hw6 - Homework 6 Discrete Math 1 Exercise 20 b p 230 Log(2n...

Info icon This preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
Homework 6 Discrete Math 1. Exercise 20, b) p. 230 Log(2n)/ Log(n)=log(2)= 1 2. Exercise 20, f) p. 230 3. (2n) 3 /n 3 =8 Time increases by 8 4. Exercise 20, g) p. 230 Increases by 2 n time 5. Exercise 4, p. 329 Let P(n) be the statement that 13 +23 +···+n3 = (n(n+1)/2)2 for the positive integer n. a) What is the statement P(1)? 1 3 = ((1(1 + 1)/2) 2 b) Show that P(1) is true, completing the basis step of the proof. 1³ = 1 (1(1+1)/2)² = (2/2)² = 1² = 1 Thus, P(1) is true c) What is the inductive hypothesis? P(n) for some positive integer n, that is, the statement 1 3 + 2 3 +· · ·+n 3 = (n(n+ 1)/2) 2 d) What do you need to prove in the inductive step? We assume that P(n) is true, then we need to show that P(n + 1) is also true. To do this, we need to derive the equation 1 3 +2 3 +· · ·+n 3 +(n+1) 3 =((n + 1)(n + 2)/2) 2 from the equation 1 3 + 2 3 +· · ·+n 3 = (n(n+ 1)/2) 2 e) Complete the inductive step, identifying where you use the inductive hypothesis. 1³ + ... + n³ + (n+1)³ = ((n+1)(n+1+1)/2)² = ((n+1)(n+2)/2)² 1³ + ... + n³ + (n+1)³ = (n(n+1)/2)² + (n+1)³ = n²(n+1)²/4 + (n+1)³ = (n+1)² [n²/4 + (n+1)] = (n+1)² [(n² + 4n + 4)/4] = (n+1)² (n+2)² / 4 = ((n+1)(n+2)/2)² f) Explain why these steps show that this formula is true whenever n is a positive integer. Using this proof, we have shown that if P(n) is true, then P(n+1) is also true Thus, P(n) is true for all positive integer n. 6. Exercise 32, p.330 Prove that 3 divides n 3 +2n whenever n is a positive integer.
Image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    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.

    Student Picture

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

  • Left Quote Icon

    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.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    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.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern