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# 11X1 T14 08 mathematical induction 1 (2011)

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### 11X1 T14 08 mathematical induction 1 (2011)

1. 1. Mathematical Induction
2. 2. Mathematical InductionStep 1: Prove the result is true for n = 1 (or whatever the first term is)
3. 3. Mathematical InductionStep 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. 4. Mathematical InductionStep 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. 5. Mathematical InductionStep 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 + 1NOTE: 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. 6. Mathematical InductionStep 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 + 1NOTE: It is important to note in your conclusion that the result is true for n = k + 1 if it is true for n = kStep 4: Since the result is true for n = 1, then the result is true for all positive integral values of n by induction
7. 7. 1e.g.i  1  3  5    2n  1  n2n  12n  1 2 2 2 2 3
8. 8. 1e.g.i  1  3  5    2n  1  n2n  12n  1 2 2 2 2 3Step 1: Prove the result is true for n = 1
9. 9. 1e.g.i  1  3  5    2n  1  n2n  12n  1 2 2 2 2 3Step 1: Prove the result is true for n = 1 LHS  12 1
10. 10. 1e.g.i  1  3  5    2n  1  n2n  12n  1 2 2 2 2 3Step 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. 11. 1e.g.i  1  3  5    2n  1  n2n  12n  1 2 2 2 2 3Step 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. 12. 1e.g.i  1  3  5    2n  1  n2n  12n  1 2 2 2 2 3Step 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. 13. 1e.g.i  1  3  5    2n  1  n2n  12n  1 2 2 2 2 3Step 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 = 1Step 2: Assume the result is true for n = k, where k is a positive integer
14. 14. 1e.g.i  1  3  5    2n  1  n2n  12n  1 2 2 2 2 3Step 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 = 1Step 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. 15. 1e.g.i  1  3  5    2n  1  n2n  12n  1 2 2 2 2 3Step 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 = 1Step 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 3Step 3: Prove the result is true for n = k + 1
16. 16. 1e.g.i  1  3  5    2n  1  n2n  12n  1 2 2 2 2 3Step 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 = 1Step 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 3Step 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. 17. Proof:
18. 18. 12  32  52    2k  1 2Proof:  12  32  52    2k  1  2k  1 2 2
19. 19. 12  32  52    2k  1 2Proof:  12  32  52    2k  1  2k  1 2 2     Sk
20. 20. 12  32  52    2k  1 2Proof:  12  32  52    2k  1  2k  1 2 2      Sk Tk 1
21. 21. 12  32  52    2k  1 2Proof:  12  32  52    2k  1  2k  1 2 2      Sk Tk 1 1  k 2k  12k  1  2k  1 2 3
22. 22. 12  32  52    2k  1 2Proof:  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. 23. 12  32  52    2k  1 2Proof:  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. 24. 12  32  52    2k  1 2Proof:  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. 25. 12  32  52    2k  1 2Proof:  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. 26. 12  32  52    2k  1 2Proof:  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. 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. 28. n 1 nii    k 1 2k  12k  1 2n  1
29. 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. 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  1Step 1: Prove the result is true for n = 1
31. 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  1Step 1: Prove the result is true for n = 1 1 LHS  1 3 1  3
32. 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  1Step 1: Prove the result is true for n = 1 1 1 LHS  RHS  1 3 2 1 1 1   3 3
33. 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  1Step 1: Prove the result is true for n = 1 1 1 LHS  RHS  1 3 2 1 1 1   3 3  LHS  RHS
34. 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  1Step 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. 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  1Step 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 = 1Step 2: Assume the result is true for n = k, where k is a positive integer
36. 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  1Step 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 = 1Step 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. 37. Step 3: Prove the result is true for n = k + 1
38. 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. 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  3Proof:
40. 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  3Proof: 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. 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  3Proof: 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. 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  3Proof: 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. 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  3Proof: 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. 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  3Proof: 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. 45.  k  1 2k  3
46. 46.  k  1 2k  3Hence the result is true for n = k + 1 if it is also true for n = k
47. 47.  k  1 2k  3 Hence the result is true for n = k + 1 if it is also true for n = kStep 4: Since the result is true for n = 1, then the result is true for all positive integral values of n by induction
48. 48.  k  1 2k  3 Hence the result is true for n = k + 1 if it is also true for n = kStep 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