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11X1 T14 10 mathematical induction 3 (2010)
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11X1 T14 10 mathematical induction 3 (2010)

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  • 1. Mathematical Induction
  • 2. Mathematical Induction e.g.v  Prove 2n  n 2 for n  4
  • 3. Mathematical Induction e.g.v  Prove 2n  n 2 for n  4 Step 1: Prove the result is true for n = 5
  • 4. Mathematical Induction e.g.v  Prove 2n  n 2 for n  4 Step 1: Prove the result is true for n = 5 LHS  25  32
  • 5. Mathematical Induction e.g.v  Prove 2n  n 2 for n  4 Step 1: Prove the result is true for n = 5 LHS  25 RHS  52  32  25
  • 6. Mathematical Induction e.g.v  Prove 2n  n 2 for n  4 Step 1: Prove the result is true for n = 5 LHS  25 RHS  52  32  25  LHS  RHS
  • 7. Mathematical Induction e.g.v  Prove 2n  n 2 for n  4 Step 1: Prove the result is true for n = 5 LHS  25 RHS  52  32  25  LHS  RHS Hence the result is true for n = 5
  • 8. Mathematical Induction e.g.v  Prove 2n  n 2 for n  4 Step 1: Prove the result is true for n = 5 LHS  25 RHS  52  32  25  LHS  RHS Hence the result is true for n = 5 Step 2: Assume the result is true for n = k, where k is a positive integer > 4 i.e. 2k  k 2
  • 9. Mathematical Induction e.g.v  Prove 2n  n 2 for n  4 Step 1: Prove the result is true for n = 5 LHS  25 RHS  52  32  25  LHS  RHS Hence the result is true for n = 5 Step 2: Assume the result is true for n = k, where k is a positive integer > 4 i.e. 2k  k 2 Step 3: Prove the result is true for n = k + 1 k 1  k  1 2 i.e. Prove : 2
  • 10. Proof:
  • 11. Proof: 2 k 1
  • 12. Proof: 2 k 1  2 2k
  • 13. Proof: 2 k 1  2 2k  2k 2
  • 14. Proof: 2 k 1  2 2k  2k 2  k2  k2
  • 15. Proof: 2 k 1  2 2k  2k 2  k2  k2  k2  k k
  • 16. Proof: 2 k 1  2 2k  2k 2  k2  k2  k2  k k  k 2  4k
  • 17. Proof: 2 k 1  2 2k  2k 2  k2  k2  k2  k k  k 2  4k  k  4
  • 18. Proof: 2 k 1  2 2k  2k 2  k2  k2  k2  k k  k 2  4k  k  4  k 2  2k  2k
  • 19. Proof: 2 k 1  2 2k  2k 2  k2  k2  k2  k k  k 2  4k  k  4  k 2  2k  2k  k 2  2k  8
  • 20. Proof: 2 k 1  2 2k  2k 2  k2  k2  k2  k k  k 2  4k  k  4  k 2  2k  2k  k 2  2k  8  k  4
  • 21. Proof: 2 k 1  2 2k  2k 2  k2  k2  k2  k k  k 2  4k  k  4  k 2  2k  2k  k 2  2k  8  k  4  k 2  2k  1
  • 22. Proof: 2 k 1  2 2k  2k 2  k2  k2  k2  k k  k 2  4k  k  4  k 2  2k  2k  k 2  2k  8  k  4  k 2  2k  1  k  1 2
  • 23. Proof: 2 k 1  2 2k  2k 2  k2  k2  k2  k k  k 2  4k  k  4  k 2  2k  2k  k 2  2k  8  k  4  k 2  2k  1  k  1 2  2  k  1 k 1 2
  • 24. Proof: 2 k 1  2 2k  2k 2  k2  k2  k2  k k  k 2  4k  k  4  k 2  2k  2k  k 2  2k  8  k  4  k 2  2k  1  k  1 2  2  k  1 k 1 2 Hence the result is true for n = k + 1 if it is also true for n = k
  • 25. Proof: 2 k 1  2 2k  2k 2  k2  k2  k2  k k  k 2  4k  k  4  k 2  2k  2k  k 2  2k  8  k  4  k 2  2k  1  k  1 2  2  k  1 k 1 2 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 = 5, then the result is true for all positive integral values of n > 4 by induction .
  • 26. Proof: 2 k 1  2 2k  2k 2  k2  k2  k2  k k  k 2  4k  k  4 Exercise 6N;  k 2  2k  2k 6 abc, 8a, 15  k 2  2k  8  k  4  k 2  2k  1  k  1 2  2  k  1 k 1 2 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 = 5, then the result is true for all positive integral values of n > 4 by induction .

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