11X1 T14 08 mathematical induction 1 (2010)

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<ul><li> 1. Mathematical Induction </li></ul> <p> 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 n2n 12n 122 22 3 8. 1 e.g.i 1 3 5 2n 1 n2n 12n 122 223 Step 1: Prove the result is true for n = 1 9. 1 e.g.i 1 3 5 2n 1 n2n 12n 122 223 Step 1: Prove the result is true for n = 1 LHS 121 10. 1 e.g.i 1 3 5 2n 1 n2n 12n 1 2 22 23 Step 1: Prove the result is true for n = 1 1LHS 12RHS 12 12 1 31 1 113 31 11. 1 e.g.i 1 3 5 2n 1 n2n 12n 1 2 22 23 Step 1: Prove the result is true for n = 1 1LHS 12RHS 12 12 1 31 1 113 31 LHS RHS 12. 1 e.g.i 1 3 5 2n 1 n2n 12n 1 2 22 23 Step 1: Prove the result is true for n = 1 1LHS 12RHS 12 12 1 31 1 113 3 1 LHS RHSHence the result is true for n = 1 13. 1 e.g.i 1 3 5 2n 1 n2n 12n 1 2 22 23 Step 1: Prove the result is true for n = 1 1LHS 12RHS 12 12 1 31 1 113 3 1 LHS RHSHence the result is true for n = 1Step 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 n2n 12n 1 2 22 23 Step 1: Prove the result is true for n = 1 1LHS 12RHS 12 12 1 31 1 113 3 1 LHS RHSHence 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 12k 1 23 15. 1 e.g.i 1 3 5 2n 1 n2n 12n 1 2 22 23 Step 1: Prove the result is true for n = 1 1LHS 12RHS 12 12 1 31 1 113 3 1 LHS RHSHence the result is true for n = 1Step 2: Assume the result is true for n = k, where k is a positive integer1 i.e. 12 32 52 2k 1 k 2k 12k 1 2 3 Step 3: Prove the result is true for n = k + 1 16. 1 e.g.i 1 3 5 2n 1 n2n 12n 1 22 2 23 Step 1: Prove the result is true for n = 1 1LHS 12RHS 12 12 1 31 1 113 3 1 LHS RHSHence the result is true for n = 1Step 2: Assume the result is true for n = k, where k is a positiveinteger 1 i.e. 12 32 52 2k 1 k 2k 12k 1 23 Step 3: Prove the result is true for n = k + 1 1i.e. Prove : 12 32 52 2k 1 k 12k 12k 3 23 17. Proof: 18. 12 32 52 2k 12 Proof: 12 32 52 2k 1 2k 122 19. 12 32 52 2k 12 Proof: 12 32 52 2k 1 2k 122 Sk 20. 12 32 52 2k 12 Proof: 12 32 52 2k 1 2k 12 2 SkTk 1 21. 12 32 52 2k 12 Proof: 12 32 52 2k 1 2k 122 Sk Tk 1 1 k 2k 12k 1 2k 1 23 22. 12 32 52 2k 12 Proof: 12 32 52 2k 1 2k 12 2 Sk Tk 11 k 2k 12k 1 2k 12 31 2k 1 k 2k 1 2k 13 23. 12 32 52 2k 1 2 Proof: 12 32 52 2k 1 2k 1 22 SkTk 11 k 2k 12k 1 2k 12 31 2k 1 k 2k 1 2k 13 1 2k 1k 2k 1 32k 1 3 24. 12 32 52 2k 1 2 Proof: 12 32 52 2k 1 2k 1 22 SkTk 11 k 2k 12k 1 2k 12 31 2k 1 k 2k 1 2k 13 1 2k 1k 2k 1 32k 1 3 2k 12k 2 k 6k 313 2k 12k 2 5k 313 25. 12 32 52 2k 1 2 Proof: 12 32 52 2k 1 2k 1 22 SkTk 11 k 2k 12k 1 2k 12 31 2k 1 k 2k 1 2k 13 1 2k 1k 2k 1 32k 1 3 2k 12k 2 k 6k 313 2k 12k 2 5k 313 1 2k 1k 12k 3 3 26. 12 32 52 2k 1 2 Proof: 12 32 52 2k 1 2k 1 2 2 SkTk 1 1 k 2k 12k 1 2k 1 231 2k 1 k 2k 1 2k 13 1 2k 1k 2k 1 32k 13 2k 12k 2 k 6k 313 2k 12k 2 5k 3131 2k 1k 12k 33 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 12k 1 2n 1 29. n 1 n ii k 1 2k 12k 1 2n 1 111 1 n 1 3 3 5 5 72n 12n 1 2n 1 30. n 1 n ii k 1 2k 12k 1 2n 1 111 1 n 1 3 3 5 5 72n 12n 1 2n 1 Step 1: Prove the result is true for n = 1 31. n 1 n ii k 1 2k 12k 1 2n 1 111 1 n 1 3 3 5 5 72n 12n 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 12k 1 2n 1 111 1 n 1 3 3 5 5 72n 12n 1 2n 1 Step 1: Prove the result is true for n = 1 11 LHS RHS 1 3 2 1 1 1 3 3 33. n 1 n ii k 1 2k 12k 1 2n 1 111 1 n 1 3 3 5 5 72n 12n 1 2n 1 Step 1: Prove the result is true for n = 1 11 LHS RHS 1 3 2 1 1 1 3 3 LHS RHS 34. n 1 n ii k 1 2k 12k 1 2n 1 111 1 n 1 3 3 5 5 72n 12n 1 2n 1 Step 1: Prove the result is true for n = 1 11 LHS RHS 1 3 2 1 1 1 3 3 LHS RHSHence the result is true for n = 1 35. n 1 n ii k 1 2k 12k 1 2n 1 111 1 n 1 3 3 5 5 72n 12n 1 2n 1 Step 1: Prove the result is true for n = 1 11 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 12k 1 2n 1 111 1 n 1 3 3 5 5 72n 12n 1 2n 1 Step 1: Prove the result is true for n = 1 11 LHS RHS 1 3 2 1 1 1 3 3 LHS RHSHence the result is true for n = 1 Step 2: Assume the result is true for n = k, where k is a positive integer1 11 1 k i.e. 1 3 3 5 5 7 2k 12k 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 111k 1 i.e. Prove : 1 3 3 5 5 72k 12k 3 2k 3 39. Step 3: Prove the result is true for n = k + 1 1 111k 1 i.e. Prove : 1 3 3 5 5 72k 12k 3 2k 3 Proof: 40. Step 3: Prove the result is true for n = k + 1 1 111 k 1 i.e. Prove : 1 3 3 5 5 72k 12k 3 2k 3 Proof:11 1 1 1 3 3 5 5 7 2k 12k 311 11 1 1 3 3 5 5 7 2k 12k 1 2k 12k 3 41. Step 3: Prove the result is true for n = k + 1 1 111 k 1 i.e. Prove : 1 3 3 5 5 72k 12k 3 2k 3 Proof:11 1 1 1 3 3 5 5 7 2k 12k 311 11 1 1 3 3 5 5 7 2k 12k 1 2k 12k 3 k1 2k 1 2k 12k 3 42. Step 3: Prove the result is true for n = k + 1 1 111 k 1 i.e. Prove : 1 3 3 5 5 72k 12k 3 2k 3 Proof:1 11 1 1 3 3 5 5 7 2k 12k 31 111 1 1 3 3 5 5 7 2k 12k 1 2k 12k 3k 1 2k 1 2k 12k 3 k 2k 3 1 2k 12k 3 43. Step 3: Prove the result is true for n = k + 1 1 111 k 1 i.e. Prove : 1 3 3 5 5 72k 12k 3 2k 3 Proof:1 11 1 1 3 3 5 5 7 2k 12k 31 111 1 1 3 3 5 5 7 2k 12k 1 2k 12k 3k 1 2k 1 2k 12k 3 k 2k 3 1 2k 12k 3 2k 2 3k 1 2k 12k 3 44. Step 3: Prove the result is true for n = k + 1 1 111 k 1 i.e. Prove : 1 3 3 5 5 72k 12k 3 2k 3 Proof:1 11 1 1 3 3 5 5 7 2k 12k 31 111 1 1 3 3 5 5 7 2k 12k 1 2k 12k 3k 1 2k 1 2k 12k 3 k 2k 3 1 2k 12k 3 2k 2 3k 1 2k 12k 3 2k 1k 1 2k 12k 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 12k 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. k 12k 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 </p>