# 11X1 T14 09 mathematical induction 2 (2010)

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• 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 = 1 123 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 = 1 123 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 = 1 123 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 = 1 123 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 = 1 123 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 1k 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 = 1 123 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 1k 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 = 1 123 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 1k 2 3P, where P is an integer Step 3: Prove the result is true for n = k + 1 i.e. Prove : k 1k 2k 3 3Q, where Q is an integer
• 11. Proof:
• 12. Proof: k 1k 2k 3
• 13. Proof: k 1k 2k 3 k k 1k 2 3k 1k 2
• 14. Proof: k 1k 2k 3 k k 1k 2 3k 1k 2 3P 3k 1k 2
• 15. Proof: k 1k 2k 3 k k 1k 2 3k 1k 2 3P 3k 1k 2 3P k 1k 2
• 16. Proof: k 1k 2k 3 k k 1k 2 3k 1k 2 3P 3k 1k 2 3P k 1k 2 3Q, where Q P k 1k 2 which is an integer
• 17. Proof: k 1k 2k 3 k k 1k 2 3k 1k 2 3P 3k 1k 2 3P k 1k 2 3Q, where Q P k 1k 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 1k 2k 3 k k 1k 2 3k 1k 2 3P 3k 1k 2 3P k 1k 2 3Q, where Q P k 1k 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 2n2 is divisible by 5
• 20. iv Prove 33n 2n2 is divisible by 5 Step 1: Prove the result is true for n = 1
• 21. iv Prove 33n 2n2 is divisible by 5 Step 1: Prove the result is true for n = 1 33 23 27 8 35
• 22. iv Prove 33n 2n2 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 2n2 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 2n2 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 2n2 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 275P 2k 2 2k 3
• 30. Proof: 33k 3 2k 3 27 33k 2k 3 275P 2k 2 2k 3 135P 27 2k 2 2k 3
• 31. Proof: 33k 3 2k 3 27 33k 2k 3 275P 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 275P 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 275P 2k 2 2k 3 135P 27 2k 2 2k 3 135P 27 2k 2 2 2k 2 135P 25 2k 2 527 P 5 2k 2
• 34. Proof: 33k 3 2k 3 27 33k 2k 3 275P 2k 2 2k 3 135P 27 2k 2 2k 3 135P 27 2k 2 2 2k 2 135P 25 2k 2 527 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 275P 2k 2 2k 3 135P 27 2k 2 2k 3 135P 27 2k 2 2 2k 2 135P 25 2k 2 527 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 275P 2k 2 2k 3 135P 27 2k 2 2k 3 135P 27 2k 2 2 2k 2 135P 25 2k 2 527 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 275P 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 2 527 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.