Homework 1 Solutions.pdf

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

• Homework Help
• 2

This preview shows 1 out of 2 pages.

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

Subscribe to view the full document.

You've reached the end of this preview.

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern