CSC_226_Midterm_4

CSC_226_Midterm_4 - the second term is divisible by 6. 8....

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7. P(n) : 6 divides - n3 n Domain : {n| n ≥ 0} Basis: P(2) : - = 23 2 6 (True) Proof: P(n + 1) : ( + ) - ( + ) n 1 3 n 1 (Inductive Hypothesis) = + + + - + n 1n2 2n 1 n 1 (Distribute) = + + + - - n3 3n2 3n 1 n 1 (Distribute) = - + + n3 n 3n2 3n (Arrange Terms) = - + ( + ) n3 n 3n n 1 (Factor Out 3n) Because P(n) is divisible by 6, the first term is as well. Also, since 3n(n+1) is divisible by 3 and either (n) or (n+1) must be an even number,
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Unformatted text preview: the second term is divisible by 6. 8. Amounts : 20, 25, 30, 35, … P(n) : . j. k(j ≥ 0 ∧ k ≥ 0 ∧ 5n = 10j + 25k) Domain : {n| n ≥ 4} Basis: P(7) : True for j = k = 1 Proof: P(n + 1) : . j’. k’(j’ ≥ 0 ∧ k’ ≥ 0 ∧ 5(n + 1) = 10j’ + 25k’) Proof By Cases: Case 1 : k ≠ 0. Then j’ = j + 3 and k’ = k – 1 Case 2 : k = 0. Then j’ = j – 2 and k’ = k + 1...
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This note was uploaded on 04/28/2009 for the course CSC 226 taught by Professor Watkins during the Spring '08 term at N.C. State.

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