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B. Mathematical Induction Applied to Sums

# 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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Section 9.4 Mathematical Induction

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.

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