Problem Solution
consecutive numbers is divisible by 2 As a result, we may conclude that the product of any three consecutive numbers is divisible by 2 Since 2,3 are prime numbers, a number that is divisible by 2 and 3 is also divisible by 6 We conclude that the product of three consecutive numbers is divisible by 6 Theorem 2: For every number n, n2 +n+1 is
Proof by Induction
Suppose that 7n-2n is divisible by 5 Our goal is to show that this implies that 7n+1-2n+1 is divisible by 5 We note that 7n+1-2n+1 = 7x7n-2x2n= 5x7n+2x7n-2x2n = 5x7n +2(7n-2n) By induction hypothesis, (7n-2n) = 5k for some integer k Hence, 7n+1-2n+1= 5x7n +2x5k = 5(7n +2k), so 7n+1-2n+1 =5 x some integer Thus, the claim follows by
Solutions to Induction Problems Fall 2009 1Let P n
CS240 Solutions to Induction Problems Fall 2009 1 Let P(n) be the statement that n < nn, where n 2 is an integer Basis step: 2 = 2 1 = 2 < 4 = 22 Inductive hypothesis: Assume k < kk for some k 2
n2 = n + 1)(2n + 1)/6 for the positive integer n
*35 Prove that n2 − 1 is divisible by 8 whenever n is an odd positive integer *36 Prove that 21 divides 4n + 1 + 52n − 1 whenever n is a positive integer *37 Prove that if n is a positive integer, then 133 divides 11n + 1 + 122n − 1
Proof by Induction
12=1, 22=4, 32=9, 42=16, (n+1)2 = n2+n+n+1 = n2+2n+1 1+3+5+7 = 42 Chapter 4 Proofs by Induction I think some intuition leaks out in every step of an induction proof — Jim Propp, talk at AMS special session, January 2000 The principle of induction and the related principle of strong induction have been introduced in the previous chapter
1 Multiples et diviseurs - WordPresscom
Remarque 2 1n’est pas premier car il n’a qu’un seul diviseur, lui-même Exemple 3 Citer 6 nombres premiers : Test de primalité Soit nun entier supérieur ou égal à 2 Si nn’admet pour diviseur aucun des nombres premiers inférieurs ou égaux à √ n, alors nest un nombre premier
Math 2260 Exam Practice Problem Solutions
2 2 2 (2x)2 = lim x1 12cos x x2 4x2 2 = lim x1 4cos 1 x = 4 where I went from the second to the third lines using L’H^opital’s Rule Since the limit of the terms is equal to 4, not zero, the series must diverge 1
Hadamard Matrices and Designs
and taking the inner products of rows 1 and 2, 1 and 3, and, 2 and 3 we get x + y - z - w = 0 x - y + z - w = 0 x - y - z + w = 0 Solving this system of equations gives, x = y = z = w = h/4 Thus, the integer h must be divisible by 4
PRINCIPLE OF MATHEMATICAL INDUCTION
22 – 1 = 4 – 1 =3 1 is divisible by 3 Assume that P(n) is true for some natural number k, i e , P(k): 22k – 1 is divisible by 3, i e , 22k – 1 = 3q, where q ∈ N Now, to prove that P(k + 1) is true, we have P(k + 1) : 22(k+1) – 1 =22k + 2 – 1 = 22k 22 – 1 = 22k 4 – 1 = 3 22k + (22k – 1)
[PDF] montrer que n(n+1)(2n+1) est divisible par 3
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[PDF] montrer que n^3-n est divisible par 3
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