1. 1.1The Principle of Mathematical Induction 1.2Divisibility Chapter Summary Case Study Mathematical Induction 1

2. P. 2 In a supermarket, some cans of coke are being piled up such that the top layer has 1 can, the second layer has 1 + 2  3 cans, the third layer has 1  2  3  6 cans, and so on. Case Study The manager of the supermarket suggests that there should be a total of n(n  1)(n  2) cans in n layers. Do you think that his formula is correct? If yes, how can you prove it? That’s easy! We can find a formula to calculate it. Let me explain it to you. How can I find the total number of cans in all layers?

3. P. 3 Consider the following sums: 1.1 The Principle of Mathematical Induction Mathematical Induction S(1)  S(2)  S(3)  S(4)  We can guess that the result of S(n): However, to prove that S(n)  is true, we have to check whether the formula is true for every positive integer n, which is impossible to do so!

4. P. 4 Instead, we can use a simple method called mathematical induction to prove that this kind of propositions is true for all positive integers n. 1.1 The Principle of Mathematical Induction Mathematical Induction Principle of Mathematical Induction Let P(n) be a proposition that involves a positive integer n. Then P(n) is true for all positive integers n if both of the following conditions are satisfied: (I)P(1) is true. (II)For any positive integer k, if P(k) is assumed to be true, then P(k  1) is also true. Now, let us apply Mathematical Induction to prove some propositions as illustrated in Example 1.1T, 1.2T and 1.3T. In other words, we only need to check two conditions, (I) and (II), in order to show that a proposition P(n) is true for all positive integers n.

5. P. 5 1.1 The Principle of Mathematical Induction Mathematical Induction Example 1.1T Prove, by mathematical induction, that for all positive integers n. Solution: For n  1, L.H.S.  1  2  2 R.H.S. ∴ The proposition is true for n  1. Next, assume the proposition is true for some positive integers k, that is, When n  k  1, L.H.S.  1  2  2  3  3  4    k(k  1)  (k  1)(k  2) Use the assumption

6. P. 6 1.1 The Principle of Mathematical Induction Mathematical Induction Example 1.1T Prove, by mathematical induction, that for all positive integers n. Solution: When n  k  1, L.H.S.  1  2  2  3  3  4    k(k  1)  (k  1)(k  2) When n  k  1, prove that L.H.S. . ∴ The proposition is true for n  k  1.  R.H.S. ∴ By the principle of mathematical induction, the proposition is true for all positive integers n. Take out the common factor

7. P. 7 1.1 The Principle of Mathematical Induction Mathematical Induction Example 1.2T Solution: For n  1, L.H.S.R.H.S. (a)Prove, by mathematical induction, that for all positive integers n. ∴ The proposition is true for n  1. Next, assume the proposition is true for some positive integers k, that is, When n  k  1, L.H.S.

8. P. 8 1.1 The Principle of Mathematical Induction Mathematical Induction Example 1.2T (a)Prove, by mathematical induction, that for all positive integers n. Solution: When n  k  1, L.H.S.  R.H.S. ∴ The proposition is true for n  k  1.

9. P. 9 1.1 The Principle of Mathematical Induction Mathematical Induction Example 1.2T (a)Prove, by mathematical induction, that for all positive integers n. (b)Hence find the value of. Solution: n  20n  9

10. P. 10 Remarks: Although mathematical induction is a very good method for proving a proposition, it may NOT be the only method in some cases. 1.1 The Principle of Mathematical Induction Mathematical Induction For instance, the proposition in Follow-up 1.2 can also be proved by using the following fact:

11. P. 11 1.1 The Principle of Mathematical Induction Mathematical Induction Example 1.3T Solution: For n  1, L.H.S.  1  2  2 R.H.S. ∴ The proposition is true for n  1. Next, assume the proposition is true for some positive integers k, that is, When n  k  1, L.H.S.  1  2  3  4  5  6    (2k  1)(2k)  (2k  1)(2k  2) (a)Prove, by mathematical induction, that for all positive integers n.

12. P. 12 1.1 The Principle of Mathematical Induction Mathematical Induction Example 1.3T Solution: When n  k  1, L.H.S.  1  2  3  4  5  6    (2k  1)(2k)  (2k  1)(2k  2)  R.H.S. ∴ The proposition is true for n  k  1. (a)Prove, by mathematical induction, that for all positive integers n.

13. P. 13 1.1 The Principle of Mathematical Induction Mathematical Induction Example 1.3T Solution: n  49 (a)Prove, by mathematical induction, that for all positive integers n. (b)Hence evaluate 2  4  4  6  6  8 ...  98  100. [Hint: You may use the fact 2  4  6 ...  2n  n(n  1) without proof.]

14. P. 14 Apart from proving propositions that involve summation, mathematical induction can also be used in many other situations. 1.2 Divisibility For example, 28 is divisible by 4 as 28  4  7. Suppose p and q are two integers, where q  0. If p  qm, where m is an integer, then p is divisible by q.

15. P. 15 1.2 Divisibility Example 1.4T Prove, by mathematical induction, that 11 n – 4 n is divisible by 7 for all positive integers n. Solution: For n  1, 11 n – 4 n  11 – 4  7, which is divisible by 7. ∴ The proposition is true for n  1. Next, assume for some positive integers k, 11 k – 4 k is divisible by 7, that is, 11 k – 4 k  7m, where m is a positive integer. When n  k  1, ∴ 11 k + 1 – 4 k + 1 is divisible by 7. ∴ The proposition is true for n  k  1.

16. P. 16 1.2 Divisibility Example 1.5T Prove, by mathematical induction, that 3  5 2n  1  2 3n  1 is divisible by 17 for all positive integers n. Solution: For n  1, 3  5 2n  1  2 3n  1  3  5 3  2 4  391, which is divisible by 17. ∴ The proposition is true for n  1. Next, assume for some positive integers k, 3  5 2k  1  2 3k  1 is divisible by 17, that is, 3  5 2k  1  2 3k  1  17m, where m is a positive integer. When n  k  1, ∴ 3  5 2(k  1)  1  2 3(k  1)  1 is divisible by 17. ∴ The proposition is true for n  k  1.

17. P. 17 1.2 Divisibility Example 1.6T Prove, by mathematical induction, that 6 n  1 – 5(n  1) – 1 is divisible by 25 for all positive integers n. Solution: For n  1, 6 n  1 – 5(n  1) – 1  6 2 – 5(2) – 1  25, which is divisible by 25. ∴ The proposition is true for n  1. Next, assume for some positive integers k, 6 k  1 – 5(k  1) – 1 is divisible by 25, that is, 6 k  1 – 5(k  1) – 1  25m, where m is a positive integer. When n  k  1, ∴ 6 (k  1)  1 – 5[(k  1)  1] – 1 is divisible by 25. ∴ The proposition is true for n  k  1.

18. P. 18 1.1 Principle of Mathematical Induction Chapter Summary A proposition P(n) is true for all positive integers n if both of the following conditions are satisfied: (I)P(1) is true. (II)For any positive integer k, if P(k) is assumed to be true, then P(k  1) is also true.

19. P. 19 1.2 Divisibility Chapter Summary Suppose p and q are two integers. p is divisible by q (q  0) if p  qm, where m is an integer.

20. Follow-up 1.1 1.1 The Principle of Mathematical Induction Mathematical Induction Prove, by mathematical induction, that for all positive integers n. Solution: For n  1, L.H.S.  1  3  3 R.H.S. ∴ The proposition is true for n  1. Next, assume the proposition is true for some positive integers k, that is, When n  k  1, L.H.S.  1  3  3  5  5  7    (2k  1)(2k  1)  (2k  1)(2k  3)

21. 1.1 The Principle of Mathematical Induction Mathematical Induction Follow-up 1.1 Prove, by mathematical induction, that for all positive integers n. Solution: When n  k  1, L.H.S.  1  3  3  5  5  7    (2k  1)(2k  1)  (2k  1)(2k  3)  R.H.S. ∴ The proposition is true for n  k  1.

22. Follow-up 1.2 1.1 The Principle of Mathematical Induction Mathematical Induction Solution: For n  1, L.H.S.R.H.S. (a)Prove, by mathematical induction, that for all positive integers n. ∴ The proposition is true for n  1. Next, assume the proposition is true for some positive integers k, that is, When n  k  1, L.H.S.

23. Follow-up 1.2 1.1 The Principle of Mathematical Induction Mathematical Induction Solution: (a)Prove, by mathematical induction, that for all positive integers n. When n  k  1, L.H.S. ∴ The proposition is true for n  k  1.  R.H.S.

24. Follow-up 1.2 1.1 The Principle of Mathematical Induction Mathematical Induction Solution: (a)Prove, by mathematical induction, that for all positive integers n. (b)Hence find the value of. n  31n  10

25. Follow-up 1.3 1.1 The Principle of Mathematical Induction Mathematical Induction Solution: For n  1, L.H.S.  1  5  5 R.H.S. ∴ The proposition is true for n  1. Next, assume the proposition is true for some positive integers k, that is, When n  k  1, L.H.S.  1  5  2  6  3  7    k(k  4)  (k  1)(k  5) (a)Prove, by mathematical induction, that for all positive integers n.

26. Follow-up 1.3 1.1 The Principle of Mathematical Induction Mathematical Induction When n  k  1, L.H.S.  1  5  2  6  3  7    k(k  4)  (k  1)(k  5)  R.H.S. ∴ The proposition is true for n  k  1. Solution: (a)Prove, by mathematical induction, that for all positive integers n.

27. Follow-up 1.3 1.1 The Principle of Mathematical Induction Mathematical Induction (a)Prove, by mathematical induction, that for all positive integers n. (b)Hence evaluate 1  6  2  7  3  8 ...  40  45. [Hint: You may use the fact 2  4  6    2n  n(n  1) without proof.] Solution: n  40

28. Follow-up 1.4 1.2 Divisibility Prove, by mathematical induction, that 19 n – 7 n is divisible by 12 for all positive integers n. Solution: For n  1, 19 n – 7 n  19 – 7  12, which is divisible by 12. ∴ The proposition is true for n  1. Next, assume for some positive integers k, 19 k – 7 k is divisible by 12, that is, 19 k – 7 k  12m, where m is a positive integer. When n  k  1, ∴ 19 k + 1 – 7 k + 1 is divisible by 12. ∴ The proposition is true for n  k  1.

29. Follow-up 1.5 1.2 Divisibility Prove, by mathematical induction, that 8  9 2n  1  2 3n  3 is divisible by 73 for all positive integers n. Solution: For n  1, 8  9 2n  1  2 3n  3  8  9 1  2 0  73, which is divisible by 73. ∴ The proposition is true for n  1. Next, assume for some positive integers k, 8  9 2k  1  2 3k  3 is divisible by 73, that is, 8  9 2k  1  2 3k  3  73m, where m is a positive integer. When n  k  1, ∴ 8  9 2(k  1)  1  2 3(k  1)  3 is divisible by 73. ∴ The proposition is true for n  k  1.

30. Follow-up 1.6 1.2 Divisibility Prove, by mathematical induction, that 7 n  1 – 6n  29 is divisible by 36 for all positive integers n. Solution: For n  1, 7 n  1 – 6n  29  7 2 – 6  29  72, which is divisible by 36. ∴ The proposition is true for n  1. Next, assume for some positive integers k, 7 k  1 – 6k  29 is divisible by 36, that is, 7 k  1 – 6k  29  36m, where m is a positive integer. When n  k  1, ∴ 7 (k  1)  1 – 6(k  1)  29 is divisible by 36. ∴ The proposition is true for n  k  1.