Hm..

Question

Hm..

abb0t · Answer

hmmmm...

anonymous · Answer

hmmmm.....

anonymous · Answer

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

kinggeorge · Answer

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.

anonymous · Answer

You are absolutely correct!

anonymous · Answer

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.

anonymous · Answer

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

anonymous · Answer

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

raden · Answer

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

anonymous · Answer

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

kinggeorge · Answer

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.