ADVANCED HIGHER MATHEMATICS
PROOF BY INDUCTION
Recall that a mathematical statement is either true or false.
Let p n be a given statement about the positive integer n.
Example
Let p n be the statement that n 2  2n is divisible by 3.
p1 is the statement that 12  2  1 is divisible by 3.
Since 12  2  1  3 , the statement p1 is true.
p 2 is the statement that 2 2  2  2 is divisible by 3.
Since 2 2  2  2  8 , the statement p 2 is false.
Hence the statement p n is not true for all positive integers n.
The method of proof by induction can be used to prove that a given statement p n is
true for all positive integers n.
STEP 1
Prove that the statement p1 is true.
This is known as the basis for the proof by induction.
STEP 2
Assume that the statement p n is true for some positive integer n.
STEP 3
Use the assumption made in STEP 2 to prove that the statement p n 1 is then also true.
This is known as the inductive step.
STEP 4
This completes the proof by induction that the statement p n is true for all positive
integers n.
The logic behind the method of proof by induction can be explained as follows.
By following the steps outlined, we have shown that:
(1)
(2)
The statement p1 is true;
If the statement p n is true, then the statement p n 1 is also true for all positive
integers n. This can be written as

p n true
p n 1 true for all positive integers n.
We have proved that the statement p1 is true, therefore by (2) above, the statement
p 2 is also true. This gives rise to the following inductive chain using (2) above:



p1 true
p 2 true
p 3 true
p 2 true;
p 3 true;
p 4 true;
and so on.
Thus the statement p n is true for all positive integers n.
Note that we must prove that the statement p1 is true to begin this chain of induction.
NOTE
The method of proof by induction can also be used to prove that a given statement p n
is true for all integers n  m as follows:
(1)
(2)
Prove that the statement p m is true.
Assuming that the statement p n is true for some integer n  m , prove that the
statement p n 1 is also true.
When attempting to prove the inductive step that p n 1 is true, it is useful to note that
n 1

k 1
n
f (k )   f (k )  f (n  1) .
k 1
Worked Example 1
n
Prove by induction that
 (2k  1)  n
2
for all positive integers n.
k 1
Solution
n
Let p n be the statement that
 (2k  1)  n
2
.
k 1
BASIS
1
p1 is the statement that
 (2k  1)  1
2
.
k 1
1
LHS   (2k  1)  2  1  1  1
k 1
RHS  12  1
Hence the statement p1 is true.
ASSUMPTION
Assume that the statement p n is true for some positive integer n,
n
i.e. assume that
 (2k  1)  n
2
.
k 1
AIM
Prove that the statement p n 1 is also true,
n 1
i.e. prove that
 (2k  1)  (n  1)
2
.
k 1
METHOD
n 1
n
k 1
k 1
 (2k  1)   (2k  1)  2(n  1)  1
 n 2  2(n  1)  1
 n  2n  2  1
 n 2  2n  1
 (n  1)( n  1)
2
 (n  1) 2
[since we have assumed that p n is true]
Hence the statement p n 1 is also true.
CONCLUSION
This completes the proof by induction that the statement p n is true for all positive
integers n.
Worked Example 2
n
Prove by induction that
k
2
k 1
1
 n(n  1)( 2n  1) for all positive integers n.
6
Solution
n
Let p n be the statement that
k
k 1
2
1
 n(n  1)( 2n  1) .
6
BASIS
1
p1 is the statement that
k
k 1
2
1
 (1)(1  1)( 2(1)  1) .
6
1
LHS   k 2  12  1
k 1
RHS 
1
1
(1)(1  1)( 2(1)  1)  (1)( 2)(3)  1
6
6
Hence the statement p1 is true.
ASSUMPTION
Assume that the statement p n is true for some positive integer n,
n
i.e. assume that
k
k 1
2
1
 n(n  1)( 2n  1) .
6
AIM
Prove that the statement p n 1 is also true,
n 1
i.e. prove that
k
k 1
2
1
 (n  1)(n  1)  12(n  1)  1
6
1
 (n  1)( n  2)( 2n  3)
6
METHOD
n 1
n
k 1
k 1
 k 2   k 2  (n  1) 2
1
n(n  1)( 2n  1)  (n  1) 2 [since we have assumed that p n is true]
6
1
 (n  1)n(2n  1)  6(n  1)
6
1
 (n  1)( 2n 2  n  6n  6)
6
1
 (n  1)( 2n 2  7 n  6)
6
1
 (n  1)( 2n  3)( n  2)
6
1
 (n  1)( n  2)( 2n  3)
6

Hence the statement p n 1 is also true.
CONCLUSION
This completes the proof by induction that the statement p n is true for all positive
integers n.
Worked Example 3
n
(a)
Prove by induction that
1
 k (k  1)  3n(n  1)(n  2)
for all positive integers
k 1
n.
20
(b)
Hence evaluate
 k (k  1) .
k 11
Solution
n
(a)
Let p n be the statement that
1
 k (k  1)  3n(n  1)(n  2) .
k 1
BASIS
1
p1 is the statement that
1
 k (k  1)  3 (1)(1  1)(1  2) .
k 1
1
LHS   k (k  1)  (1)(1  1)  2
k 1
1
1
RHS  (1)(1  1)(1  2)  (1)( 2)(3)  2
3
3
Hence the statement p1 is true.
ASSUMPTION
Assume that the statement p n is true for some positive integer n,
n
i.e. assume that
1
 k (k  1)  3n(n  1)(n  2) .
k 1
AIM
Prove that the statement p n 1 is also true,
n 1
i.e. prove that
 k (k  1)  3 (n  1)(n  1)  1(n  1)  2
1
k 1
1
 (n  1)( n  2)( n  3)
3
METHOD
n 1
n
k 1
k 1
 k (k  1)   k (k  1)  (n  1)(n  1)  1
1
 n(n  1)( n  2)  (n  1)( n  2)
3
[since we have assumed that
p n is true]
1
 (n  1)n(n  2)  3(n  2)
3
1
 (n  1)( n 2  2n  3n  6)
3
1
 (n  1)( n 2  5n  6)
3
1
 (n  1)( n  2)( n  3)
3
Hence the statement p n 1 is also true.
CONCLUSION
This completes the proof by induction that the statement p n is true for all
positive integers n.
(b)
20
20
10
k 11
k 1
k 1
 k (k  1)   k (k  1)   k (k  1)
n
Now
1
 k (k  1)  3n(n  1)(n  2) .
k 1
20
1
1
 k (k  1)  3 (20)(20  1)(20  2)  3 (20)(21)(22)  3080
k 1
10
1
1
 k (k  1)  3 (10)(10  1)(10  2)  3 (10)(11)(12)  440
k 1
20
Hence
 k (k  1)  3080  440  2640 .
k 11
Worked Example 4
Prove by induction that
2  5  8  ....  (3n  1) 
1
n(3n  1)
2
for all positive integers n.
Solution
Let p n be the statement that 2  5  8  ....  (3n  1) 
n
i.e. that
1
n(3n  1) ,
2
1
 (3k  1)  2 n(3n  1) .
k 1
BASIS
1
p1 is the statement that
1
 (3k  1)  2 (1)(3(1)  1) .
k 1
1
LHS   (3k  1)  3  1  1  2
k 1
RHS 
1
1
(1)(3(1)  1)  (1)( 4)  2
2
2
Hence the statement p1 is true.
ASSUMPTION
Assume that the statement p n is true for some positive integer n,
n
i.e. assume that
1
 (3k  1)  2 n(3n  1) .
k 1
AIM
Prove that the statement p n 1 is also true,
n 1
i.e. prove that
 (3k  1)  2 (n  1)3(n  1)  1
1
k 1

1
(n  1)(3n  4)
2
METHOD
n 1
n
k 1
k 1
 (3k  1)   (3k  1)  3(n  1)  1
1
n(3n  1)  3(n  1)  1
2
1
 n(3n  1)  3n  2
2
1
 n(3n  1)  6n  4
2
1
 (3n 2  n  6n  4)
2
1
 (3n 2  7 n  4)
2
1
 (3n  4)( n  1)
2
1
 (n  1)(3n  4)
2

[since we have assumed that p n is true]
Hence the statement p n 1 is also true.
CONCLUSION
This completes the proof by induction that the statement p n is true for all positive
integers n.
[Note that it is convenient to express the statement p n using

notation.]
Worked Example 5
n
Prove by induction that
1
n
 k (k  1)  n  1 for all positive integers n.
k 1
Solution
n
Let p n be the statement that
1
n
 k (k  1)  n  1 .
k 1
BASIS
1
p1 is the statement that
1
1
 k (k  1)  1  1 .
k 1
1
1
1
1


1(1  1) 2
k 1 k (k  1)
LHS  
RHS 
1
1

11 2
Hence the statement p1 is true.
ASSUMPTION
Assume that the statement p n is true for some positive integer n,
n
i.e. assume that
1
n
 k (k  1)  n  1 .
k 1
AIM
Prove that the statement p n 1 is also true,
n 1
i.e. prove that
1
n 1
n 1
 k (k  1)  (n  1)  1  n  2 .
k 1
METHOD
n 1
n
1
1
1




(n  1)(n  1)  1
k 1 k (k  1)
k 1 k (k  1)
n
1


[since we have assumed that p n is true]
n  1 (n  1)( n  2)
n(n  2)
1


(n  1)( n  2) (n  1)( n  2)
n ( n  2)  1

(n  1)( n  2)
n 2  2n  1
(n  1)( n  2)
(n  1)( n  1)

(n  1)( n  2)
n 1

n2

Hence the statement p n 1 is also true.
CONCLUSION
This completes the proof by induction that the statement p n is true for all positive
integers n.
n
[The result
1
n
 k (k  1)  n  1
can also be proved directly by expressing
k 1
partial fractions.
It can be shown that
n
Hence
1
1
1
1
 
.
k (k  1) k k  1
n
1
1 
 k (k  1)    k  k  1 
k 1
k 1
1 1
 
1 2
1 1
 
2 3
1 1
 
3 4
1 1
 
4 5
 ........
1
1
 
n n 1
(k  1)
( k  2)
(k  3)
( k  4)
( k  n)
1
in
k (k  1)
1
[note that the terms cancel in diagonal pairs as shown]
n 1
n 1
1


n 1 n 1
n 11

n 1
n

]
n 1
1
Worked Example 6
Prove by induction that 4 n  1 is divisible by 3 for all positive integers n.
Solution
Let p n be the statement that 4 n  1 is divisible by 3.
BASIS
p1 is the statement that 41  1 is divisible by 3.
Now 41  1  3 , which is divisible by 3, hence the statement p1 is true.
ASSUMPTION
Assume that the statement p n is true for some positive integer n,
i.e. assume that 4 n  1 is divisible by 3.
AIM
Prove that the statement p n 1 is also true,
i.e. prove that 4 n 1  1 is divisible by 3.
METHOD
4 n1  1  4  4 n  1
 4 (4 n  1)  1  1
 4(4 n  1)  4  1
 4(4 n  1)  3


Now 4 n  1 is divisible by 3 (since we have assumed that p n is true) and 3 is
divisible by 3, hence 4 n 1  1 is divisible by 3.
Hence the statement p n 1 is also true.
CONCLUSION
This completes the proof by induction that the statement p n is true for all positive
integers n.
[The inductive step can also be proved as follows:
4 n1  1  4  4 n  1
 (3  1)4 n  1
 3  4n  4n  1
 3  4 n  (4 n  1)
Now 3  4 n is divisible by 3 and 4 n  1 is divisible by 3 (since we have assumed that
p n is true), hence 4 n 1  1 is divisible by 3.]
Worked Example 7
 2 1
 .
Let A  
  1 0
n 1 n 
 for all positive integers n.
Prove by induction that A n  
  n 1  n
Solution
n 1 n 
 .
Let p n be the statement that A n  
  n 1  n
BASIS
1  1 1 
 .
p1 is the statement that A1  
  1 1  1
 2 1
 .
LHS  A1  A  
  1 0
1  1 1   2 1 
  

RHS  
  1 1  1   1 0 
Hence the statement p1 is true.
ASSUMPTION
Assume that the statement p n is true for some positive integer n,
n 1 n 
 .
i.e. assume that A n  
  n 1  n
AIM
Prove that the statement p n 1 is also true,
n  1   n  2 n  1
 (n  1)  1
  
 .
i.e. prove that A n 1  
  (n  1) 1  (n  1)    n  1  n 
METHOD
n 
 2 1  n  1

 [since we have assumed that p n is true]
A n 1  AA n  
  1 0   n 1  n 
2n  1(1  n) 
 2(n  1)  1(n)

 
  1(n  1)  0(n)  1n  0(1  n) 
 2n  2  n 2n  1  n 

 
 n 
  n 1
 n  2 n  1

 
 n 1  n 
Hence the statement p n 1 is also true.
CONCLUSION
This completes the proof by induction that the statement p n is true for all positive
integers n.
Worked Example 7
Let z  cos   i sin  .
Prove by induction that z n  cos n  i sin n for all positive integers n.
[Recall that this is de Moivre's theorem.]
Solution
Let p n be the statement that z n  cos n  i sin n .
BASIS
p1 is the statement that z 1  cos 1  i sin 1 .
LHS  z 1  z  cos   i sin 
RHS  cos1  i sin 1  cos  i sin 
Hence the statement p1 is true.
ASSUMPTION
Assume that the statement p n is true for some positive integer n,
i.e. assume that z n  cos n  i sin n .
AIM
Prove that the statement p n 1 is also true,
i.e. prove that z n 1  cos(n  1)  i sin( n  1) .
METHOD
z n 1  zz n  (cos   i sin  )(cos n  i sin n )
[since we have assumed that p n is
true]
 cos  cos n  i sin  cos n  i cos  sin n  i 2 sin  sin n
 cos  cos n  sin  sin n  i (sin  cos n  cos  sin n )
 cos(  n )  i sin(   n )
 cos( n  1)  i sin( n  1)
Hence the statement p n 1 is also true.
CONCLUSION
This completes the proof by induction that the statement p n is true for all positive
integers n.
Worked Example 8
dn
( xe x )  ( x  n)e x for all positive integers n.
Prove by induction that
n
dx
Solution
Let p n be the statement that
dn
( xe x )  ( x  n)e x .
n
dx
BASIS
p1 is the statement that
LHS 
d1
( xe x )  ( x  1)e x .
1
dx
d1
d
( xe x ) 
( xe x )
1
dx
dx
d
d
 x  (e x )  e x  ( x )
dx
dx
x
x
 x  e  e 1
 ( x  1)e x
= RHS
[using the product rule]
Hence the statement p1 is true.
ASSUMPTION
Assume that the statement p n is true for some positive integer n,
i.e. assume that
dn
( xe x )  ( x  n)e x .
dx n
AIM
Prove that the statement p n 1 is also true,
i.e. prove that
d n 1
( xe x )  ( x  n  1)e x .
n 1
dx
METHOD
d n 1
d  dn
x
x 
(
xe
)

 n ( xe )
n 1
dx  dx
dx

d
 ( x  n)e x  [since we have assumed that p n is true]
dx
d
d
 ( x  n)  (e x )  e x  ( x  n) [using the product rule]
dx
dx
x
x
 ( x  n)  e  e  1
 ( x  n  1)e x
Hence the statement p n 1 is also true.
CONCLUSION
This completes the proof by induction that the statement p n is true for all positive
integers n.