Use Theorem and the result of exercise 10 to prove that if p is any prime number with p ? 5, then the sum of squares of any p consecutive integers is divisible by p.Theorem 1Sum of the First n IntegersFor all integers n ? 1, Proof (by mathematical induction):Let the property P(n) be the equation Show that P(1) is true:To establish P(1), we must show that But the left-hand side of this equation is 1 and the right-hand side is also. Hence P(1) is true.Show that for all integers k? 1, if P(k) is true then P(k + 1) is also true:[Suppose that P(k) is true for a particular but arbitrarily chosen integer k ? 1.That is:] Suppose that k is any integer with k ? 1 such that [We must show that P(k + 1) is true. That is:] We must show that or, equivalently, that [We will show that the left-hand side and the right-hand side of P(k + 1) are equal to the same quantity and thus are equal to each other.]The left-hand side of P(k + 1) is And the right-hand side of P(k + 1) is Thus the two sides of P(k + 1) are equal to the same quantity and so they are equal to each other. Therefore the equation P(k + 1) is true [as was to be shown].[Since we have proved both the basis step and the inductive step, we conclude that thetheorem is true.]

# Use Theorem and the result of exercise 10 to prove that if

## Solution for problem 37E Chapter 5.2

Discrete Mathematics with Applications | 4th Edition

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Discrete Mathematics with Applications | 4th Edition

Get Full SolutionsThe full step-by-step solution to problem: 37E from chapter: 5.2 was answered by , our top Math solution expert on 07/19/17, 06:34AM. This textbook survival guide was created for the textbook: Discrete Mathematics with Applications , edition: 4. The answer to “Use Theorem and the result of exercise 10 to prove that if p is any prime number with p ? 5, then the sum of squares of any p consecutive integers is divisible by p.Theorem 1Sum of the First n IntegersFor all integers n ? 1, Proof (by mathematical induction):Let the property P(n) be the equation Show that P(1) is true:To establish P(1), we must show that But the left-hand side of this equation is 1 and the right-hand side is also. Hence P(1) is true.Show that for all integers k? 1, if P(k) is true then P(k + 1) is also true:[Suppose that P(k) is true for a particular but arbitrarily chosen integer k ? 1.That is:] Suppose that k is any integer with k ? 1 such that [We must show that P(k + 1) is true. That is:] We must show that or, equivalently, that [We will show that the left-hand side and the right-hand side of P(k + 1) are equal to the same quantity and thus are equal to each other.]The left-hand side of P(k + 1) is And the right-hand side of P(k + 1) is Thus the two sides of P(k + 1) are equal to the same quantity and so they are equal to each other. Therefore the equation P(k + 1) is true [as was to be shown].[Since we have proved both the basis step and the inductive step, we conclude that thetheorem is true.]” is broken down into a number of easy to follow steps, and 244 words. Since the solution to 37E from 5.2 chapter was answered, more than 232 students have viewed the full step-by-step answer. Discrete Mathematics with Applications was written by and is associated to the ISBN: 9780495391326. This full solution covers the following key subjects: true, hand, SIDE, show, integers. This expansive textbook survival guide covers 131 chapters, and 5076 solutions.

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Use Theorem and the result of exercise 10 to prove that if