5.4 mathematical induction

Mathematical Induction
Mathematical induction is a method for verifying
infinitely many (related) mathematical statements.

In this section, we use induction to verify formulas
for sums and inequalities.

If we add consecutive odd numbers starting from 1:

1 = 1

1 = 1
1 + 3 = 4

1 = 1
1 + 3 = 4
1 + 3 + 5 = 9

1 = 1
1 + 3 = 4
1 + 3 + 5 = 9
1 + 3 + 5 + 7 = 16

1 = 1
1 + 3 = 4
1 + 3 + 5 = 9
1 + 3 + 5 + 7 = 16
1 + 3 + 5 + 7 + 9 = 25 = 52
5 odd numbers

1 = 1
1 + 3 = 4
1 + 3 + 5 = 9
1 + 3 + 5 + 7 = 16
1 + 3 + 5 + 7 + 9 = 25 = 52
1 + 3 + 5 + 7 + 9 + 11 = 36 = 62
5 odd numbers
6 odd numbers

1 = 1
1 + 3 = 4
1 + 3 + 5 = 9
1 + 3 + 5 + 7 = 16
We observe the following pattern:
the sum of first n odd numbers is n2.
1 + 3 + 5 + 7 + 9 = 25 = 52
1 + 3 + 5 + 7 + 9 + 11 = 36 = 62
5 odd numbers
6 odd numbers

1 = 1
1 + 3 = 4
1 + 3 + 5 = 9
1 + 3 + 5 + 7 = 16
We observe the following pattern:
the sum of first n odd numbers is n2.
Using (2k – 1) for the odd numbers, we summarize
this pattern as "(2k – 1) = n2 for n = 1, 2, 3,... "k=1
n
1 + 3 + 5 + 7 + 9 = 25 = 52
1 + 3 + 5 + 7 + 9 + 11 = 36 = 62
5 odd numbers
6 odd numbers

Two observations about the above formula.

I. The formula actually consists of infinitely many
statements (assertions), one for each number n,

i.e. we claim that all the following statements are true.

S1: The sum of the first one odd number is 12 or that
1 = 12.

1 = 12.
S2: The sum of the first two odd numbers is 22 or that
1 + 3 = 22.

1 = 12.
1 + 3 = 22.
S3: The sum of the first three odd numbers is 32 or that
1 + 3 + 5 = 32.

1 = 12.
1 + 3 = 22.
S3: The sum of the first three odd numbers is 32 or that
1 + 3 + 5 = 32.
..
.
Sn: The sum of the first n odd numbers is n2 or that
1 + 3 + 5 + … + (2n – 1) = n2.
..

II. We may verify the first few statements easily,
but it is impossible to check all of them one by one.

Mathematical induction is a method of verifying all of
them without actually checking every one of them.

If infinitely many dominos are lined up such that

2. they are close enough so the fall of the any domino
would cause next one to fall,

1. the first domino falls,
would cause next one to fall,

would cause next one to fall, then all of them would fall.

(The Domino Principle) Mathematical Induction

Given infinitely many statements S1, S2, S3,.. such that
I. S1 is true,

I. S1 is true,
II. if SN being true would imply that the next statement
SN+1 must also be true,

I. S1 is true,
II. if SN being true would imply that the next statement
SN+1 must also be true,
then all the statements S1, S2, S3,.. are true.

To use induction to verify infinitely many statements
S1, S2, S3,.. ,

S1, S2, S3,.. ,
Example A.
Verify that (2k – 1) = n2 for all natural numbers n.
k=1
n

S1, S2, S3,.. , always
0. declare that an induction argument is used,
Example A.
k=1
n

Example A.
k=1
n
We will use induction. (declaring the methodology)

1. verify the first statement S1 is true,
Example A.
k=1
n

Example A.
k=1
I. Verify its true when n = 1. (i.e S1 is true )
n

Example A.
k=1
n
If n = 1, we've (2k – 1) = 1k=1
1

Example A.
k=1
n
If n = 1, we've (2k – 1) = 1 = 12
.k=1
1

Example A.
k=1
n
If n = 1, we've (2k – 1) = 1 = 12
. Done.k=1
1

2. assume the N'th statement SN is true, use this to
verify that as a consequence, SN+1 must also be true.
Example A.
k=1
n
If n = 1, we've (2k – 1) = 1 = 12
. Done.k=1
1
These steps are established format when invoking the
mathematical induction and they should be followed.

II. Assume the formula works for n = N.

(This means that (2k – 1) = 1 + 3 + .. + (2N – 1) = N2 is true.)k=1
N

N
From this, verify the formula will work for n = N+1.

N
(i.e. to show (2k – 1) = (N+1)2 )
k=1
N+1

N
For n = N+1,
(i.e. to show (2k – 1) = (N+1)2 )
k=1
N+1

N
For n = N+1, the left hand sum is
(2k – 1)k=1
N+1
(i.e. to show (2k – 1) = (N+1)2 )
k=1
N+1

N
(2k – 1) = 1 + 3 + .. + (2N – 1) + [2(N+1) – 1]k=1
N+1
(i.e. to show (2k – 1) = (N+1)2 )
k=1
N+1

N
(2k – 1) = 1 + 3 + .. + (2N – 1) + [2(N+1) – 1]k=1
N+1
by the induction assumation this sum is N2,
(i.e. to show (2k – 1) = (N+1)2 )
k=1
N+1

N
(2k – 1) = 1 + 3 + .. + (2N – 1) + [2(N+1) – 1]k=1
N+1
by the induction assumation this sum is N2, hence
= N2 + [2(N+1) – 1]
(i.e. to show (2k – 1) = (N+1)2 )
k=1
N+1

N
(2k – 1) = 1 + 3 + .. + (2N – 1) + [2(N+1) – 1]k=1
N+1
= N2 + [2(N+1) – 1]
= N2 + 2N + 1
(i.e. to show (2k – 1) = (N+1)2 )
k=1
N+1

N
(2k – 1) = 1 + 3 + .. + (2N – 1) + [2(N+1) – 1]k=1
N+1
= N2 + [2(N+1) – 1]
= N2 + 2N + 1
= (N + 1)2 Done.
(i.e. to show (2k – 1) = (N+1)2 )
k=1
N+1

N
(2k – 1) = 1 + 3 + .. + (2N – 1) + [2(N+1) – 1]k=1
N+1
= N2 + [2(N+1) – 1]
= N2 + 2N + 1
= (N + 1)2 Done.
(i.e. to show (2k – 1) = (N+1)2 )
k=1
N+1
Therefore (2k – 1) = n2 for all natural numbers n.k=1
n

Example B.
Verify that n + 1 < 2n for all numbers n =1, 2, 3,…

Example B.
We use induction to verify this.

I. Verify its true when n = 1.
Example B.

If n = 1,
Example B.

If n = 1, we check that its true that 1 + 1 < 21.
Example B.

Example B.
II. Assume the inequality is true for n = N,

Example B.
II. Assume the inequality is true for n = N, that is,
it's true that N + 1 < 2N ,

Example B.
it's true that N + 1 < 2N , then verify that it has to be
true when n = N + 1,

Example B.
true when n = N + 1, i.e (N + 1) + 1 < 2N+1.

Example B.
true when n = N + 1, i.e (N + 1) + 1 < 2N+1.
When n = N + 1, the left hand side is
(N + 1) + 1

Example B.
true when n = N + 1, i.e (N + 1) + 1 < 2N+1.
(N + 1) + 1
by the induction assumation this is < 2N

Example B.
true when n = N + 1, i.e (N + 1) + 1 < 2N+1.
(N + 1) + 1 < 2N + 1

Example B.
true when n = N + 1, i.e (N + 1) + 1 < 2N+1.
(N + 1) + 1 < 2N + 1 < 2N + 2N

Example B.
true when n = N + 1, i.e (N + 1) + 1 < 2N+1.
(N + 1) + 1 < 2N + 1 < 2N + 2N < 2*2N

Example B.
it's true that N + 1 < 2N, then verify that it has to be
true when n = N + 1, i.e (N + 1) + 1 < 2N+1.
(N + 1) + 1 < 2N + 1 < 2N + 2N < 2*2N < 2N+1. Done.

In mathematics, it’s always the case that after we
observe a certain pattern first, then the induction
method is used to verify our observation.

In other words, only after the pattern of a formula is
observed yet unproven, then the induction method is
deployed to verify the insight.

In other words, only after the pattern of a formula is
observed yet unproven, then the induction method is
deployed to verify the insight.
The induction–arguments themselves do not help us
in spotting the formulas.

Exercise A.
Write the following sums in the  notation.
(Do not find the sums). For example, using k as the index,
1 + 2 + 3 + 4 .. + N is k.
k=1
N
1. Given N, the sum 12 + 22 + 32 + 42 .. + N2
2. Given K, the sum 1(2) + 2(3) + 3(4) + 4(5) .. + K(K+1)
4. Given D, the sum 3(5/2)2 + 3(6/2)2 + 3(7/2)2 + .. + 3(D/2)2
3. Given M, the sum 3(4) + 5(6) .. + (2M – 1)(2M)
5. Given D, the sum
(5/6)2 + (6/7)2 + (7/8)2 + .. + [(2D – 1)/(2D)]2
6. Given T, the sum 2(34) + 2(36) + 2(38) +.. + 2(32T)
7. Given N, the sum
1,000(e0.03) + 1,000(e0.03)2 + 1,000(e0.03)3 .. + 1,000(e0.03)N – 1

C. Verify that each of the following formulas is true for all
positive n using mathematical induction by completing
the following steps:
a. Tell the readers that an induction argument is coming.
b. Using k as the index, show the formula holds for n = 1.
c. Assume the formula works for the case n = N – 1
and write this “true” statement in the expanded form.
Using this fact to show the statement has to be true
for the next case when n = N
3. 1 + 2 + 3 + 4 .. + n = n (n + 1)/2
2. 1 + 3 + 5 + .. + (2n – 1) = n2
4. 2 + 6 + 10 + ... + (4n – 2) = 2n2
1. 2 + 4 + 6 + .. + 2n = n(n + 1)
5. 13 + 23 + 33 + 43 .. + n3 = n2(n + 1) 2 / 4

D. Depending on the forms of the problems, the inductive
steps maybe phrased differently.
Verify each of the following formula is true for all
positive N using mathematical induction by completing
b. Show the formula holds for N = 1.
c. Assuming the formula works for the case for N – 1
and write this “true” statement in the expanded form.
Using this fact to show the statement has to be true for the
next case for N.
6. 𝑘=1
𝑁
2 𝑘 = 2N+1 – 2
7. 𝑘=1
𝑁
𝑘(𝑘 + 1) = N(N + 1)(N + 2)/3
8. 𝑘=1
𝑁.
1/[k(k + 1)] = 1 – 1/(N + 1)
9. 𝑘=1
𝑁.
2/[k(k + 2)] = 1 – 2/(N + 2)

E. Verify each of the following formula is true for all
positive N using mathematical induction by completing
b. Show the statement is true for N = 1.
c. Assuming the statement is true for the case N – 1
and write down this “true” statement.
Then use this fact to establish that the next case for N,
the statement must also be true.
1. 3N - 1 is divisible by 2.
4. 2N3 – 3N2 + N is divisible by 6.

(Answers to the odd problems) Exercise A.
k=1
N
1.  k2
k=2
M
3.  (2k – 1)(2k)
k=3
D
5.  [(2k – 1)/(2k)]2
k=2
N
7.  1,000(e0.03)k – 1
Exercise B.
1. 3 + 5 + … + [2(k – 1) +1] + [2k +1]
3. 2 + 5 + … + [3(n – 2) –1] + [3n –1]
5. 13 + 16 + … + [3(N – 2) –20] + [3(N – 1) –20]
7. – 7 – 10 – … – [2 – 3(2K– 2)] + [2 – 3(2K– 1)]
Exercise C.
1. We will use induction
I. Verify it’s true when n = 1
If n = 1 we have 2k = 2(1) = 2 = 1(1 + 1)k=1
1

II. Assume the formula works for n = N – 1
i.e., 2k = 2 + 4 + 6 + … + 2(N – 1) = (N – 1)(N)k=1
N – 1
III. Verify that the formula works for n = N
For n = N, we have 2k =  2k + 2N = (N – 1)(N) + 2Nk=1
N
k=1
N – 1
= N2 – N + 2N
= N2 + N = N(N + 1)
If n = 1 we have k = 1 = 1(1 + 1)/2k=1
1
i.e., k = 1 + 2 + … + (N – 1) = (N – 1)(N)/2k=1
N – 1
For n = N, we have  k =  k + N = (N – 1)N/2 + Nk=1
N
k=1
N – 1
= (N2 – N + 2N)/2
= (N2 + N)/2 = N(N + 1)/2

If n = 1 we have k3 = 13 = 1 = 12(1 + 1)2/4k=1
1
i.e., k3 = 13 + 23 + … + (N – 1)3 = (N – 1)2N2/4k=1
N – 1
For n = N, we have  k3 =  k3 + N3 = (N – 1)2N2/4 + N3
k=1
N
k=1
N – 1
= (N4 – 2N3 + N2 + 4N3)/4
= N2(N2 + 2N + 1)/4
= (N4 + 2N3 + N2)/4
= N2(N + 1)2/4
If n = 1 we have  k(k + 1) = 1(1 + 1) = 2 = 1(1 + 1) (1 + 2)/3k=1
1
Exercise D.

i.e.,  k (k + 1) = 2 + 6 + … + (N – 1)N = (N – 1)(N)(N + 1)/3k=1
N – 1
For n = N, we have  k(k + 1) =  k(k + 1) + N(N + 1)k=1
N
k=1
N – 1
= (N – 1)(N)(N + 1)/3 + N(N + 1)
= [(N – 1)(N)(N + 1) + 3N(N + 1)]/3
= N[(N – 1)(N + 1) + 3(N + 1)]/3
= N[N2 – 1 + 3N + 3]/3
= N[N2 3N + 2]/3 = N(N+1)(N+2)/3
9. The statements is not true, because when n = 1, we have
𝑘=1
1.
2/[k(k + 2)] = 2/[1(1+2)] = 2/3 = 1 – 2/(1 + 2)

If n = 1 we have 31 – 1 = 2, which is divisible by 2.
i.e. 3N-1 – 1 is divisible by 2.
For n = N, we have 3N – 1 = 3(3N-1 – 1) + 2
Exercise E.
divisible by 2 divisible by 2
If n = 1 we have 71 – 1 = 6, which is divisible by 6.
i.e. 7N-1 – 1 is divisible by 6.
For n = N, we have 7N – 1 = 7(7N-1 – 1) + 7

5.4 mathematical induction

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5.4 mathematical induction