Problem 1: Prove that for every integer n > 7, (n – 2)! > n² . (Hint: using the requirement n >7 as well as the regular parts of your induction hypothesiscan be useful.)Problem 2: Define a sequence rn where ro9, ri = 12, r2 = -6, and for integers n > 3, rn = 7rn-3. Prove that for all nonnegative integers n, 3 rn (thatis, each term in the sequence is divisible by 3).

We use induction hypothesis to prove the result.

Answered: Problem 1: Prove that for every integer…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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4. Let P(n) be the statement that a postage of ŉ cents can be formed using just 4-cent stamps and 7-cent stamps. The parts Page 363 of this exercise outline a strong induction proof that P(n) is true for all integers n ≥ 18. a) Show that the statements P(18), P(19), P(20), and P(21) are true, completing the basis step of a proof by strong induction that P(n) is true for all integers n ≥ 18. b) What is the inductive hypothesis of a proof by strong induction that P(n) is true for all integers n ≥ 18? c) What do you need to prove in the inductive step of a proof that P(n) is true for all integers n ≥ 18? d) Complete the inductive step for k ≥ 21. e) Explain why these steps show that P(n) is true for all integers n ≥ 18.
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n>1 Problem 4. Prove by induction that, for every geq2 (1-)(1--)-(1-)- n +1 1 - 32 42 n2 2n
Given that P(n) is the equation 1. (1!) + 2 · (2!) + · · · + n · an integer such that n > 1, we will prove that P(n) is true for all n > 1 by induction. 6. (п!) — (п + 1)! — 1, where n is (а) Base case: i. Write P(1). ii. Show that P(1) is true. In this case, this requires showing that a left-hand side is equal to a right-hand side. (b) Inductive hypothesis: Let k > 1 be a natural number. Assume P(k) is true. Write P(k). (c) Inductive step: i. Write P(k + 1). ii. Use the assumption that P(k) is true to prove that P(k +1) is true.
Problem 12 Use Mathematical Induction to prove that 6 n³-n whenever n is a positive integer.
(6) Use congruences to prove the following. They were proven by induction in Prob lem 16 of Section 6.3. (a) 6|n(n² + 5). (b) 5 (n5 -n). (c) 7 (n7 -n).
5. Use ordinary induction to prove that for every positive integer n, n-n is a multiple of 6. Only proofs by induction are accepted.
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Problem 1: Prove that for every integer n > 7, (n – 2)! > n² . (Hint: using the requirement n > 7 as well as the regular parts of your induction hypothesis can be useful.)