Chapter Five
Mathematical Induction
Mathematical induction is a method of proof that has been
useful in every area of mathematics. It is used to prove
statements that assert some thing is true foe every integer value
of the natural numbers N , you need mathematical induction.
Principle of Mathematical Induction.
Let
S (n )
be a statement. The mathematical induction steps
are:
1. Prove that the statement S (1) is true.
2. Suppose that the statement S (k ) is true for all positive
integers k  1 . This is called the induction hypothesis step.
3. Prove that statement S (k  1) is true for all positive integers
k 1.
Then the statement S (n ) is true for all positive integers n N .
Example 5.1. Use mathematical induction to prove that
n
i  1 2  3 
n 
i 1
n (n  1)
2
for all integers n  1 , and then find, 2  4  6  500 .
Solution. For each positive integer n  1 , let the statement
S (n ) : 1  2  3 
Step 1.
S (1) : 1 
1(1  1)
2
. Thus
n 
S (1)
n (n  1)
2
is true.
Step 2. (Inductive step): Suppose that
S (k ) : 1  2  3 
k 
is true for all positive integers
k 1.
k (k  1)
2
Step 3. We want to prove that
S (k  1) : 1  2  3 
 k  ( k  1) 
(k  1)((k  1)  1)
2
48
We have
1 2  3 
k 
k (k  1)
2
Then
k (k  1)
 (k  1)
2
k (k  1)  2(k  1)

2
(k  1)(k  2)

2
(k  1)((k  1)  1)

2
(k  1)((k  1)  1)
 k  ( k  1) 
is
2
1 2  3 
Hence,
S (k  1) : 1  2  3 
 k  (k  1) 
true and by
the principle of mathematical induction, the statement
S (n ) : 1  2  3 
is true for all positive integers
We have
n 
n (n  1)
2
n 1.
S (5) : 1  2  3  4  5  15
5(5  1) 5(6)

 15  S (5)
2
2
2  4  6  500  2(1  2  3   250)
250(251)
 2(
)
2
 62750
Example 5.2. Use mathematical induction to prove that
n
 (2i  1) 1  3  5 
 (2n  1)  n 2
i 1
for all integers n  1 .
Solution. For each positive integer
n  1 , let the statement
S (n ) : 1  3  5   (2n  1)  n 2
Step 1.
S (1) : 1  12
. Thus
S (1)
is true.
Step 2. (Inductive step): Suppose that
S (k ) : 1  3  5 
is true for all positive integers
 (2k  1)  k 2
k 1.
Step 3. We want to prove that
S (k  1) : 1  3  5 
 (2k  1)  (2(k  1)  1)  (k  1) 2
49
We have
S (k ) : 1  3  5 
 (2k  1)  k 2
Then
1 3  5 
 (2k  1)  (2( k  1)  1)  k 2  (2( k  1)  1)
 k 2  2k  2  1
 k 2  2k  1
 (k  1) 2
Hence, S (k  1) : 1  3  5   (2k 1)  (2(k  1) 1)  (k  1)2 is true,
and by the principle of mathematical induction, the statement
S (n ) : 1  3  5 
is true for all positive integers
We have
 (2n  1)  n 2
n 1.
S (5) : 1  3  5  7  9  25  52
Example 5.3. Use mathematical induction to prove that
n
i
2
12  22  32 
 n2 
i 1
n (n  1)(2n  1)
6
for all integers n  1 .
Solution. For each positive integer
S (n ) : 12  22  32 
Step 1.
S (1) : 12 
1(2)(3)
1
6
n  1 , let the statement
n (n  1)(2n  1)
 n2 
6
. Thus
S (1)
is true.
Step 2. (Inductive step): Suppose that
S (k ) : 12  22  32 
k (k  1)(2k  1)
6
k 1.
k 2 
is true for all positive integers
Step 3. We want to prove that
S (k  1) : 12  22  32 
 k 2  ( k  1) 2 
We have
S (k ) : 12  22  32 
k 2 
(k  1)((k  1)  1)(2( k  1)  1)
6
k (k  1)(2k  1)
6
Then
50
12  22  32 
k (k  1)(2k  1)
 ( k  1) 2
6
k (k  1)(2k  1)  6( k  1) 2

6
(k  1)[k (2k  1)  6( k  1)]

6
2
(k  1)(2k  k  6k  6)

6
2
(k  1)(2k  7 k  6)

6
(k  1)(k  2)(2k  3)

6
(k  1)((k  1)  1)(2( k  1)  1)

6
 k 2  ( k  1) 2 
Hence,
S (k  1) : 12  22  32 
 k 2  ( k  1) 2 
(k  1)((k  1)  1)(2( k  1)  1)
6
is true and by the principle of mathematical induction, the
statement
S (n ) : 12  22  32 
is true for all positive integers
We have
 n2 
n (n  1)(2n  1)
6
n 1.
S (3) : 12  22  33  14 
3(4)(7)
6
Example 5.4. Use mathematical induction to prove that 22n 1 is
divisible by 3 , that is 3 (22n 1) for all integers n  1 .
Solution. For each positive integer n  1 , let the statement
S (n ) : 3 (22n 1)
Step 1.
S (1) : 22(1)  1  4  1  3
. Thus
3 (22(1)  1) ,
and
S (1)
is true.
Step 2. (Inductive step): Suppose that
S (k ) : 3 (22k 1)
is true for all positive integers
k 1.
Step 3. We want to prove that
S (k  1) : 3 (22( k 1) 1)
51
But,
22( k 1)  1  22 k  2  1
 22 k .22  1
 22 k (3  1)  1
 22 k .3  (2 2 k  1)
Now, 22 k .3 is divisible by 3 , and 22k 1 is divisible by 3 by
step 2. Hence, the sum of 22 k .3 and 22k 1 is divisible by 3 .
Therefore, S (k  1) : 3 (22(k 1) 1) is true and 3 (22n 1) , and the
statement
S (n ) : 3 (22n 1)
is true for all integers
n 1.
 In general, the principle mathematical induction, can be
formulated as:
Let S (n ) be a statement. The mathematical induction steps
are:
1. Prove that the statement S (a ) is true for a particular integer a .
2. Suppose that the statement S (k ) is true for all integers k  a .
This is called the induction hypothesis step.
3. Prove that statement S (k  1) is true for all integers k  a .
Then the statement S (n ) is true for all integers n  a .
Example 5.5. Use mathematical induction to prove that 2n  1  2n
for all integers n  3 .
Solution. For each positive integer n  3 , let the statement
S ( n ) : 2n  1  2 n
Step 1.
S (3) : 2(3)+1  6  1  7  23  8
. Thus
S (3)
is true.
Step 2. (Inductive step): Suppose that
S ( k ) : 2k  1  2 k
is true for all positive integers
k  3.
Step 3. We want to prove that
S (k  1) : 2(k  1)  1  2k 1
We have
52
2(k  1)  1  2k  2  1
 2k  1  2
 2k  2, since S (k ) is true
 2k  2k , since k  3, 2  2k
 2k (1  1)
 2k 1
Thus the statement
S (k  1) : 2(k  1)  1  2k 1
is true. Hence
S ( n ) : 2n  1  2 n
is true for all
n  3.
Example 5.6. Use mathematical induction to prove that
n
 i (i  1) 2  6  12 
 n (n  1) 
i 1
for all integers n  1 .
Solution. For each positive integer
n
S (n ) :
 i (i  1) 2  6  12 
n (n  1)(n  2)
3
n 1,
let the statement
 n (n  1) 
i 1
Step 1.
S (1) : 1(1  1)  1(2)  2 
n (n  1)(n  2)
3
1(1  1)(1  2) 1(2)(3)

3
3
. Thus
S (1)
is
true.
Step 2. (Inductive step): Suppose that
S (k ) : 2  6  12 
 k (k  1) 
is true for all positive integers
k (k  1)(k  2)
3
k 1.
Step 3. We want to prove that
S (k  1) : 2  6  12 
(k  1)((k  1)  1)(( k  1)  2)
3
(k  1)(k  2)(k  3)

3
 ( k  1)(( k  1)  1) 
We have
S (k ) : 2  6  12 
 k (k  1) 
k (k  1)(k  2)
3
Then
53
2  6  12 
k (k  1)(k  2)
 (k  1)(k  2)
3
k (k  1)(k  2)  3(k  1)(k  2)

3
(k  1)(k  2)(k  3)

3
 k (k  1)  (k  1)((k  1)  1) 
Hence,
S (k  1) : 2  6  12 
(k  1)((k  1)  1)(( k  1)  2)
3
(k  1)(k  2)(k  3)

3
 ( k  1)(( k  1)  1) 
is true and by the principle of mathematical induction, the
statement
n
S (n ) :
 i (i  1) 2  6  12 
 n (n  1) 
i 1
is true for all positive integers
We have
n (n  1)(n  2)
3
n 1.
S (3) : 2  6  12  20 
3(4)(5)
3
Example 5.7. Use mathematical induction to prove that
for all integers n  4 .
Solution. For each positive integer n  4 , let the statement
2n  n !
S (n ) : 2n  n !
Step 1.
S (4) : 24  16  4!  4.3.2.1  24
. Thus
S (3)
is true.
Step 2. (Inductive step): Suppose that
S ( k ) : 2k  k !
is true for all positive integers
k  4.
Step 3. We want to prove that
S (k  1) : 2k 1  (k  1)!
We have
2k 1  2k .2
 k !(2) since S (k ) is true
 k !(k  1), since k  4, 2  (k  1)
 (k  1)!
54
Thus the statement
S (k  1) : 2k 1  (k  1)!
is true. Hence
S (n ) : 2n  n !
is true for all
n  4.
Example 5.8. Use mathematical induction to prove that
n
 (4i  2) 2  6  10 
 (4n  2)  2n 2
i 1
for all integers n  1 .
Solution. For each positive integer
n 1,
n
S (n ) :
 (4i  2) 2  6  10 
let the statement
 (4n  2)  2n 2
i 1
Step 1.
S (1) : 2=2(1) 2
. Thus
S (1)
is true.
Step 2. (Inductive step): Suppose that
S (k ) : 2  6  10 
is true for all positive integers
 (4k  2)  2k 2
k 1.
Step 3. We want to prove that
S (k  1) : 2  6  10 
 (4k  2)  (4(k  1)  2)  2(k  1) 2
We have
S (k ) : 2  6  10 
 (4k  2)  2k 2
Then
2  6  10 
 (4k  2)  (4( k  1)  2)  2k 2  (4k  4  2)
 2k 2  4k  2
 2(k 2  2k  1)
 2(k  1) 2
Hence, S (k  1) : 2  6  10   (4k  2)  (4(k  1)  2)  2(k  1)2 is true
and by the principle of mathematical induction, the statement
n
S (n ) :
 (4i  2) 2  6  10 
 (4n  2)  2n 2
i 1
is true for all positive integers
We have
n 1.
S (5) : 2  6  10  14  18  50  2.52
55
Example 5.9. Use mathematical induction to prove that
n
 r i 1  r  r 2  r 3 
rn 
i 0
for all integers n  0 .
Solution. For each integer
n  0,
let the statement
n
 r i 1  r  r 2  r 3 
S (n ) :
r n 1  1
;r 1
r 1
rn 
i 0
Step 1.
0
S (0) :
ri 
i 0
r 01 -1
 1, r  1
r -1
. Thus
r n 1  1
;r 1
r 1
S (0)
is true.
Step 2. (Inductive step): Suppose that
k
S (k ) :
 r i 1  r  r 2  r 3 
rk 
i 0
is true for all positive integers
r k 1  1
;r 1
r 1
k 1.
Step 3. We want to prove that
r ( k 1) 1  1
r 1
( k 1) 1
r
1

;r 1
r 1
 r k  r k 1 
1 r  r 2  r 3 
We have
k
S (k ) :
r
i
1  r  r  r 
2
3
i 0
r k 1  1
r 
;r 1
r 1
k
Then
1 r  r 2  r 3 
Hence,
r k 1  1 k 1
 r ;r  1
r 1
(r k 1  1)  r k 1 (r  1)

r 1
k 1
r  1  r k  2  r k 1

r 1
k 2
r 1

r 1
( k 1) 1
r
1

r 1
r k 2  1
k 1
r 
; r  1 is
r 1
 r k  r k 1 
S (k  1) : 1  r  r 2  r 3 
true and
by the principle of mathematical induction, the statement
n
S (n ) :
 r i 1  r  r 2  r 3 
i 0
is true for all positive integers
rn 
r n 1  1
;r 1
r 1
n  0.
56
We have
1  3  32  33 
 37 
37 1  1 38  1 6561  1 6560



 3280
3 1
3 1
2
2
Example 5.10. Use mathematical induction to prove that
dn
(x 1 )  (1) n n !x  ( n 1)
n
dx
for all integers n  1 .
Solution. For each integer
n  0 , let the statement
dn
S (n ) :
(x 1 )  (1) n n !x  ( n 1)
n
dx
Step 1.
S (1) :
d
(x 1 )  ( 1)11!x  (11)  x 2
dx
. Thus
S (1)
is true.
Step 2. (Inductive step): Suppose that
S (k ) :
is true for all positive
dk
(x 1 )  (1) k k !x  ( k 1)
k
dx
integers k  1 .
Step 3. We want to prove that
S (k  1) :
d k 1
(x 1 )  (1) k 1 (k  1)!x  ( k  2)
k 1
dx
We have
S (k ) :
Then
dk
(x 1 )  (1) k k !x  ( k 1)
k
dx
d k 1
d dk
(x 1 ) 
[
(x 1 )]
k

1
k
dx
dx
dx
d
(k  1)

[(1)k k !x
]
dx
(k  2)
 (1)k k !((k  1))x
 (1)k k !(1)(k  1)x
(k  2)
(k  2)
 (1)k  1 (k  1)!x
1Hence,
d k 1
S (k  1) :
(x 1 )  (1) k 1 (k  1)!x  ( k  2)
k 1
dx
is true and by the
principle of mathematical induction, the statement
dn
(x 1 )  (1) n n !x  ( n 1)
n
dx
integers n  1
S (n ) :
is true for all positive
57
Problems.
Use mathematical induction to prove:
n 2 (n  1)2
, for all integers n  1 .
4
i 1
n
1
1
1
1
1
n



 

, for all integers

1.2 2.3 3.4
n (n  1) n (n  1)
i 1 i (i  1)
n
1.  i 3 13  23  33 
2.
 n3 
n 1.
n
3.  2i 2  2.12  2.22  2.32 
 2n 2  n 3  n ,
for all integers
n 1.
i 1
n
4.  (2i 1)
i 1
2
1  3  5 
2
2
2
n (4n 2  1)
,
 (2n  1) 
4
3
for all integers
n 1.
7n  1
is divisible by 6 , that is 6 (7 n  1) for all integers n  1 .
n 3  n is divisible by 3 , that is 3 (n 3  n ) for all integers n  1 .
7 n  2n is divisible by 5 , that is 5 (7n  2n ) for all integers
n 1.
8. 3n  1 is divisible by 2 , for all integers n  1 .
n
9.  3i  3  6  9   3n  3n (n  1) , for all integers n  1 .
5.
6.
7.
2
i 1
10.
n
 (2n  3)  5  7  9 
 (2n  3)  n (n  4)
, for all integers
i 1
n 1.
58