Introduction to Mathematical Induction!!

2
(i) S(1), S(2), S(3) is true.
S(4), S(5), S(6).......have not
been proved to be true
Class Practice 1:
(iii) If S(k) is true, S(k + 1) is true
Can we be sure that S(1) is
true?

3
.
(v) S(2) is true. Also if S(k) is true,
then S(k + 2) is true
S(2) is true then S(4) is true
…
The statement is true for all
positive even number n.

4
Sequence: A set of number with ordering.
E.g. (i) 1, 4, 7, 10, 13, …, 3n –
2, …
(ii) 32, 16, 8, 4, 2, 1, …,
…
6
n
2 

For (i), the first term is 1
the second term is 4
the third term is 7, and
so on.
3n – 2 is called the
general term of the

5
Series: Sum of terms of sequences
For the sequence, 1, 4, 7, 10, …, 3n – 2, …
we can obtain the following series:
1
1 + 4
1 + 4 + 7
1 + 4 + 7 + 10
1 + 4 + 7 + 10 + … + (3n – 2)
1 + 4 + 7 + 10 + … + (3n – 2) + (3(n + 1) – 2)
1 term
2 terms
3 terms
4 terms
n terms
n + 1 terms
Sum of first

6
Example 3.3 (Another presentation):
When n = 1
L.H.S = 1
)
1
(
2 
R.H.S = 12
= 1
Since L.H.S = R.H.S,
the statement is true when n = 1.
= 1
This part is to prove the
“starting point”.
Only one term
on LHS when
n = 1

7
Assume the statement is true when n = k,
that is, 2
)
1
2
(
...
5
3
1 k
k 





]
1
)
1
(
2
[
)
1
2
(
...
5
3
1 






 k
k
)
1
2
(
)
1
2
(
...
5
3
1 





 k
k
1
2
2

 k
k
2
)
1
( 
k
When n = k + 1,
L.H.S =
=
= R.H.S
Since L.H.S = R.H.S, the statement is
true when n = k + 1.
=
Applying
the
assumption
=
where k is some positive integers.
k + 1 terms
in total

8
By the principle of mathematical
induction, the statement is true
for all natural numbers n.
NOT integers / real numbers

9
Extra Example 1:
Prove by mathematical induction that
for all positive integers n.
2
2
3
3
3
3
)
1
(
4
1
...
3
2
1 




 n
n
n

10
Solution to Extra Example 1 (a):
When n = 1
Since L.H.S = R.H.S,
L.H.S = 13
= 1
1
)
1
1
(
)
1
(
4
1 2
2


R.H.S =

11
Assume the statement is true when n = k,
that is,
When n = k + 1,
=
=
where k is some positive integers.
2
2
3
3
3
3
)
1
(
4
1
...
3
2
1 




 k
k
k
L.H.S =
3
3
3
3
3
)
1
(
...
3
2
1 




 k
k
3
2
2
)
1
(
)
1
(
4
1


 k
k
k





 

 )
1
(
4
1
)
1
( 2
2
k
k
k
Rather than
expand, use
factorization
is simpler.

12
= 




 

 )
1
(
4
1
)
1
( 2
2
k
k
k





 


4
)
1
(
4
4
1
)
1
( 2
2 k
k
k
=
 
4
4
)
1
(
4
1 2
2


 k
k
k
=
To factorize
the ¼ and
make the
(k+2)2
term.
2
2
)
2
(
)
1
(
4
1

 k
k
=

13
R.H.S =
2
2
)
1
1
(
)
1
(
4
1


 k
k
=
2
2
)
2
(
)
1
(
4
1

 k
k
Since L.H.S = R.H.S, the statement is
true when n = k + 1.
By the principle of mathematical
induction, the statement is true
for all positive integers n.

14
Steps in MI:
(1) Prove the statement is true
when n = 1
(2) Assume the statement is
true when n = k
(3) Prove the statement is true
when n = k + 1,with the
help of the assumption
(4) WRITE CONCLUSION!!!

15
Technique in MI:
In the part to prove the statement is true
when n = k +1 assuming the statement
is true when n = k,
You’d better write down the LHS and
RHS of the statement for n = k + 1 in
advance~! ^^

16
Sometimes, for convenience, we will let
S(n) be the statement. For example, we
can let
Hence, we have
2
)
1
2
(
...
5
3
1
:
)
( n
n
n
S 





2
1
1
:
)
1
( 
S
2
2
3
1
:
)
2
( 

S
2
3
5
3
1
:
)
3
( 


S
2
)
1
2
(
..
.
5
3
1
:
)
( k
k
k
S 





2
)
1
(
)
1
2
(
)
1
2
(
..
.
5
3
1
:
)
1
( 








 k
k
k
k
S

17
Extra Example 1 (b):
Use the result in (a),
2
2
3
3
3
3
)
1
(
4
1
...
3
2
1 




 n
n
n
Find the value of
(i)
(ii)
3
3
3
3
40
...
23
22
21 



3
3
3
3
50
...
6
4
2 




18
Solution to Extra Example 1 (b):
3
3
3
3
40
...
23
22
21 



(i)
)
21
)(
20
(
4
1
)
41
)(
40
(
4
1 2
2
2
2

=
)
20
...
3
2
1
(
)
40
...
3
2
1
(
3
3
3
3
3
3
3
3









=
Target: To change the given sum into a
form that can apply the formula in (a)
2
2
3
3
3
3
)
1
(
4
1
...
3
2
1 




 k
k
k
= 628300

19
(ii
)
3
3
3
3
)
25
2
(
...
)
3
2
(
)
2
2
(
2 






3
3
3
3
50
...
6
4
2 



=
= )
25
...
3
2
1
(
2 3
3
3
3
3









 
 2
2
26
25
4
1
8
=
= 845000
Try to make use
of factorization!!

20
Solution to Class Practice 2 7(b)
We have proved the following
identities:
)
2
)(
1
(
3
1
)
1
(
...
4
3
3
2
2
1 









 n
n
n
n
n
)
1
(
2
1
...
3
2
1 




 n
n
n
To find the sum of )
2
(
...
5
3
4
2
3
1 






 n
n
Compare:
1 3 + 2  4 + 3  5 + … + n  (n +
2)
and
1 2 + 2  3 + 3  4 + … + n  (n +
1)
Difference = 1

21
1 3 + 2  4 + 3  5 + … + n  (n + 2)
1 (2 + 1) + 2  (3 + 1) + 3  (4 + 1) + … + n  (n + 1
1)
=
[1 3 + 2  4 + 3  5 + … + n  (n + 2)]
+
(1 + 2 + 3 + … + n)
=
= n (n + 1) (n + 2)
1
3
n (n + 1)
1
2
+
= n (n + 1) ( (n + 2)
1
2
+ )
1
3
n (n + 1) (2n + 7)
1
6
=

22
Divisibility:
An integer a is said to be divisible by another integer b if
or a = bk
a
b
= k
where k is another integer.
It is meaningless to say a number is divisible by
a non-integer.

23
Example 3.5 (Another presentation):
When n = 1, 3
1
)
1
(
4 3


Since 3 is divisible by 3,
There is no LHS /
RHS in this proof.
Assume the statement is true for some positive integers k
that is,
where t is an integer.
t is an integer,
but not any
real number
4k3
– k = 3t
4k3
= 3t + k

24
When n = k + 1,
4(k + 1)3
– (k + 1)
= (k + 1)[4(k + 1)2
– 1]
= (k + 1)[4(k2
+ 2k + 1) – 1]
= (k + 1)(4k2
+ 8k + 3)
= 4k3
+ 12k2
+ 11k + 3
= 3t + k + 12k2
+ 11k + 3
= 3t + 12k2
+ 12k + 3
= 3(t + 4k2
+ 4k + 1)
Using assumption:
4k3
= 3t + k
t, k and k2
are integers
and the sum are also
an integer.
 The statement is true when n = k + 1 assuming the
statement is true when n = k.

25
 By the principle of mathematical
induction, the statement is true for all
positive integers n.

26
Class Practice 3 (2):
Prove by mathematical induction that 4n
+ 15n – 1 i
s divisible by 3 for all positive integers n.
Let S(n) be the statement “4n
+ 15n – 1 is divisible
by 3”, where n is a positive integer.
Don’t just let S(n) be “4n
+ 15n – 1”!
This is not a statement, but a functi
on only.

27
When n = 1,
41
+ 15(1) – 1 = 18 = 3  6
 S(1) is true.
Assume S (k) is true, that is,
4k
+ 15k – 1 = 3t or 4k
= 3t –15k + 1
where t is an integer.
When n = k + 1,
4k + 1
+ 15(k + 1) – 1
= 4  4k
+ 15k + 15 – 1
= 4  (3t – 15k + 1) + 15k +14
= 12t – 45k + 18
= 3  (4t – 15k + 6)
 S (k + 1) is true.
4a + b
= 4a
 4b
Remember to factorize 3 out.

28
 By the principle of mathematical
induction, the statement is true for all
positive integers n.

Introduction to Mathematical Induction!!

More Related Content

Similar to Introduction to Mathematical Induction!! (20)

Recently uploaded (20)

Introduction to Mathematical Induction!!