A-BLTZMC11_967-1052-hr
8-10-2008
14:59
Page 1003
Section 11.4 Mathematical Induction
8. Find the sum of the first ten terms of the sequence:
5, 10, 20, 40, Á .
9. Find the sum of the first 50 terms of the sequence:
16. Express 0.45 as a fraction in lowest terms.
17. Express the sum using summation notation. Use i for the
index of summation.
1
2
3
18
+ + + Á +
3
4
5
20
- 2, 0, 2, 4, Á .
10. Find the sum of the first ten terms of the sequence:
- 20, 40, - 80, 160, Á .
11. Find the sum of the first 100 terms of the sequence:
4, - 2, - 8, -14, Á .
In Exercises 12–15, find each indicated sum.
4
50
12. a 1i + 421i - 12
13. a 13i - 22
6
3 i
14. a a b
i=1 2
2 i-1
15. a a - b
5
i=1
i=1
18. A skydiver falls 16 feet during the first second of a dive,
48 feet during the second second, 80 feet during the third
second, 112 feet during the fourth second, and so on. Find the
distance that the skydiver falls during the 15th second and
the total distance the skydiver falls in 15 seconds.
19. If the average value of a house increases 10% per year, how
much will a house costing $120,000 be worth in 10 years?
Round to the nearest dollar.
q
i=1
Section
11.4 Mathematical Induction
Objectives
�
�
1003
After ten years of work, Princeton
University’s Andrew Wiles proved
Fermat’s Last Theorem.
Understand the principle of
mathematical induction.
Prove statements using
mathematical induction.
P
ierre de Fermat (1601–1665)
was a lawyer who enjoyed
studying mathematics. In a
margin of one of his books, he
claimed that no positive integers
satisfy
x n + y n = zn
if n is an integer greater than or equal to 3.
If n = 2, we can find positive integers satisfying xn + yn = zn, or x 2 + y2 = z2:
32 + 4 2 = 52.
However, Fermat claimed that no positive integers satisfy
x3 + y3 = z3,
x4 + y4 = z4,
x5 + y5 = z5,
and so on. Fermat claimed to have a proof of his conjecture, but added, “The margin
of my book is too narrow to write it down.” Some believe that he never had a proof
and intended to frustrate his colleagues.
In 1994, 40-year-old Princeton math professor Andrew Wiles proved Fermat’s
Last Theorem using a principle called mathematical induction. In this section, you
will learn how to use this powerful method to prove statements about the positive
integers.
�
Understand the principle
of mathematical induction.
The Principle of Mathematical Induction
How do we prove statements using mathematical induction? Let’s consider an
example. We will prove a statement that appears to give a correct formula for the
sum of the first n positive integers:
Sn : 1 + 2 + 3 + Á + n =
n1n + 12
2
.
A-BLTZMC11_967-1052-hr
8-10-2008
14:59
Page 1004
1004 Chapter 11 Sequences, Induction, and Probability
We can verify Sn: 1 + 2 + 3 + Á + n =
n1n + 12
2
for, say, the first four positive
integers. If n = 1, the statement S1 is
Take the first
term on the left.
?
1=
1(1+1)
2
Substitute 1 for
n on the right.
1#2
1ⱨ
2
1 = 1 ✓. This true statement shows
that S1 is true.
If n = 2, the statement S2 is
Add the first two
terms on the left.
?
1+2=
2(2+1)
2
Substitute 2 for
n on the right.
2#3
3ⱨ
2
3 = 3 ✓. This true statement shows
that S2 is true.
If n = 3, the statement S3 is
Add the first three
terms on the left.
?
1+2+3=
3(3+1)
2
Substitute 3 for
n on the right.
3#4
6ⱨ
2
6 = 6 ✓. This true statement shows
that S3 is true.
Finally, if n = 4, the statement S4 is
Add the first four
terms on the left.
?
1+2+3+4=
4(4+1)
2
Substitute 4 for
n on the right.
4#5
10 ⱨ
2
10 = 10 ✓. This true statement shows
that S4 is true.
This approach does not prove that the given statement Sn is true for every
positive integer n. The fact that the formula produces true statements for
n = 1, 2, 3, and 4 does not guarantee that it is valid for all positive integers n. Thus,
we need to be able to verify the truth of Sn without verifying the statement for each
and every one of the positive integers.
A legitimate proof of the given statement Sn involves a technique called
mathematical induction.
A-BLTZMC11_967-1052-hr
8-10-2008
14:59
Page 1005
Section 11.4 Mathematical Induction
1005
The Principle of Mathematical Induction
Let Sn be a statement involving the positive integer n. If
1. S1 is true, and
2. the truth of the statement Sk implies the truth of the statement Sk + 1 , for
every positive integer k,
then the statement Sn is true for all positive integers n.
The principle of mathematical induction can be illustrated using an unending
line of dominoes, as shown in Figure 11.7. If the first domino is pushed over, it
knocks down the next, which knocks down the next, and so on, in a chain reaction.
To topple all the dominoes in the infinite sequence, two conditions must be satisfied:
Figure 11.7 Falling dominoes
illustrate the principle of
mathematical induction.
1. The first domino must be knocked down.
2. If the domino in position k is knocked down, then the domino in position
k + 1 must be knocked down.
If the second condition is not satisfied, it does not follow that all the dominoes will
topple. For example, suppose the dominoes are spaced far enough apart so that a
falling domino does not push over the next domino in the line.
The domino analogy provides the two steps that are required in a proof by
mathematical induction.
The Steps in a Proof by Mathematical Induction
Let Sn be a statement involving the positive integer n. To prove that Sn is true for
all positive integers n requires two steps.
Step 1
Show that S1 is true.
Step 2 Show that if Sk is assumed to be true, then Sk + 1 is also true, for every
positive integer k.
Notice that to prove Sn , we work only with the statements S1 , Sk , and Sk + 1 .
Our first example provides practice in writing these statements.
EXAMPLE 1
Writing S1 , Sk , and Skⴙ1
For the given statement Sn , write the three statements S1 , Sk , and Sk + 1 .
a. Sn: 1 + 2 + 3 + Á + n =
n1n + 12
b. Sn: 12 + 2 2 + 32 + Á + n2 =
2
n1n + 1212n + 12
6
Solution
a. We begin with
Sn: 1 + 2 + 3 + Á + n =
n1n + 12
2
.
Write S1 by taking the first term on the left and replacing n with 1 on the right.
S1: 1 =
111 + 12
2
A-BLTZMC11_967-1052-hr
8-10-2008
14:59
Page 1006
1006 Chapter 11 Sequences, Induction, and Probability
Sn: 1 + 2 + 3 + Á + n =
n1n + 12
2
Write Sk by taking the sum of the first k terms on the left and replacing n with
k on the right.
The statement for part (a)
(repeated)
Sk: 1 + 2 + 3 + Á + k =
k1k + 12
2
Write Sk + 1 by taking the sum of the first k + 1 terms on the left and replacing
n with k + 1 on the right.
Sk + 1 : 1 + 2 + 3 + Á + 1k + 12 =
Sk + 1 : 1 + 2 + 3 + Á + 1k + 12 =
1k + 1231k + 12 + 14
2
1k + 121k + 22
Simplify on the
right.
2
b. We begin with
Sn: 12 + 2 2 + 32 + Á + n2 =
n1n + 1212n + 12
6
.
Write S1 by taking the first term on the left and replacing n with 1 on the right.
S1 : 12 =
111 + 1212 # 1 + 12
6
n1n + 1212n + 12
Using Sn: 12 + 2 2 + 32 + Á + n2 =
, we write Sk by taking
6
the sum of the first k terms on the left and replacing n with k on the right.
Sk : 12 + 2 2 + 32 + Á + k2 =
k1k + 1212k + 12
6
Write Sk + 1 by taking the sum of the first k + 1 terms on the left and replacing
n with k + 1 on the right.
Sk + 1: 12 + 2 2 + 32 + Á + 1k + 122 =
Sk + 1 : 12 + 2 2 + 32 + Á + 1k + 122 =
Check Point
1
1k + 1231k + 12 + 14321k + 12 + 14
6
1k + 121k + 2212k + 32
6
Simplify on the
right.
For the given statement Sn , write the three statements S1 , Sk ,
and Sk + 1 .
a. 2 + 4 + 6 + Á + 2n = n1n + 12
b. 13 + 2 3 + 33 + Á + n3 =
n21n + 122
4
Always simplify Sk + 1 before trying to use mathematical induction to prove
that Sn is true. For example, consider
Sn: 12 + 32 + 52 + Á + 12n - 122 =
n12n - 1212n + 12
3
.
Begin by writing Sk + 1 as follows:
Sk±1: 12+32+52+. . .+[2(k+1)-1]2
The sum of the
first k + 1 terms
=
(k+1)[2(k+1)-1][2(k+1)+1]
.
3
Replace n with k + 1 on
the right side of Sn.
A-BLTZMC11_967-1052-hr
8-10-2008
14:59
Page 1007
Section 11.4 Mathematical Induction
Now simplify both sides of the equation.
Sk + 1 : 12 + 32 + 52 + Á + 12k + 2 - 122 =
Prove statements using
mathematical induction.
1k + 1212k + 2 - 1212k + 2 + 12
3
1k + 1212k + 1212k + 32
Sk + 1: 12 + 32 + 52 + Á + 12k + 122 =
�
1007
3
Proving Statements about Positive Integers
Using Mathematical Induction
Now that we know how to find S1 , Sk , and Sk + 1 , let’s see how we can use these
statements to carry out the two steps in a proof by mathematical induction. In
Examples 2 and 3, we will use the statements S1 , Sk , and Sk + 1 to prove each of the
statements Sn that we worked with in Example 1.
EXAMPLE 2
Proving a Formula by Mathematical Induction
Use mathematical induction to prove that
n1n + 12
1 + 2 + 3 + Á + n =
2
for all positive integers n.
Solution
Step 1 Show that S1 is true. Statement S1 is
1 =
111 + 12
2
.
Simplifying on the right, we obtain 1 = 1. This true statement shows that S1 is true.
Step 2 Show that if Sk is true, then Skⴙ1 is true. Using Sk and Sk + 1 from Example 1(a),
show that the truth of Sk ,
1 + 2 + 3 + Á + k =
implies the truth of Sk + 1 ,
1 + 2 + 3 + Á + 1k + 12 =
k1k + 12
2
,
1k + 121k + 22
2
.
We will work with Sk . Because we assume that Sk is true, we add the next consecutive
integer after k—namely, k + 1—to both sides.
1 + 2 + 3 + Á + k =
k1k + 12
2
k(k+1)
1+2+3+. . . +k+(k+1)=
+(k+1)
2
We do not have to write this k
because k is understood to be
the integer that precedes k + 1.
1 + 2 + 3 + Á + 1k + 12 =
1 + 2 + 3 + Á + 1k + 12 =
1 + 2 + 3 + Á + 1k + 12 =
k1k + 12
2
1k + 12
2
21k + 12
+
1k + 22
1k + 121k + 22
2
2
This is Sk , which we
assume is true.
Add k + 1 to both
sides of the
equation.
Write the right side
with a common
denominator of 2.
Factor out the
common factor k + 1
2
on the right.
This final result is the
statement Sk + 1 .
A-BLTZMC11_967-1052-hr
8-10-2008
14:59
Page 1008
1008 Chapter 11 Sequences, Induction, and Probability
We have shown that if we assume that Sk is true and we add k + 1 to both sides of
Sk , then Sk + 1 is also true. By the principle of mathematical induction, the statement
Sn , namely,
1 + 2 + 3 + Á + n =
n1n + 12
2
,
is true for every positive integer n.
2
Check Point
Use mathematical induction to prove that
2 + 4 + 6 + Á + 2n = n1n + 12
for all positive integers n.
Proving a Formula by Mathematical Induction
EXAMPLE 3
Use mathematical induction to prove that
12 + 2 2 + 32 + Á + n2 =
n1n + 1212n + 12
6
for all positive integers n.
Solution
Step 1 Show that S1 is true. Statement S1 is
12 =
111 + 1212 # 1 + 12
6
.
1#2#3
. Further simplification on the right gives the
6
statement 1 = 1. This true statement shows that S1 is true.
Simplifying, we obtain 1 =
Step 2 Show that if Sk is true, then Skⴙ1 is true. Using Sk and Sk + 1 from
Example 1(b), show that the truth of
k1k + 1212k + 12
Sk: 12 + 2 2 + 32 + Á + k2 =
6
implies the truth of
Sk + 1 : 12 + 2 2 + 32 + Á + 1k + 122 =
1k + 121k + 2212k + 32
6
.
We will work with Sk . Because we assume that Sk is true, we add the square of the
next consecutive integer after k, namely, 1k + 122, to both sides of the equation.
12 + 2 2 + 32 + Á + k2 =
k1k + 1212k + 12
6
12 + 2 2 + 32 + Á + k2 + 1k + 122 =
k1k + 1212k + 12
1 + 2 + 3 + Á + 1k + 122 =
k1k + 1212k + 12
2
2
2
This is Sk , assumed to be true. We must
work with this and show Sk + 1 is true.
6
=
1k + 12
=
1k + 12
6
6
6
+ 1k + 122
61k + 122
+
6
Add 1k + 122 to both sides.
It is not necessary to write k2 on the
left. Express the right side with the least
common denominator, 6.
3k12k + 12 + 61k + 124
Factor out the common factor
12k2 + 7k + 62
Multiply and combine like terms.
k + 1
.
6
A-BLTZMC11_967-1052-hr
8-10-2008
14:59
Page 1009
Section 11.4 Mathematical Induction
=
1k + 12
=
1k + 121k + 2212k + 32
6
1k + 2212k + 32
6
1009
Factor 2k2 + 7k + 6.
This final statement is Sk + 1 .
We have shown that if we assume that Sk is true, and we add 1k + 122 to both sides of
Sk , then Sk + 1 is also true. By the principle of mathematical induction, the statement
Sn , namely,
12 + 2 2 + 32 + Á + n2 =
n1n + 1212n + 12
6
,
is true for every positive integer n.
Check Point
3
Use mathematical induction to prove that
3
3
3
1 + 2 + 3 + Á + n3 =
n21n + 122
4
for all positive integers n.
Example 4 illustrates how mathematical induction can be used to prove
statements about positive integers that do not involve sums.
EXAMPLE 4
Using the Principle of Mathematical Induction
Prove that 2 is a factor of n2 + 5n for all positive integers n.
Solution
Step 1 Show that S1 is true. Statement S1 reads
2 is a factor of 12 + 5 # 1.
Simplifying the arithmetic, the statement reads
2 is a factor of 6.
This statement is true: that is, 6 = 2 # 3. This shows that S1 is true.
Step 2 Show that if Sk is true, then Skⴙ1 is true. Let’s write Sk and Sk + 1:
Sk :
Sk + 1 :
2 is a factor of k2 + 5k.
2 is a factor of 1k + 122 + 51k + 12.
We can rewrite statement Sk + 1 by simplifying the algebraic expression in the statement
as follows:
(k+1)2+5(k+1)=k2+2k+1+5k+5=k2+7k+6.
Use the formula (A + B)2 = A2 + 2AB + B2.
Statement Sk + 1 now reads
2 is a factor of k2 + 7k + 6.
We need to use statement Sk—that is, 2 is a factor of k2 + 5k—to prove statement
Sk + 1 . We do this as follows:
k2+7k+6=(k2+5k)+(2k+6)=(k2+5k)+2(k+3).
We assume that 2 is
a factor of k2 + 5k because
we assume Sk is true.
Factoring the last two
terms shows that 2 is
a factor of 2k + 6.
A-BLTZMC11_967-1052-hr
8-10-2008
14:59
Page 1010
1010 Chapter 11 Sequences, Induction, and Probability
k2+7k+6=(k2+5k)+(2k+6)=(k2+5k)+2(k+3)
We assume that 2 is
a factor of k2 + 5k because
we assume Sk is true.
Factoring the last two
terms shows that 2 is
a factor of 2k + 6.
We’ve repeated the equation from the bottom of the previous page. The voice
balloons show that 2 is a factor of k2 + 5k and of 21k + 32. Thus, if Sk is true, 2 is a
factor of the sum 1k2 + 5k2 + 21k + 32, or of k2 + 7k + 6. This is precisely
statement Sk + 1 . We have shown that if we assume that Sk is true, then Sk + 1 is also
true. By the principle of mathematical induction, the statement Sn , namely 2 is a
factor of n2 + 5n, is true for every positive integer n.
Check Point
4
Prove that 2 is a factor of n2 + n for all positive integers n.
Exercise Set 11.4
Practice Exercises
In Exercises 1–4, a statement Sn about the positive integers is given.
Write statements S1 , S2 , and S3 , and show that each of these
statements is true.
1. Sn: 1 + 3 + 5 + Á + 12n - 12 = n2
3n - 1
18. 1 + 3 + 32 + Á + 3n - 1 =
2
19. 2 + 4 + 8 + Á + 2 n = 2 n + 1 - 2
20.
1
1
1
1
1
+ + + Á + n = 1 - n
2
4
8
2
2
n1n + 52
2. Sn: 3 + 4 + 5 + Á + 1n + 22 =
2
21. 1 # 2 + 2 # 3 + 3 # 4 + Á + n1n + 12 =
n1n + 121n + 22
3
3. Sn: 2 is a factor of n2 - n.
4. Sn: 3 is a factor of n3 - n.
22. 1 # 3 + 2 # 4 + 3 # 5 + Á + n1n + 22 =
In Exercises 5–10, a statement Sn about the positive integers is given.
Write statements Sk and Sk + 1 , simplifying statement Sk + 1 completely.
n1n + 1212n + 72
6
23.
1
1
1
n
1
+ # + # + Á +
=
1#2
2 3
3 4
n1n + 12
n + 1
24.
1
1
1
1
n
+ # + # + Á +
=
2#3
3 4
4 5
1n + 121n + 22
2n + 4
5. Sn: 4 + 8 + 12 + Á + 4n = 2n1n + 12
n1n + 52
6. Sn: 3 + 4 + 5 + Á + 1n + 22 =
2
Á
+ 14n - 12 = n12n + 12
7. Sn: 3 + 7 + 11 +
8. Sn: 2 + 7 + 12 + Á + 15n - 32 =
n15n - 12
2
2
9. Sn: 2 is a factor of n - n + 2.
Practice Plus
In Exercises 25–34, use mathematical induction to prove that each
statement is true for every positive integer n.
25. 2 is a factor of n2 - n.
10. Sn: 2 is a factor of n2 - n.
26. 2 is a factor of n2 + 3n.
In Exercises 11–24, use mathematical induction to prove that each
statement is true for every positive integer n.
27. 6 is a factor of n1n + 121n + 22.
11. 4 + 8 + 12 + Á + 4n = 2n1n + 12
28. 3 is a factor of n1n + 121n - 12.
12. 3 + 4 + 5 + Á + 1n + 22 =
n1n + 52
13. 1 + 3 + 5 + Á + 12n - 12 = n2
14. 3 + 6 + 9 + Á + 3n =
i=1
2
n
30. a 7 # 8i = 818n - 12
i=1
3n1n + 12
31. n + 2 7 n
2
15. 3 + 7 + 11 + Á + 14n - 12 = n12n + 12
16. 2 + 7 + 12 + Á + 15n - 32 =
n15n - 12
17. 1 + 2 + 2 + Á + 2 n - 1 = 2 n - 1
2
n
29. a 5 # 6i = 616n - 12
2
32. If 0 6 x 6 1, then 0 6 xn 6 1.
33. 1ab2n = anbn
a n an
34. a b = n
b
b
A-BLTZMC11_967-1052-hr
8-10-2008
14:59
Page 1011
Section 11.4 Mathematical Induction
Writing in Mathematics
35. Explain how to use mathematical induction to prove that a
statement is true for every positive integer n.
36. Consider the statement Sn given by
n2 - n + 41 is prime.
Although S1 , S2 , Á , S40 are true, S41 is false. Verify that S41 is
false. Then describe how this is illustrated by the dominoes in
the figure. What does this tell you about a pattern, or
formula, that seems to work for several values of n?
Some statements are false for the first few positive integers, but true
for some positive integer m on. In these instances, you can prove Sn
for n Ú m by showing that Sm is true and that Sk implies Sk + 1 when
k 7 m. Use this extended principle of mathematical induction to
prove that each statement in Exercises 41–42 is true.
41. Prove that n2 7 2n + 1 for n Ú 3. Show that the formula is
true for n = 3 and then use step 2 of mathematical induction.
42. Prove that 2 n 7 n2 for n Ú 5. Show that the formula is true
for n = 5 and then use step 2 of mathematical induction.
In Exercises 43–44, find S1 through S5 and then use the pattern to
make a conjecture about Sn . Prove the conjectured formula for
Sn by mathematical induction.
43. Sn:
S35
S36
S37
S38
S39
S41
S40
S42
1011
1
1
1
1
+
+
+ Á +
= ?
4
12
24
2n1n + 12
44. Sn: a 1 -
1
1
1
1
b a1 - b a 1 - b Á a 1 b = ?
2
3
4
n + 1
Group Exercise
Critical Thinking Exercises
Make Sense? In Exercises 37–40, determine whether each
statement makes sense or does not make sense, and explain
your reasoning.
37. I use mathematical induction to prove that statements are
true for all real numbers n.
38. I begin proofs by mathematical induction by writing Sk and
Sk + 1 , both of which I assume to be true.
39. When a line of falling dominoes is used to illustrate the
principle of mathematical induction, it is not necessary for all
the dominoes to topple.
40. This triangular arrangement of 36 circles illustrates that
1 + 2 + 3 + Á + n =
is true for n = 8.
n1n + 12
2
45. Fermat’s most notorious theorem, described in the section
opener on page 1003, baffled the greatest minds for more
than three centuries. In 1994, after ten years of work,
Princeton University’s Andrew Wiles proved Fermat’s
Last Theorem. People magazine put him on its list of “the
25 most intriguing people of the year,” the Gap asked him
to model jeans, and Barbara Walters chased him for an
interview. “Who’s Barbara Walters?” asked the bookish
Wiles, who had somehow gone through life without a
television.
Using the 1993 PBS documentary “Solving Fermat:
Andrew Wiles” or information about Andrew Wiles on the
Internet, research and present a group seminar on what
Wiles did to prove Fermat’s Last Theorem, problems along
the way, and the role of mathematical induction in the proof.
Preview Exercises
Exercises 46–48 will help you prepare for the material covered in
the next section. Each exercise involves observing a pattern in the
expanded form of the binomial expression 1a + b2n.
1a + b21 = a + b
1a + b22 = a2 + 2ab + b2
1a + b23 = a3 + 3a2 b + 3ab2 + b3
1a + b24 = a4 + 4a3 b + 6a2 b2 + 4ab3 + b4
1a + b25 = a5 + 5a4 b + 10a3 b2 + 10a2b3 + 5ab4 + b5
46. Describe the pattern for the exponents on a.
47. Describe the pattern for the exponents on b.
48. Describe the pattern for the sum of the exponents on the
variables in each term.