The document uses mathematical induction to prove that 2n is greater than n^2 for all n greater than 4. It shows that the statement holds for n=5. It then assumes the statement is true for some integer k greater than 4, and proves that it must also be true for k+1. Therefore, by the principle of mathematical induction, the statement is true for all integers n greater than 4.

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 .