Here's the question you clicked on:
LogicalApple
Hm..
\[a _{1} = 1\] \[a _{n+1} = (\frac{ n+2 }{ n })a _{n}, n \ge 1\] Find \[a _{30}\] Careful . .
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.
You are absolutely correct!
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.
The way this particular recursive relation is set up, almost everything cancels.
At some point I will understand this... I have a few things before it to catch up on.
it is a triangle number : 1, 3, 6, 10, 15, 21, 28, ...., n/2 * (n+1) so, a30 = 30/2 (30+1) = 465
That's true -- you can derive that the recursive formula generates triangular numbers.
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.