotient of
two factorials.
13. 9
14. 12
15. 6 7 8
16. 9 10 11 12
41. (A) Find the largest term of the sequence a0, a1, a2,
. . . , a10 to three decimal places, where
10
ak (0.6)10k(0.4)k.
k
(B) According to the binomial formula, what is the sum of
the series a0 a1 a2 . . . a10?
42. (A) Find the largest term of the sequence a0, a1, a2,
. . . , a10 to three decimal places, where
10
ak (0.3)10k(0.7)k.
k
(B) According to the binomial formula, what is the sum of
the series a0 a1 a2 . . . a10?
B
Evaluate each expression in Problems 17–22.
139
16
20. 8
17.
125
100
21. 97 18.
147
100
22. 3 19.
23. Find the smallest positive integer n such that n! produces an
overflow error on your calculator.
24. Find the smallest positive integer n such that
2n
pron
duces an overflow error on your calculator.
Expand Problems 25–30 using the binomial formula.
25. (2x 3y)3
26. (3u 2v)3
27. (x 2)4
28. (3p q)4
29. (2x y)5
30. (2x y)6
In Problems 31–38, find the indicated term in each
expansion.
31. (u v)15; seventh term
32. (a b)12; fifth term
33. (2m n)10; ninth term
34. (x 3y)13; third term
35. (w 3)20; fifth term
36. (2x 5y)8; sixth term
37. (3x 2y)8; sixth term
38. (2p 3q)7; fourth term
In Problems 39–42, use a graphing utility to graph each
sequence and to display it in table form.
39. Find the number of terms of the sequence
0 , 1 , 2 , . . . , 20
20
20
20
20
that are greater than one-half of the largest term.
40. Find the number of terms of the sequence
400, 401, 402, . . . , 4040
that are greater than one-half of the largest term.
C
43. Evaluate (1.01)10 to four decimal places, using the binomial
formula. [Hint: Let 1.01 1 0.01.]
44. Evaluate (0.99)6 to four decimal places, using the binomial
formula.
In Problems 45–48, determine whether the statement is true
or false. If true, explain why. If false, give a counterexample.
45.
nr n n r
46. If the positive integer n is divisible by a prime p, then
is divisible by p for 1 r n 1.
nr
pr is divisible by p for 1 r p 1.
2p
48. If p is an odd prime, then is not divisible by p.
p
k
k
k1
49. Show that: r1
r
r k
k1
50. Show that: 0
0 k
k1
51. Show that: k
k 1
n
52. Show that is given by the recursion formula
r
nr n rr 1 r n 1
n
where 1.
0
47. If p is a prime, then
53. Write 2n (1 1)n and expand, using the binomial formula to obtain
2n 0 1 2 . . . n
n
n
n
n
8-5
Multiplication Principle, Permutations, and Combinations
589
1
54. Can you guess what the next two rows in Pascal’s triangle,
shown at right, are? Compare the numbers in the triangle
with the binomial coefficients obtained with the binomial
formula.
1 1
1 2 1
1 3 3 1
1 4 6 4 1
SECTION
8-5
Multiplication Principle, Permutations,
and Combinations
• Multiplication Principle
• Permutations
• Combinations
We may expand the binomial form (a b)n in two steps: first, expand into a sum of
2n terms, each with coefficient 1; second, group together those terms in which b
appears to the same power, obtaining the sum of the n 1 terms of the binomial formula. For example,
(a b)3 (a b)(a b)2 (a b)(aa ab ba bb)
aaa aab aba abb baa bab bba bbb
a 3a b 3ab b
3
2
2
3
Step 1
Step 2
Consider the term aba of step 1: The first a comes from the first factor of a b, the
b comes from the second factor of a b, and the final a from the third factor. There3
fore,
3, the coefficient of a2b in step 2, is the number of ways of choosing
1
b from exactly one of the three factors of a b in (a b)3.
52
In the same way,
2,598,960 is the number of ways of choosing b from
5
exactly five of the 52 factors of a b in (a b)52. Analogously, 2,598,960 is
the number of 5-card hands which can be chosen from a standard 52-card deck. In
this section we study such counting techniques that are related to the sequence
n
n
n
n
,
,
,...,
, and we develop important counting tools that form the foun0
1
2
n
dation of probability theory.
• Multiplication
We start with an example.
Principle
EXAMPLE 1
Combined Outcomes
Suppose we flip a coin and then throw a single die (see Fig. 1). What are the possible combined outcomes?