Homework 1 Solutions.pdf - ECE 3020A Homework 1 Spring 2018...

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1 ECE 3020A Homework 1, Spring 2018 Solution Problem 1. 1-Basis: Let 𝑛 = 0 . Then, 𝐶(𝑛) = 0 by definition, and 1 4 ( 𝑛 4 + 2𝑛 3 + 𝑛 2 ) = 0. So 𝐶(𝑛) = 1 4 ( 𝑛 4 + 2𝑛 3 + 𝑛 2 ) when 𝑛 = 0. 2- Induction step: Assume true for 𝑛 = 𝑘 where 𝑘 ≥ 0 and prove true for 𝑛 = 𝑘 + 1. We can use this series of equalities: 𝐶(𝑘 + 1) = 1 3 + 2 3 + 3 3 + ⋯ + 𝑘 3 + (𝑘 + 1) 3 by definition = 𝐶(𝑘) + (𝑘 + 1) 3 𝐶(𝑘) = 1 3 + ⋯ + (𝑘) 3 = 1 4 (𝑛 4 + 2𝑛 3 + 𝑛 2 ) + (𝑘 + 1) 3 ) induction hypothesis = 1 4 ((𝑛 + 1) 4 + 2(𝑛 + 1) 3 + (𝑛 + 1) 2 ) QED Problem 2. 1-Basis: For: 𝑛 = 1 1! = 1, 2 1−1 = 2 0 = 1 , 1 ≥ 1 2- Induction step: We assume true for 𝑛 = 𝑘, i.e., assume that 𝑘! ≥ 2 𝑘−1 . We show true for 𝑛 = 𝑘 , that is, we show (𝑘 + 1)! ≥ 2 𝑘 . (𝑘 + 1)! = (𝑘 + 1)𝑘! ≥ (𝑘 + 1)2 𝑘−1 (by inductive hypothesis) Since 𝑘 ≥ 1, 𝑘 + 1 ≥ 2
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