B. Mathematical Induction Applied to Sums
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797
add next term akϩ1
>
a1 ϩ a2 ϩ a3 ϩ p ϩ akϪ1 ϩ ak ϩ akϩ1 ϩ p ϩ an–1 ϩ an
c
c
sum of k terms
Sk
sum of k ϩ 1 terms
Skϩ1
Now, let’s return to the sum 1 ϩ 3 ϩ 5 ϩ 7 ϩ p ϩ 12n Ϫ 12. This is an arithmetic series with a1 ϭ 1, d ϭ 2, and nth term an ϭ 2n Ϫ 1. Using the sum formula
for an arithmetic sequence, an alternative formula for this sum can be established.
Sn ϭ
ϭ
ϭ
n1a1 ϩ an 2
2
n11 ϩ 2n ؊ 12
2
No matter how distant the city or
how many relay stations are
involved, if the generating plant is
working and the kth station relays
to the (k ϩ 1)st station, the city will
get its power.
substitute 1 for a1 and 2n؊1 for an
n12n2
ϭ n2
WORTHY OF NOTE
summation formula for an arithmetic sequence
2
simplify
result
This shows that the sum of the first n positive odd integers is given by Sn ϭ n2. As a
check we compute S5 ϭ 1 ϩ 3 ϩ 5 ϩ 7 ϩ 9 ϭ 25 and compare to S5 ϭ 52 ϭ 25✓.
We also note S6 ϭ 62 ϭ 36, and S5 ϩ a6 ϭ 25 ϩ 11 ϭ 36, showing S6 ϭ S5 ϩ a6.
For more on this relationship, see Exercises 19 through 24. While it may seem
simplistic now, showing S5 ϩ a6 ϭ S6 and Sk ϩ akϩ1 ϭ Skϩ1 (in general) is a critical
component of a proof by induction.
Unfortunately, general summation formulas for many sequences cannot be established from known formulas. In addition, just because a formula works for the first few
values of n, we cannot assume that it will hold true for all values of n (there are infinitely many). As an illustration, the formula an ϭ n2 Ϫ n ϩ 41 yields a prime number
for every natural number n from 1 to 40, but fails to yield a prime for n ϭ 41. This
helps demonstrate the need for a more conclusive proof, particularly when a relationship appears to be true, and can be “verified” in a finite number of cases, but whether
it is true in all cases remains in doubt.
Proof by induction is based
on a relatively simple idea. To
help understand how it works,
consider n relay stations that are
used to transport electricity from a
generating plant to a distant city.
If we know the generating plant is
operating, and if we assume that
the kth relay station (any station
(k ϩ 1)st
kth
Generating plant
in the series) is making the transrelay
relay
fer to the 1k ϩ 12st station (the
next station in the series), then we’re sure the city will have electricity.
This idea can be applied mathematically as follows. Consider the statement, “The
sum of the first n positive even integers is n2 ϩ n.” In other words,
2 ϩ 4 ϩ 6 ϩ 8 ϩ p ϩ 2n ϭ n2 ϩ n. We can certainly verify the statement for the
first few even numbers:
112 2 ϩ 1 ϭ 2
The first even number is 2 and p
122 2 ϩ 2 ϭ 6
The sum of the first two even numbers is 2 ϩ 4 ϭ 6 and p
The sum of the first three even numbers is
132 2 ϩ 3 ϭ 12
2 ϩ 4 ϩ 6 ϭ 12 and p
The sum of the first four even numbers is
2 ϩ 4 ϩ 6 ϩ 8 ϭ 20 and p
142 2 ϩ 4 ϭ 20
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While we could continue this process for a very long time (or even use a computer),
no finite number of checks can prove a statement is universally true. To prove the statement true for all positive integers, we use a reasoning similar to that applied in the relay
stations example. If we are sure the formula works for n ϭ 1 (the generating station is operating), and if the truth of n ϭ k implies that n ϭ k ϩ 1 is true [the kth relay station is
transferring electricity to the 1k ϩ 12st station], then the statement is true for all n (the city
will get its electricity). The case where n ϭ 1 is called the base case of an inductive proof,
and the assumption that the formula is true for n ϭ k is called the induction hypothesis.
When the induction hypothesis is applied to a sum formula, we attempt to show that
Sk ϩ akϩ1 ϭ Skϩ1. Since k and k ϩ 1 are arbitrary, the statement must be true for all n.
Mathematical Induction Applied to Sums
Let Sn be a sum formula involving positive integers.
If
1. S1 is true, and
2. the truth of Sk implies that Skϩ1 is true,
then Sn must be true for all positive integers n.
WORTHY OF NOTE
To satisfy our finite minds, it might
help to show that Sn is true for the
first few cases, prior to extending
the ideas to the infinite case.
EXAMPLE 2
ᮣ
Both parts 1 and 2 must be verified for the proof to be complete. Since the process
requires the terms Sk, akϩ1, and Skϩ1, we will usually compute these first.
Proving a Statement Using Mathematical Induction
Use induction to prove that the sum of the first n perfect squares is given by
n1n ϩ 12 12n ϩ 12
.
1 ϩ 4 ϩ 9 ϩ 16 ϩ 25 ϩ p ϩ n2 ϭ
6
Solution
ᮣ
Given an ϭ n2 and Sn ϭ
n1n ϩ 12 12n ϩ 12
6
, the needed components are p
akϩ1 ϭ 1k ϩ 12 2
For an ϭ n2: ak ϭ k2
and
n1n ϩ 12 12n ϩ 12
k1k ϩ 12 12k ϩ 12
1k ϩ 12 1k ϩ 2212k ϩ 32
: Sk ϭ
For Sn ϭ
and Skϩ1 ϭ
6
6
6
1. Show Sn is true for n ϭ 1.
Sn ϭ
S1 ϭ
n1n ϩ 12 12n ϩ 12
1122 132
sum formula
6
base case: n ϭ 1
6
ϭ 1✓
result checks, the first term is 1
2. Assume Sk is true,
1 ϩ 4 ϩ 9 ϩ 16 ϩ p ϩ k2 ϭ
k1k ϩ 1212k ϩ 12
6
and use it to show the truth of Skϩ1 follows. That is,
1k ϩ 121k ϩ 2212k ϩ 32
⎫
⎪
⎪
⎪
⎪
⎪
⎪
⎬
⎪
⎪
⎪
⎪
⎪
⎪
⎭
⎫
⎬
⎭
1 ϩ 4 ϩ 9 ϩ 16 ϩ p ϩ k2 ϩ 1k ϩ 12 2 ϭ
induction hypothesis: Sk is true
6
Sk
akϩ1
Skϩ1
Working with the left-hand side, we have
1 ϩ 4 ϩ 9 ϩ 16 ϩ p ϩ k2 ϩ 1k ϩ 12 2
⎫
⎪
⎪
⎪
⎪
⎪
⎪
⎬
⎪
⎪
⎪
⎪
⎪
⎪
⎭
k1k ϩ 12 12k ϩ 12
ϭ
ϭ
ϩ 1k ϩ 12
6
k1k ϩ 12 12k ϩ 12 ϩ 61k ϩ 12 2
6
2
induction hypothesis: substitute
k1k ϩ 1212k ϩ 12
for 1 ϩ 4 ϩ 9 ϩ 16 ϩ 25 ϩ p ϩ k2
common denominator
6
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ϭ
1k ϩ 12 3 k12k ϩ 12 ϩ 61k ϩ 12 4
factor out k ϩ 1
6
1k ϩ 12 3 2k2 ϩ 7k ϩ 64
ϭ
multiply and combine terms
6
1k ϩ 121k ϩ 2212k ϩ 32
ϭ
799
factor the trinomial, result is Skϩ1
6
Since the truth of Skϩ1 follows from Sk, the formula is true for all n.
B. You’ve just seen how
we can apply the principle of
mathematical induction to
sum formulas involving
natural numbers
Now try Exercises 27 through 38
ᮣ
C. The General Principle of Mathematical Induction
Proof by induction can be used to verify many other kinds of relationships involving a
natural number n. In this regard, the basic principles remain the same but are stated
more broadly. Rather than using Sn to represent a sum, we will use Pn to represent any
proposed statement or relationship we might wish to verify. This broadens the scope
of the proof and makes it more widely applicable, while maintaining its connection to
the sum formulas verified earlier.
The General Principle of Mathematical Induction
Let Pn be a statement involving natural numbers.
If
1. P1 is true, and
2. the truth of Pk implies that Pkϩ1 is also true
then Pn must be true for all natural numbers n.
EXAMPLE 3
ᮣ
Proving a Statement Using the General Principle of Mathematical Induction
Use the general principle of mathematical induction to show the statement Pn is
true for all natural numbers n. Pn: 2n Ն n ϩ 1
Solution
ᮣ
The statement Pn is defined as 2n Ն n ϩ 1. This means that Pk is represented by
2k Ն k ϩ 1 and Pkϩ1 by 2kϩ1 Ն k ϩ 2.
1. Show Pn is true for n ϭ 1:
Pn:
P1:
2n Ն n ϩ 1
21 Ն 1 ϩ 1
2 Ն 2✓
given statement
base case: n ϭ 1
true
Although not a part of the formal proof, a table of values can help to illustrate
the relationship we’re trying to establish. It appears that the statement is true.
n
1
2
3
4
5
2n
2
4
8
16
32
nϩ1
2
3
4
5
6
2. Assume that Pk is true.
Pk:
2k Ն k ϩ 1
induction hypothesis
and use it to show the truth of Pkϩ1. That is,
Pkϩ1:
2kϩ1 Ն 1k ϩ 12 ϩ 1
Նkϩ2
Begin by working with the left-hand side of the inequality, 2kϩ1.
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2kϩ1 ϭ 212k 2
Ն 21k ؉ 12
Ն 2k ϩ 2
properties of exponents
induction hypothesis: substitute k ϩ 1 for 2k
(symbol changes since k ϩ 1 is less than or equal to 2k)
distribute
Since k is a positive integer, 2kϩ1 Ն 2k ϩ 2 Ն k ϩ 2,
showing 2kϩ1 Ն k ϩ 2.
WORTHY OF NOTE
Note there is no reference to an, ak,
or ak+1 in the statement of the
general principle of mathematical
induction.
EXAMPLE 4
Since the truth of Pkϩ1 follows from Pk, the formula is true for all n.
Now try Exercises 39 through 42
ᮣ
ᮣ
Proving Divisibility Using Mathematical Induction
Let Pn be the statement, “4n Ϫ 1 is divisible by 3 for all positive integers n.” Use
mathematical induction to prove that Pn is true.
Solution
ᮣ
If a number is evenly divisible by three, it can be written as the product of 3 and
some positive integer we will call p.
1. Show Pn is true for n ϭ 1:
Pn: 4n Ϫ 1 ϭ 3p
P1: 4112 Ϫ 1 ϭ 3p
3 ϭ 3p ✓
given statement, p ʦ ޚ
substitute 1 for n
statement is true for n ϭ 1
2. Assume that Pk is true.
Pk:
4k Ϫ 1 ϭ 3p
4k ϭ 3p ϩ 1
induction hypothesis
isolate 4k
and use it to show the truth of Pkϩ1. That is,
Pkϩ1:
4kϩ1 Ϫ 1 ϭ 3q for q ʦ ޚis also true.
Beginning with the left-hand side we have:
4kϩ1 Ϫ 1 ϭ 4 # 4k Ϫ 1
ϭ 4 # 13p ؉ 12 Ϫ 1
ϭ 12p ϩ 3
ϭ 314p ϩ 12 ϭ 3q
properties of exponents
induction hypothesis: substitute 3p ϩ 1 for 4k
distribute and simplify
factor
The last step shows 4
Ϫ 1 is divisible by 3. Since the original statement is
true for n ϭ 1, and the truth of Pk implies the truth of Pkϩ1, the statement,
“4n Ϫ 1 is divisible by 3” is true for all positive integers n.
kϩ1
Now try Exercises 43 through 47
C. You’ve just seen how
we can apply the principle of
mathematical induction to
general statements involving
natural numbers
ᮣ
We close this section with some final notes. Although the base step of a proof by
induction seems trivial, both the base step and the induction hypothesis are necessary
1
1
parts of the proof. For example, the statement n 6
is false for n ϭ 1, but true for
3
3n
all other positive integers. Finally, for a fixed natural number p, some statements are
false for all n 6 p, but true for all n Ն p. By modifying the base case to begin at p, we
can use the induction hypothesis to prove the statement is true for all n greater than p.
For example, n 6 13n2 is false for n 6 4, but true for all n Ն 4.