quiz 8 - positive integer p k 1 k 1 2 5 k 1 2 Expanding and...

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CSE 260 QUIZ-9– Mathematical Induction (20 minutes) NAME: 1. Let p(n) be the proposition p(n): 2+4+6+8+ . . .+2n=n(n+1) (a) What are i. (2 points) p(1): p (1) : 2 = 1(1 + 1) ii. (2 points) p(2): p (2) : 2 + 4 = 2(2 + 1) iii. (2 points) p(n+1): p ( n + 1) : 2 + 4 + 6 + 8 + ... + 2 n + 2( n + 1) = ( n + 1)( n + 2) (b) (10 points) Prove p(n) n ( n > 1) by mathematical induction. p ( n ) : 2 + 4 + ... + 2 n = n ( n + 1) Basis Step: p (1) : 2 = 1(1 + 1) true? Left hand side of the equality= 2 Right hand side of the equality= 1.2 Therefore, the equality is true. Induction Step: Show p ( k ) p ( k + 1) for 1 k n p ( k + 1) : 2 + 4 + ... + 2 k + 2( k + 1) = ( k + 1)( k + 2) Lefthand side of the equality: 2 + 4 + ... + 2 k + 2( k + 1) = k ( k + 1) + 2( k + 1) by the hypothesis =( k + 1)( k + 2) This equal to the right hand side of the equality. 2. Consider the proposition p ( n ) : n 2 + 5 n + 2 is odd (a) (10 points) Prove that the truth of p(k) implies the truth of p(k+1) when k is a
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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....
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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.

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