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Mathematical Induction

1. Mathematical Induction. Case Study. 1.1 The Principle of Mathematical Induction. 1.2 Divisibility. Chapter Summary. How can I find the total number of cans in all layers?. That’s easy! We can find a formula to calculate it. Let me explain it to you.

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Mathematical Induction

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

  2. How can I find the total number of cans in all layers? That’s easy! We can find a formula to calculate it. Let me explain it to you. The manager of the supermarket suggests that there should be a total of n(n 1)(n 2) cans in n layers. Case Study 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. Do you think that his formula is correct? If yes, how can you prove it?

  3. 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! 1.1 The Principle of Mathematical Induction Consider the following sums: S(1)  S(2)  S(3)  S(4)  We can guess that the result of S(n):

  4. 1.1 The Principle of Mathematical Induction Instead,wecanuseasimplemethodcalledmathematical induction to prove that this kind of propositions is true for all positive integers n. 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. 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. Now, let us apply Mathematical Induction to prove some propositions as illustrated in Example 1.1T, 1.2T and 1.3T.

  5. Use the assumption 1.1 The Principle of 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)

  6. When n k  1, prove that L.H.S.  . Take out the common factor 1.1 The Principle of 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)  R.H.S. ∴ The proposition is true for n k  1. ∴ By the principle of mathematical induction, the proposition is true for all positive integers n.

  7. (a) Prove, by mathematical induction, that for all positive integers n. 1.1 The Principle of Mathematical Induction Example 1.2T Solution: For n 1, L.H.S. 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.

  8. 1.1 The Principle of 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. n 20 n 9 1.1 The Principle of Mathematical Induction Example 1.2T (a) Prove, by mathematical induction, that for all positive integers n. (b) Hence find the value of . Solution:

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

  11. 1.1 The Principle of Mathematical Induction Example 1.3T (a) 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  3  4  5  6    (2k  1)(2k)  (2k  1)(2k  2)

  12. 1.1 The Principle of Mathematical Induction Example 1.3T (a) Prove, by mathematical induction, that for all positive integers n. 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.

  13. n 49 n 49 1.1 The Principle of Mathematical Induction Example 1.3T (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.] Solution:

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

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

  16. 1.2 Divisibility Example 1.5T Prove, by mathematical induction, that 3  52n  1 23n  1is divisible by 17 for all positive integers n. Solution: For n 1, 3  52n  1 23n  1 3  53 24  391, which is divisible by 17. ∴ The proposition is true for n 1. Next, assume for some positive integers k, 3  52k  1 23k  1is divisible by 17, that is, 3  52k  1 23k  1 17m, where m is a positive integer. When n k  1, ∴ 3  52(k  1)  1 23(k  1)  1is divisible by 17. ∴ The proposition is true for n k  1.

  17. 1.2 Divisibility Example 1.6T Prove, by mathematical induction, that 6n  1 – 5(n  1) – 1 is divisible by 25 for all positive integers n. Solution: For n 1, 6n  1 – 5(n  1) – 1  62 – 5(2) – 1  25, which is divisible by 25. ∴ The proposition is true for n 1. Next, assume for some positive integers k, 6k1–5(k1)–1 is divisible by 25, that is, 6k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. Chapter Summary 1.1 Principle of Mathematical Induction 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. Chapter Summary 1.2 Divisibility Suppose p and q are two integers. p is divisible by q (q 0) if pqm, where m is an integer.

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