The document discusses mathematical induction and provides two examples of using it to prove statements. It first proves that the expression nn+1(n+2) is divisible by 3 for all positive integers n. It shows the basis step for n=1 and inductive step, assuming true for n=k and proving for n+1. The second example proves 33n+2n+2 is divisible by 5 for all n, following the same process of basis and inductive steps.

1. Mathematical Induction

2. Mathematical Induction
    
e.g. iii Prove n n  1 n  2 is divisible by 3

3. Mathematical Induction
    
e.g. iii Prove n n  1 n  2 is divisible by 3
Step 1: Prove the result is true for n = 1

4. Mathematical Induction
    
e.g. iii Prove n n  1 n  2 is divisible by 3
Step 1: Prove the result is true for n = 1
123
6

5. Mathematical Induction
    
e.g. iii Prove n n  1 n  2 is divisible by 3
Step 1: Prove the result is true for n = 1
123
6 which is divisible by 3

6. Mathematical Induction
    
e.g. iii Prove n n  1 n  2 is divisible by 3
Step 1: Prove the result is true for n = 1
123
6 which is divisible by 3
Hence the result is true for n = 1

7. Mathematical Induction
    
e.g. iii Prove n n  1 n  2 is divisible by 3
Step 1: Prove the result is true for n = 1
123
6 which is divisible by 3
Hence the result is true for n = 1
Step 2: Assume the result is true for n = k, where k is a positive
integer

8. Mathematical Induction
    
e.g. iii Prove n n  1 n  2 is divisible by 3
Step 1: Prove the result is true for n = 1
123
6 which is divisible by 3
Hence the result is true for n = 1
Step 2: Assume the result is true for n = k, where k is a positive
integer
i.e. k k  1k  2  3P, where P is an integer

9. Mathematical Induction
    
e.g. iii Prove n n  1 n  2 is divisible by 3
Step 1: Prove the result is true for n = 1
123
6 which is divisible by 3
Hence the result is true for n = 1
Step 2: Assume the result is true for n = k, where k is a positive
integer
i.e. k k  1k  2  3P, where P is an integer
Step 3: Prove the result is true for n = k + 1

10. Mathematical Induction
    
e.g. iii Prove n n  1 n  2 is divisible by 3
Step 1: Prove the result is true for n = 1
123
6 which is divisible by 3
Hence the result is true for n = 1
Step 2: Assume the result is true for n = k, where k is a positive
integer
i.e. k k  1k  2  3P, where P is an integer
Step 3: Prove the result is true for n = k + 1
i.e. Prove : k  1k  2k  3  3Q, where Q is an integer

11. Proof:

12. Proof:
k  1k  2k  3

13. Proof:
k  1k  2k  3
 k k  1k  2  3k  1k  2

14. Proof:
k  1k  2k  3
 k k  1k  2  3k  1k  2
 3P  3k  1k  2

15. Proof:
k  1k  2k  3
 k k  1k  2  3k  1k  2
 3P  3k  1k  2
 3P  k  1k  2

16. Proof:
k  1k  2k  3
 k k  1k  2  3k  1k  2
 3P  3k  1k  2
 3P  k  1k  2
 3Q, where Q  P  k  1k  2 which is an integer

17. Proof:
k  1k  2k  3
 k k  1k  2  3k  1k  2
 3P  3k  1k  2
 3P  k  1k  2
 3Q, where Q  P  k  1k  2 which is an integer
Hence the result is true for n = k + 1 if it is also true for n = k

18. Proof:
k  1k  2k  3
 k k  1k  2  3k  1k  2
 3P  3k  1k  2
 3P  k  1k  2
 3Q, where Q  P  k  1k  2 which is an integer
Hence the result is true for n = k + 1 if it is also true for n = k
Step 4: Since the result is true for n = 1, then the result is true for
all positive integral values of n by induction.

19. iv Prove 33n  2n2 is divisible by 5

20. iv Prove 33n  2n2 is divisible by 5
Step 1: Prove the result is true for n = 1

21. iv Prove 33n  2n2 is divisible by 5
Step 1: Prove the result is true for n = 1
33  23
 27  8
 35

22. iv Prove 33n  2n2 is divisible by 5
Step 1: Prove the result is true for n = 1
33  23
 27  8
 35 which is divisible by 5

23. iv Prove 33n  2n2 is divisible by 5
Step 1: Prove the result is true for n = 1
33  23
 27  8
 35 which is divisible by 5
Hence the result is true for n = 1

24. iv Prove 33n  2n2 is divisible by 5
Step 1: Prove the result is true for n = 1
33  23
 27  8
 35 which is divisible by 5
Hence the result is true for n = 1
Step 2: Assume the result is true for n = k, where k is a positive
integer
i.e. 33k  2k 2  5P, where P is an integer

25. iv Prove 33n  2n2 is divisible by 5
Step 1: Prove the result is true for n = 1
33  23
 27  8
 35 which is divisible by 5
Hence the result is true for n = 1
Step 2: Assume the result is true for n = k, where k is a positive
integer
i.e. 33k  2k 2  5P, where P is an integer
Step 3: Prove the result is true for n = k + 1
i.e. Prove : 33k 3  2k 3  5Q, where Q is an integer

26. Proof:

27. Proof: 33k 3  2k 3

28. Proof: 33k 3  2k 3
 27  33k  2k 3

29. Proof: 33k 3  2k 3
 27  33k  2k 3
 275P  2k 2   2k 3

30. Proof: 33k 3  2k 3
 27  33k  2k 3
 275P  2k 2   2k 3
 135P  27  2k 2  2k 3

31. Proof: 33k 3  2k 3
 27  33k  2k 3
 275P  2k 2   2k 3
 135P  27  2k 2  2k 3
 135P  27  2k 2  2  2k 2

32. Proof: 33k 3  2k 3
 27  33k  2k 3
 275P  2k 2   2k 3
 135P  27  2k 2  2k 3
 135P  27  2k 2  2  2k 2
 135P  25  2k 2

33. Proof: 33k 3  2k 3
 27  33k  2k 3
 275P  2k 2   2k 3
 135P  27  2k 2  2k 3
 135P  27  2k 2  2  2k 2
 135P  25  2k 2
 527 P  5  2k 2 

34. Proof: 33k 3  2k 3
 27  33k  2k 3
 275P  2k 2   2k 3
 135P  27  2k 2  2k 3
 135P  27  2k 2  2  2k 2
 135P  25  2k 2
 527 P  5  2k 2 
 5Q, where Q  27 P  5  2k 2 which is an integer

35. Proof: 33k 3  2k 3
 27  33k  2k 3
 275P  2k 2   2k 3
 135P  27  2k 2  2k 3
 135P  27  2k 2  2  2k 2
 135P  25  2k 2
 527 P  5  2k 2 
 5Q, where Q  27 P  5  2k 2 which is an integer
Hence the result is true for n = k + 1 if it is also true for n = k

36. Proof: 33k 3  2k 3
 27  33k  2k 3
 275P  2k 2   2k 3
 135P  27  2k 2  2k 3
 135P  27  2k 2  2  2k 2
 135P  25  2k 2
 527 P  5  2k 2 
 5Q, where Q  27 P  5  2k 2 which is an integer
Hence the result is true for n = k + 1 if it is also true for n = k
Step 4: Since the result is true for n = 1, then the result is true for
all positive integral values of n by induction.

37. Proof: 33k 3  2k 3
 27  33k  2k 3
 275P  2k 2   2k 3
 135P  27  2k 2  2k 3 Exercise 6N;
3bdf, 4 ab(i), 9
 135P  27  2k 2  2  2k 2
 135P  25  2k 2
 527 P  5  2k 2 
 5Q, where Q  27 P  5  2k 2 which is an integer
Hence the result is true for n = k + 1 if it is also true for n = k
Step 4: Since the result is true for n = 1, then the result is true for
all positive integral values of n by induction.