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# 11X1 T14 10 mathematical induction 3 (2010)

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

1. 1. Mathematical Induction
2. 2. Mathematical Induction e.g.v  Prove 2n  n 2 for n  4
3. 3. Mathematical Induction e.g.v  Prove 2n  n 2 for n  4 Step 1: Prove the result is true for n = 5
4. 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. 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. 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. 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. 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. 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. 10. Proof:
11. 11. Proof: 2 k 1
12. 12. Proof: 2 k 1  2 2k
13. 13. Proof: 2 k 1  2 2k  2k 2
14. 14. Proof: 2 k 1  2 2k  2k 2  k2  k2
15. 15. Proof: 2 k 1  2 2k  2k 2  k2  k2  k2  k k
16. 16. Proof: 2 k 1  2 2k  2k 2  k2  k2  k2  k k  k 2  4k
17. 17. Proof: 2 k 1  2 2k  2k 2  k2  k2  k2  k k  k 2  4k  k  4
18. 18. Proof: 2 k 1  2 2k  2k 2  k2  k2  k2  k k  k 2  4k  k  4  k 2  2k  2k
19. 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. 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. 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. 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. 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. 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. 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. 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 .