## LogicalApple Hm.. one year ago one year ago

1. abb0t

hmmmm...

2. patdistheman

hmmmm.....

3. LogicalApple

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

4. KingGeorge

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

You are absolutely correct!

6. patdistheman

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

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

8. patdistheman

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

10. LogicalApple

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

11. KingGeorge

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.