1. Mathematical Induction

2. Mathematical Induction
Step 1: Prove the result is true for n = 1 (or whatever the first term
is)

3. Mathematical Induction
Step 1: Prove the result is true for n = 1 (or whatever the first term
is)
Step 2: Assume the result is true for n = k, where k is a positive
integer (or another condition that matches the question)

4. Mathematical Induction
Step 1: Prove the result is true for n = 1 (or whatever the first term
is)
Step 2: Assume the result is true for n = k, where k is a positive
integer (or another condition that matches the question)
Step 3: Prove the result is true for n = k + 1

5. Mathematical Induction
Step 1: Prove the result is true for n = 1 (or whatever the first term
is)
Step 2: Assume the result is true for n = k, where k is a positive
integer (or another condition that matches the question)
Step 3: Prove the result is true for n = k + 1
NOTE: It is important to note in your conclusion that the result is
true for n = k + 1 if it is true for n = k

6. Mathematical Induction
Step 1: Prove the result is true for n = 1 (or whatever the first term
is)
Step 2: Assume the result is true for n = k, where k is a positive
integer (or another condition that matches the question)
Step 3: Prove the result is true for n = k + 1
NOTE: It is important to note in your conclusion that the result is
true for n = k + 1 if it is 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

7. 1
e.g.i  1  3  5    2n  1  n2n  12n  1
2 2 2 2
3

8. 1
e.g.i  1  3  5    2n  1  n2n  12n  1
2 2 2 2
3
Step 1: Prove the result is true for n = 1

9. 1
e.g.i  1  3  5    2n  1  n2n  12n  1
2 2 2 2
3
Step 1: Prove the result is true for n = 1
LHS  12
1

10. 1
e.g.i  1  3  5    2n  1  n2n  12n  1
2 2 2 2
3
Step 1: Prove the result is true for n = 1
1
LHS  12 RHS  12  12  1
3
1
1
 113
3
1

11. 1
e.g.i  1  3  5    2n  1  n2n  12n  1
2 2 2 2
3
Step 1: Prove the result is true for n = 1
1
LHS  12 RHS  12  12  1
3
1
1
 113
3
1
 LHS  RHS

12. 1
e.g.i  1  3  5    2n  1  n2n  12n  1
2 2 2 2
3
Step 1: Prove the result is true for n = 1
1
LHS  12 RHS  12  12  1
3
1
1
 113
3
1
 LHS  RHS
Hence the result is true for n = 1

13. 1
e.g.i  1  3  5    2n  1  n2n  12n  1
2 2 2 2
3
Step 1: Prove the result is true for n = 1
1
LHS  12 RHS  12  12  1
3
1
1
 113
3
1
 LHS  RHS
Hence the result is true for n = 1
Step 2: Assume the result is true for n = k, where k is a positive
integer

14. 1
e.g.i  1  3  5    2n  1  n2n  12n  1
2 2 2 2
3
Step 1: Prove the result is true for n = 1
1
LHS  12 RHS  12  12  1
3
1
1
 113
3
1
 LHS  RHS
Hence the result is true for n = 1
Step 2: Assume the result is true for n = k, where k is a positive
integer
1
i.e. 12  32  52    2k  1  k 2k  12k  1
2
3

15. 1
e.g.i  1  3  5    2n  1  n2n  12n  1
2 2 2 2
3
Step 1: Prove the result is true for n = 1
1
LHS  12 RHS  12  12  1
3
1
1
 113
3
1
 LHS  RHS
Hence the result is true for n = 1
Step 2: Assume the result is true for n = k, where k is a positive
integer
1
i.e. 12  32  52    2k  1  k 2k  12k  1
2
3
Step 3: Prove the result is true for n = k + 1

16. 1
e.g.i  1  3  5    2n  1  n2n  12n  1
2 2 2 2
3
Step 1: Prove the result is true for n = 1
1
LHS  12 RHS  12  12  1
3
1
1
 113
3
1
 LHS  RHS
Hence the result is true for n = 1
Step 2: Assume the result is true for n = k, where k is a positive
integer
1
i.e. 12  32  52    2k  1  k 2k  12k  1
2
3
Step 3: Prove the result is true for n = k + 1
1
i.e. Prove : 12  32  52    2k  1  k  12k  12k  3
2
3

17. Proof:

18. 12  32  52    2k  1
2
Proof:
 12  32  52    2k  1  2k  1
2 2

19. 12  32  52    2k  1
2
Proof:
 12  32  52    2k  1  2k  1
2 2
   
Sk

20. 12  32  52    2k  1
2
Proof:
 12  32  52    2k  1  2k  1
2 2
    
Sk Tk 1

21. 12  32  52    2k  1
2
Proof:
 12  32  52    2k  1  2k  1
2 2
    
Sk Tk 1
1
 k 2k  12k  1  2k  1
2
3

22. 12  32  52    2k  1
2
Proof:
 12  32  52    2k  1  2k  1
2 2
    
Sk Tk 1
1
 k 2k  12k  1  2k  1
2
3
1 
 2k  1 k 2k  1  2k  1
3 

23. 12  32  52    2k  1
2
Proof:
 12  32  52    2k  1  2k  1
2 2
    
Sk Tk 1
1
 k 2k  12k  1  2k  1
2
3
1 
 2k  1 k 2k  1  2k  1
3 
1
 2k  1k 2k  1  32k  1
3

24. 12  32  52    2k  1
2
Proof:
 12  32  52    2k  1  2k  1
2 2
    
Sk Tk 1
1
 k 2k  12k  1  2k  1
2
3
1 
 2k  1 k 2k  1  2k  1
3 
1
 2k  1k 2k  1  32k  1
3
 2k  12k 2  k  6k  3
1
3
 2k  12k 2  5k  3
1
3

25. 12  32  52    2k  1
2
Proof:
 12  32  52    2k  1  2k  1
2 2
    
Sk Tk 1
1
 k 2k  12k  1  2k  1
2
3
1 
 2k  1 k 2k  1  2k  1
3 
1
 2k  1k 2k  1  32k  1
3
 2k  12k 2  k  6k  3
1
3
 2k  12k 2  5k  3
1
3
1
 2k  1k  12k  3
3

26. 12  32  52    2k  1
2
Proof:
 12  32  52    2k  1  2k  1
2 2
    
Sk Tk 1
1
 k 2k  12k  1  2k  1
2
3
1 
 2k  1 k 2k  1  2k  1
3 
1
 2k  1k 2k  1  32k  1
3
 2k  12k 2  k  6k  3
1
3
 2k  12k 2  5k  3
1
3
1
 2k  1k  12k  3
3
Hence the result is true for n = k + 1 if it is also true for n = k

27. Step 4: Since the result is true for n = 1, then the result is true for
all positive integral values of n by induction

28. n
1 n
ii   
k 1 2k  12k  1 2n  1

29. n
1 n
ii   
k 1 2k  12k  1 2n  1
1 1 1 1 n
   
1 3 3  5 5  7 2n  12n  1 2n  1

30. n
1 n
ii   
k 1 2k  12k  1 2n  1
1 1 1 1 n
   
1 3 3  5 5  7 2n  12n  1 2n  1
Step 1: Prove the result is true for n = 1

31. n
1 n
ii   
k 1 2k  12k  1 2n  1
1 1 1 1 n
   
1 3 3  5 5  7 2n  12n  1 2n  1
Step 1: Prove the result is true for n = 1
1
LHS 
1 3
1

3

32. n
1 n
ii   
k 1 2k  12k  1 2n  1
1 1 1 1 n
   
1 3 3  5 5  7 2n  12n  1 2n  1
Step 1: Prove the result is true for n = 1
1 1
LHS  RHS 
1 3 2 1
1 1
 
3 3

33. n
1 n
ii   
k 1 2k  12k  1 2n  1
1 1 1 1 n
   
1 3 3  5 5  7 2n  12n  1 2n  1
Step 1: Prove the result is true for n = 1
1 1
LHS  RHS 
1 3 2 1
1 1
 
3 3
 LHS  RHS

34. n
1 n
ii   
k 1 2k  12k  1 2n  1
1 1 1 1 n
   
1 3 3  5 5  7 2n  12n  1 2n  1
Step 1: Prove the result is true for n = 1
1 1
LHS  RHS 
1 3 2 1
1 1
 
3 3
 LHS  RHS
Hence the result is true for n = 1

35. n
1 n
ii   
k 1 2k  12k  1 2n  1
1 1 1 1 n
   
1 3 3  5 5  7 2n  12n  1 2n  1
Step 1: Prove the result is true for n = 1
1 1
LHS  RHS 
1 3 2 1
1 1
 
3 3
 LHS  RHS
Hence the result is true for n = 1
Step 2: Assume the result is true for n = k, where k is a positive
integer

36. n
1 n
ii   
k 1 2k  12k  1 2n  1
1 1 1 1 n
   
1 3 3  5 5  7 2n  12n  1 2n  1
Step 1: Prove the result is true for n = 1
1 1
LHS  RHS 
1 3 2 1
1 1
 
3 3
 LHS  RHS
Hence the result is true for n = 1
Step 2: Assume the result is true for n = k, where k is a positive
integer
1 1 1 1 k
i.e.    
1 3 3  5 5  7 2k  12k  1 2k  1

37. Step 3: Prove the result is true for n = k + 1

38. Step 3: Prove the result is true for n = k + 1
1 1 1 1 k 1
i.e. Prove :    
1 3 3  5 5  7 2k  12k  3 2k  3

39. Step 3: Prove the result is true for n = k + 1
1 1 1 1 k 1
i.e. Prove :    
1 3 3  5 5  7 2k  12k  3 2k  3
Proof:

40. Step 3: Prove the result is true for n = k + 1
1 1 1 1 k 1
i.e. Prove :    
1 3 3  5 5  7 2k  12k  3 2k  3
Proof:
1 1 1 1
  
1 3 3  5 5  7 2k  12k  3
1 1 1 1 1
    
1 3 3  5 5  7 2k  12k  1 2k  12k  3

41. Step 3: Prove the result is true for n = k + 1
1 1 1 1 k 1
i.e. Prove :    
1 3 3  5 5  7 2k  12k  3 2k  3
Proof:
1 1 1 1
  
1 3 3  5 5  7 2k  12k  3
1 1 1 1 1
    
1 3 3  5 5  7 2k  12k  1 2k  12k  3
k 1
 
2k  1 2k  12k  3

42. Step 3: Prove the result is true for n = k + 1
1 1 1 1 k 1
i.e. Prove :    
1 3 3  5 5  7 2k  12k  3 2k  3
Proof:
1 1 1 1
  
1 3 3  5 5  7 2k  12k  3
1 1 1 1 1
    
1 3 3  5 5  7 2k  12k  1 2k  12k  3
k 1
 
2k  1 2k  12k  3
k 2k  3  1

2k  12k  3

43. Step 3: Prove the result is true for n = k + 1
1 1 1 1 k 1
i.e. Prove :    
1 3 3  5 5  7 2k  12k  3 2k  3
Proof:
1 1 1 1
  
1 3 3  5 5  7 2k  12k  3
1 1 1 1 1
    
1 3 3  5 5  7 2k  12k  1 2k  12k  3
k 1
 
2k  1 2k  12k  3
k 2k  3  1

2k  12k  3
2k 2  3k  1

2k  12k  3

44. Step 3: Prove the result is true for n = k + 1
1 1 1 1 k 1
i.e. Prove :    
1 3 3  5 5  7 2k  12k  3 2k  3
Proof:
1 1 1 1
  
1 3 3  5 5  7 2k  12k  3
1 1 1 1 1
    
1 3 3  5 5  7 2k  12k  1 2k  12k  3
k 1
 
2k  1 2k  12k  3
k 2k  3  1

2k  12k  3
2k 2  3k  1

2k  12k  3

2k  1k  1
2k  12k  3

45. 
k  1
2k  3

46. 
k  1
2k  3
Hence the result is true for n = k + 1 if it is also true for n = k

47. 
k  1
2k  3
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

48. 
k  1
2k  3
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
Exercise 6N; 1 ace etc, 10(polygon), 13