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LogicalAppleBest 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 . .
 one year ago

KingGeorgeBest ResponseYou've already chosen the best response.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.
 one year ago

LogicalAppleBest ResponseYou've already chosen the best response.0
You are absolutely correct!
 one year ago

patdisthemanBest ResponseYou've already chosen the best response.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.
 one year ago

LogicalAppleBest ResponseYou've already chosen the best response.0
The way this particular recursive relation is set up, almost everything cancels.
 one year ago

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

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

LogicalAppleBest ResponseYou've already chosen the best response.0
That's true  you can derive that the recursive formula generates triangular numbers.
 one year ago

KingGeorgeBest ResponseYou've already chosen the best response.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.
 one year ago
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