Quantcast

A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

LogicalApple

  • 2 years ago

Hm..

  • This Question is Closed
  1. abb0t
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    hmmmm...

  2. patdistheman
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    hmmmm.....

  3. LogicalApple
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    \[a _{1} = 1\] \[a _{n+1} = (\frac{ n+2 }{ n })a _{n}, n \ge 1\] Find \[a _{30}\] Careful . .

  4. KingGeorge
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    I'm going to say that it's 465. I have not proved this yet, and I must go now. However, I'm pretty sure about it, and I'll be back in a couple hours to prove it.

  5. LogicalApple
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    You are absolutely correct!

  6. patdistheman
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Well I may not be able to provide the answer to this question in any reasonable time frame at the moment but it's clear I got the best answer lol.

  7. LogicalApple
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    The way this particular recursive relation is set up, almost everything cancels.

  8. patdistheman
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    At some point I will understand this... I have a few things before it to catch up on.

  9. RadEn
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    it is a triangle number : 1, 3, 6, 10, 15, 21, 28, ...., n/2 * (n+1) so, a30 = 30/2 (30+1) = 465

  10. LogicalApple
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    That's true -- you can derive that the recursive formula generates triangular numbers.

    1 Attachment
  11. KingGeorge
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    It is indeed the triangular numbers. And this is a proof by induction that \(a_n\) is the \(n\)-th triangular number. Base case: \(a_1=1\), so we're done. Assume true for some \(a_k\), \(k\in\mathbb{Z}^+\). Then,\[a_k=\frac{k(k+1)}{2}\]So \[a_{k+1}=\frac{(k+2)k(k+1)}{2k}=\frac{(k+1)(k+2)}{2}\]Which is the \(k+1\)-th triangular number, so we're done.

  12. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.