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LogicalApple
 2 years ago
Best ResponseYou've already chosen the best response.0\[a _{1} = 1\] \[a _{n+1} = (\frac{ n+2 }{ n })a _{n}, n \ge 1\] Find \[a _{30}\] Careful . .

KingGeorge
 2 years ago
Best ResponseYou've already chosen the best response.2I'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.

LogicalApple
 2 years ago
Best ResponseYou've already chosen the best response.0You are absolutely correct!

patdistheman
 2 years ago
Best ResponseYou've already chosen the best response.1Well 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.

LogicalApple
 2 years ago
Best ResponseYou've already chosen the best response.0The way this particular recursive relation is set up, almost everything cancels.

patdistheman
 2 years ago
Best ResponseYou've already chosen the best response.1At some point I will understand this... I have a few things before it to catch up on.

RadEn
 2 years ago
Best ResponseYou've already chosen the best response.0it is a triangle number : 1, 3, 6, 10, 15, 21, 28, ...., n/2 * (n+1) so, a30 = 30/2 (30+1) = 465

LogicalApple
 2 years ago
Best ResponseYou've already chosen the best response.0That's true  you can derive that the recursive formula generates triangular numbers.

KingGeorge
 2 years ago
Best ResponseYou've already chosen the best response.2It 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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