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Unformatted text preview: COMP 170 Tutorial Induction Problem 1: Use contradiction to prove that 1 · 2 + 2 · 3 + ··· + n ( n + 1) = n ( n + 1)( n + 2) 3 for all integers n ≥ 1 . Problem 1: Use contradiction to prove that 1 · 2 + 2 · 3 + ··· + n ( n + 1) = n ( n + 1)( n + 2) 3 for all integers n ≥ 1 . Solution: Start by denoting the problem statement as p ( n ) . Problem 1: Use contradiction to prove that 1 · 2 + 2 · 3 + ··· + n ( n + 1) = n ( n + 1)( n + 2) 3 for all integers n ≥ 1 . Solution: Start by denoting the problem statement as p ( n ) . Assume p ( n ) is false for some integer n ≥ 1 choose k to be the smallest n that makes it false. Problem 1: Use contradiction to prove that 1 · 2 + 2 · 3 + ··· + n ( n + 1) = n ( n + 1)( n + 2) 3 for all integers n ≥ 1 . Solution: Start by denoting the problem statement as p ( n ) . Assume p ( n ) is false for some integer n ≥ 1 choose k to be the smallest n that makes it false. p (1) is true (since 2 = 2 ), so k > 1 . Problem 1: Use contradiction to prove that 1 · 2 + 2 · 3 + ··· + n ( n + 1) = n ( n + 1)( n + 2) 3 for all integers n ≥ 1 . Solution: Start by denoting the problem statement as p ( n ) . Assume p ( n ) is false for some integer n ≥ 1 choose k to be the smallest n that makes it false. p (1) is true (since 2 = 2 ), so k > 1 . Since k is the smallest n that makes p ( n ) false, p ( k 1) must be true, so 1 · 2 + 2 · 3 + ··· + ( k 1) k = ( k 1) k ( k + 1) 3 . Problem 1: Use contradiction to prove that 1 · 2 + 2 · 3 + ··· + n ( n + 1) = n ( n + 1)( n + 2) 3 for all integers n ≥ 1 . Adding k ( k + 1) to both sides of p ( k 1) gives 1 · 2 + 2 · 3 + ··· + ( k 1) k + k ( k + 1) = ( k 1) k ( k + 1) 3 + k ( k + 1) = ( k 1) k ( k + 1) + 3 k ( k + 1) 3 = k ( k + 1)(( k 1) + 3) 3 = k ( k + 1)( k + 2) 3 Problem 1: Use contradiction to prove that 1 · 2 + 2 · 3 + ··· + n ( n + 1) = n ( n + 1)( n + 2) 3 for all integers n ≥ 1 . Adding k ( k + 1) to both sides of p ( k 1) gives 1 · 2 + 2 · 3 + ··· + ( k 1) k + k ( k + 1) = ( k 1) k ( k + 1) 3 + k ( k + 1) = ( k 1) k ( k + 1) + 3 k ( k + 1) 3 = k ( k + 1)(( k 1) + 3) 3 = k ( k + 1)( k + 2) 3 Thus p ( k ) is true, contradicting the assumption p ( k ) is false Problem 1: Use contradiction to prove that 1 · 2 + 2 · 3 + ··· + n ( n + 1) = n ( n + 1)( n + 2) 3 for all integers n ≥ 1 . Adding k ( k + 1) to both sides of p ( k 1) gives 1 · 2 + 2 · 3 + ··· + ( k 1) k + k ( k + 1) = ( k 1) k ( k + 1) 3 + k ( k + 1) = ( k 1) k ( k + 1) + 3 k ( k + 1) 3 = k ( k + 1)(( k 1) + 3) 3 = k ( k + 1)( k + 2) 3 Thus p ( k ) is true, contradicting the assumption p ( k ) is false Thus, p ( n ) must be true for all integers n ≥ 1 . Problem 2: Use induction to prove that 1 · 2 + 2 · 3 + ··· + n ( n + 1) = n ( n + 1)( n + 2) 3 for all integers n ≥ 1 . Solution: Denote the statement to proven by p ( n ) ....
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This note was uploaded on 08/25/2010 for the course COMP COMP170 taught by Professor M.j.golin during the Spring '10 term at HKUST.
 Spring '10
 M.J.Golin
 Computer Science

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