Unformatted text preview: positive integer p ( k + 1) : ( k + 1) 2 + 5( k + 1) + 2 Expanding and regrouping, we get k 2 + 2 k + 1 + 5 k + 5 + 2 = ( k 2 + 5 k + 2) + 2( k + 3) ( k 2 + 5 k + 2) is odd by the hypothesis and 2( k + 3) is even. Therefore, their sum is odd. (b) (4 points) For which values of n is p(n) actually true? What is the moral of this exercise. p ( n ) is not true for any values of n. That is, n 2 + 5 n + 2 is even for all values of n. Show n 2 + 5 n + 2 is even ∀ n ≥ assume n is even: n 2 is even, 5n is even. Therefore, n 2 + 5 n + 2 is even. assume n is odd: n 2 is odd, 5n is odd. Therefore, n 2 + 5 n + 2 is (odd+odd+2) which is even. Moral: Basis step could not be found. That is why p(n) could not be proved....
View
Full
Document
This note was uploaded on 03/30/2008 for the course CSE 260 taught by Professor Saktipramanik during the Spring '08 term at Michigan State University.
 Spring '08
 SaktiPramanik

Click to edit the document details