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Unformatted text preview: Homework 2 Solutions 1. (1.10) Prove (2 n + 1) + (2 n + 3) + (2 n + 5) + + (4 n 1) = 3 n 2 for all n N , n 1. Solution: Base step: When n = 1, we get 3 = 3, which is true. Inductive step: Suppose (2 k + 1) + (2 k + 3) + (2 k + 5) + + (4 k 1) = 3 k 2 . Well prove it holds for k + 1. (2( k + 1) + 1) + + (4( k + 1) 1) = (2 k + 3) + + (4 k 1) + (4 k + 1) + (4 k + 3) = 3 k 2 (2 k + 1) + (4 k + 1) + (4 k + 3) = 3( k + 1) 2 2. Use induction to prove Bernoullis inequality: If 1 + x > 0, then (1 + x ) n 1 + nx for all n N . Solution: Base step: When n = 0, we get 1 = 1, which is true. Inductive step: Suppose (1 + x ) k 1 + kx . (1 + x ) k +1 = (1 + x )(1 + x ) k (1 + x )(1 + kx ) = 1 + ( k + 1) x + kx 2 1 + ( k + 1) x 3. Polyas paradox. Find the error in the following proof by induction. Claim: All horses are the same color. Proof: Base case: If theres only one horse, then all horses are the same color. Inductive step: Suppose that any set of n horses have the same color. Now suppose we have a set of n + 1 horses. Take all but the first horse, i.e. horses { 2 , 3 ,...,n +1 } . This is a set of n horses, so they each have the same color by the inductive hypothesis. Now, take all but the last horse, i.e. horses { 1 , 2 ,...,n } ....
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 Fall '10
 GIESEKER,D.
 Algebra

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