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sol7 - CSE 21 Solutions Problem Set 7 1 Proof Initial step...

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CSE 21: Solutions - Problem Set 7 1. Proof. Initial step: For n = 1, the left-hand size is 1 3 = 1 and the right-hand side is ( 1+1 2 ) 2 = 1. Then the equality holds. Inductive Hypothesis: Suppose it is true for n = t , i.e., 1 3 + 2 3 + · · · + t 3 = t + 1 2 2 . We prove that it’s true for n = t + 1, i.e., 1 3 + 2 3 + · · · + t 3 + ( t + 1) 3 = t + 2 2 2 . Inductive step: 1 3 + 2 3 + · · · + t 3 + ( t + 1) 3 = ( 1 3 + 2 3 + · · · + t 3 ) + ( t + 1) 3 = t + 1 2 2 + ( t + 1) 3 Inductive hypothesis = ( t + 1) t 2 2 + ( t + 1) 3 = ( t + 1) 2 4 ( t 2 + 4( t + 1) ) = ( t + 1) 2 4 ( t + 2) 2 = ( t + 2)( t + 1) 2 2 = t + 2 2 2 2. Initial step: For n = 1, the left-hand size is 1 2 - 1 2 · 0 = 1, and the right-hand side is ( - 1) 2 = 1. So the equality holds. Inductive hypothesis: Suppose it’s true for n = k , i.e., F 2 k - F k +1 F k - 1 = ( - 1) k +1 . We prove that it’s true for n = k + 1, i.e., F 2 k +1 - F k +2 F k = ( - 1) k +2 . 1
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Inductive step: F 2 k +1 - F k +2 F k = F 2 k +1 - ( F k +1 + F k ) F k (by definition) = F 2 k +1 - F k +1 F k - F 2 k = F k +1 ( F k +1 - F k ) - F 2 k = F k +1 F k - 1 - F 2 k (by definition) = - ( - 1) k +1 (inductive hypothesis) = ( - 1) k +2 3. We show three different solutions.
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