CHAPTER 5

SEQUENCES,
MATHEMATICAL
INDUCTION, AND
RECURSION

Copyright © Cengage Learning. All rights reserved.

SECTION 5.2

Mathematical Induction I

Copyright © Cengage Learning. All rights reserved.

Mathematical Induction I
Mathematical induction is one of the more recently
developed techniques of proof in the history of
mathematics.

It is used to check conjectures about the outcomes of

processes that occur repeatedly and according to definite
patterns.

In general, mathematical induction is a method for proving

that a property defined for integers n is true for all values of
n that are greater than or equal to some initial integer.

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Mathematical Induction I

The validity of proof by mathematical induction is generally

taken as an axiom. That is why it is referred to as the
principle of mathematical induction rather than as a
theorem.
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Mathematical Induction I
Proving a statement by mathematical induction is a
two-step process. The first step is called the basis step,
and the second step is called the inductive step.

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Mathematical Induction I
The following example shows how to use mathematical
induction to prove a formula for the sum of the first n
integers.

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Example 1 – Sum of the First n Integers
Use mathematical induction to prove that

Solution:
To construct a proof by induction, you must first identify the
property P(n). In this case, P(n) is the equation

[To see that P(n) is a sentence, note that its subject is “the
sum of the integers from 1 to n” and its verb is “equals.”]
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Example 1 – Solution cont’d

In the basis step of the proof, you must show that the
property is true for n = 1, or, in other words that P(1) is true.

Now P(1) is obtained by substituting 1 in place of n in P(n).

The left-hand side of P(1) is the sum of all the successive
integers starting at 1 and ending at 1. This is just 1. Thus
P(1) is

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Example 1 – Solution cont’d

Of course, this equation is true because the right-hand side

is

which equals the left-hand side.

In the inductive step, you assume that P(k) is true, for a

particular but arbitrarily chosen integer k with k ≥ 1. [This
assumption is the inductive hypothesis.]

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Example 1 – Solution cont’d

You must then show that P(k + 1) is true. What are P(k)
and P(k + 1)? P(k) is obtained by substituting k for every n
in P(n).

Thus P(k) is

Similarly, P(k + 1) is obtained by substituting the quantity

(k + 1) for every n that appears in P(n).

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Example 1 – Solution cont’d

Thus P(k + 1) is

or, equivalently,

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Example 1 – Solution cont’d

Now the inductive hypothesis is the supposition that P(k) is

true. How can this supposition be used to show that
P(k + 1) is true? P(k + 1) is an equation, and the truth of an
equation can be shown in a variety of ways.

One of the most straightforward is to use the inductive

hypothesis along with algebra and other known facts to
transform separately the left-hand and right-hand sides
until you see that they are the same.

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Example 1 – Solution cont’d

In this case, the left-hand side of P(k + 1) is

1 + 2 +· · ·+ (k + 1),
which equals
(1 + 2 +· · ·+ k) + (k + 1)

But by substitution from the inductive hypothesis,

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Example 1 – Solution cont’d

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Example 1 – Solution cont’d

So the left-hand side of P(k + 1) is .

Now the right-hand side of P(k + 1) is

by multiplying out the numerator.

Thus the two sides of P(k + 1) are equal to each other, and
so the equation P(k + 1) is true.

This discussion is summarized as follows:

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Example 1 – Solution cont’d

Proof (by mathematical induction):

Let the property P(n) be the equation

Show that P(1) is true:

To establish P(1), we must show that

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Example 1 – Solution cont’d

But the left-hand side of this equation is 1 and the

right-hand side is

also. Hence P(1) is true.

Show that for all integers k ≥ 1, if P(k) is true then
P(k + 1) is also true:

[Suppose that P(k) is true for a particular but arbitrarily

chosen integer k ≥ 1.That is:] Suppose that k is any integer
with k ≥ 1 such that

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Example 1 – Solution cont’d

[We must show that P(k + 1) is true. That is:] We must

show that

or, equivalently, that

[We will show that the left-hand side and the right-hand
side of P(k + 1) are equal to the same quantity and thus are
equal to each other.]

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Example 1 – Solution cont’d

The left-hand side of P(k + 1) is

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Example 1 – Solution cont’d

And the right-hand side of P(k + 1) is

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Example 1 – Solution cont’d

Thus the two sides of P(k + 1) are equal to the same

quantity and so they are equal to each other. Therefore the
equation P(k + 1) is true [as was to be shown].

[Since we have proved both the basis step and the

inductive step, we conclude that the theorem is true.]

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Mathematical Induction I

For example, writing expresses the

sum 1 + 2 + 3 +· · ·+ n in closed form.

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Example 2 – Applying the Formula for the Sum of the First n Integers

a. Evaluate 2 + 4 + 6 +· · ·+ 500.

b. Evaluate 5 + 6 + 7 + 8 +· · ·+ 50.

c. For an integer h ≥ 2, write 1 + 2 + 3 +· · ·+ (h – 1) in

closed form.

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Example 2 – Solution
a.

b.

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Example 2 – Solution cont’d

c.

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Mathematical Induction I
In a geometric sequence, each term is obtained from the
preceding one by multiplying by a constant factor.

If the first term is 1 and the constant factor is r, then the

sequence is 1, r, r 2, r 3, . . . , r n, . . . .

The sum of the first n terms of this sequence is given by the

formula

for all integers n ≥ 0 and real numbers r not equal to 1.

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Mathematical Induction I
The expanded form of the formula is

and because r 0 = 1 and r 1 = r, the formula for n ≥ 1 can be

rewritten as

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Example 3 – Sum of a Geometric Sequence

Prove that , for all integers n ≥ 0 and all real

numbers r except 1.

Solution:
In this example the property P(n) is again an equation,
although in this case it contains a real variable r:

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Example 3 – Solution cont’d

Because r can be any real number other than 1, the proof

begins by supposing that r is a particular but arbitrarily
chosen real number not equal to 1.

Then the proof continues by mathematical induction on n,

starting with n = 0.

In the basis step, you must show that P(0) is true; that is,
you show the property is true for n = 0.

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Example 3 – Solution cont’d

So you substitute 0 for each n in P(n):

In the inductive step, you suppose k is any integer with

k ≥ 0 for which P(k) is true; that is, you suppose the
property is true for n = k.

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Example 3 – Solution cont’d

So you substitute k for each n in P(n):

Then you show that P(k + 1) is true; that is, you show the
property is true for n = k + 1.

So you substitute k + 1 for each n in P(n):

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Example 3 – Solution cont’d

Or, equivalently,

In the inductive step for this proof we use another common

technique for showing that an equation is true:

We start with the left-hand side and transform it

step-by-step into the right-hand side using the inductive
hypothesis together with algebra and other known facts.

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Example 3 – Solution cont’d

Proof (by mathematical induction):

Suppose r is a particular but arbitrarily chosen real number
that is not equal to 1, and let the property P(n) be the
equation

We must show that P(n) is true for all integers n ≥ 0. We do

this by mathematical induction on n. 33
Example 3 – Solution cont’d

Show that P(0) is true:

To establish P(0), we must show that

The left-hand side of this equation is r 0 = 1 and the

right-hand side is

also because r 1 = r and r ≠ 1. Hence P(0) is true.

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Example 3 – Solution cont’d

Show that for all integers k ≥ 0, if P(k) is true then

P(k + 1) is also true:
[Suppose that P(k) is true for a particular but arbitrarily
chosen integer k ≥ 0. That is:]
Let k be any integer with k ≥ 0, and suppose that

[We must show that P(k + 1) is true. That is:] We must

show that

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Example 3 – Solution cont’d

Or, equivalently, that

[We will show that the left-hand side of P(k + 1) equals the
right-hand side.] The left-hand side of P(k + 1) is

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Example 3 – Solution cont’d

which is the right-hand side of P(k + 1) [as was to be

shown.]
[Since we have proved the basis step and the inductive
step, we conclude that the theorem is true.]
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Proving an Equality

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Proving an Equality
The proofs of the basis and inductive steps in Examples 1
and 3 illustrate two different ways to show that an equation
is true:

(1) transforming the left-hand side and the right-hand side

independently until they are seen to be equal, and
(2) transforming one side of the equation until it is seen to
be the same as the other side of the equation.

Sometimes people use a method that they believe proves

equality but that is actually invalid.

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Proving an Equality
For example, to prove the basis step for Theorem 5.2.3,
they perform the following steps:

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Proving an Equality
The problem with this method is that starting from a
statement and deducing a true conclusion does not prove
that the statement is true.

A true conclusion can also be deduced from a false

statement. For instance, the steps below show how to
deduce the true conclusion that 1 = 1 from the false
statement that 1 = 0:

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Proving an Equality

When using mathematical induction to prove formulas, be

sure to use a method that avoids invalid reasoning, both for
the basis step and for the inductive step.

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Deducing Additional Formulas

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Deducing Additional Formulas
The formula for the sum of a geometric sequence can be
thought of as a family of different formulas in r, one for
each real number r except 1.

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Example 4 – Applying the Formula for the Sum of a Geometric Sequence

In each of (a) and (b) below, assume that m is an integer

that is greater than or equal to 3. Write each of the sums in
closed form.
a.
b.

Solution:

a.

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Example 4 – Solution cont’d

b.

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Deducing Additional Formulas
As with the formula for the sum of the first n integers, there
is a way to think of the formula for the sum of the terms of a
geometric sequence that makes it seem simple and
intuitive. Let

Then

and so

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Deducing Additional Formulas
But

Equating the right-hand sides of equations (5.2.1) and

(5.2.2) and dividing by r – 1 gives

This derivation of the formula is attractive and is quite

convincing. However, it is not as logically airtight as the
proof by mathematical induction.

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Deducing Additional Formulas
To go from one step to another in the previous calculations,
the argument is made that each term among those
indicated by the ellipsis (. . .) has such-and-such an
appearance and when these are canceled such-and-such
occurs.

But it is impossible actually to see each such term and

each such calculation, and so the accuracy of these claims
cannot be fully checked.

With mathematical induction it is possible to focus exactly

on what happens in the middle of the ellipsis and verify
without doubt that the calculations are correct.
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