11X1 T14 09 mathematical induction 2 (2011)

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<ul><li> 1. Mathematical Induction</li></ul> <p> 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 = 1123 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 = 1123 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 = 1123 6which 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 = 1123 6which 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 = 1123 6which 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 integeri.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 = 1123 6which 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 integeri.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 = 1123 6which 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 integeri.e. k k 1k 2 3P, where P is an integer Step 3: Prove the result is true for n = k + 1i.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 integerHence 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 integerHence 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 forall positive integral values of n by induction. 19. iv Prove 33n 2n2 is divisible by 5 20. iv Prove 33n 2n2 is divisible by 5Step 1: Prove the result is true for n = 1 21. iv Prove 33n 2n2 is divisible by 5Step 1: Prove the result is true for n = 133 23 27 8 35 22. iv Prove 33n 2n2 is divisible by 5Step 1: Prove the result is true for n = 133 23 27 8 35 which is divisible by 5 23. iv Prove 33n 2n2 is divisible by 5Step 1: Prove the result is true for n = 133 23 27 8 35 which is divisible by 5Hence the result is true for n = 1 24. iv Prove 33n 2n2 is divisible by 5Step 1: Prove the result is true for n = 133 23 27 8 35 which is divisible by 5Hence the result is true for n = 1Step 2: Assume the result is true for n = k, where k is a positiveintegeri.e. 33k 2k 2 5P, where P is an integer 25. iv Prove 33n 2n2 is divisible by 5Step 1: Prove the result is true for n = 133 23 27 8 35 which is divisible by 5Hence the result is true for n = 1Step 2: Assume the result is true for n = k, where k is a positiveintegeri.e. 33k 2k 2 5P, where P is an integerStep 3: Prove the result is true for n = k + 1i.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 integerHence 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 integerHence 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 forall 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 3Exercise 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 integerHence 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 forall positive integral values of n by induction.</p>