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LogicalApple

  • 3 years ago

Hm..

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  1. abb0t
    • 3 years ago
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    hmmmm...

  2. patdistheman
    • 3 years ago
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    hmmmm.....

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

  4. KingGeorge
    • 3 years ago
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    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
    • 3 years ago
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    You are absolutely correct!

  6. patdistheman
    • 3 years ago
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    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
    • 3 years ago
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    The way this particular recursive relation is set up, almost everything cancels.

  8. patdistheman
    • 3 years ago
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    At some point I will understand this... I have a few things before it to catch up on.

  9. RadEn
    • 3 years ago
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    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
    • 3 years ago
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    That's true -- you can derive that the recursive formula generates triangular numbers.

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  11. KingGeorge
    • 3 years ago
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    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.

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