Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

For the set $\{3,6,11,18,27,38,...\}$, the $(n+1)^\text{th}$ term, which I'll call $a_{n+1}$, is: $$a_{n+1} = a_n + 2n+1$$ How can I write this set in set-builder notation? My best guess doesn't seem quite right: $$\{a_n : n \in \mathbb{N}, a_1=3, a_{n+1}=a_n + 2n+1\}$$ Also, my reference to $\mathbb{N}$ assumes $0 \notin \mathbb{N}$, but maybe that's the less common interpretation of $\mathbb{N}$?

Thanks for the help! As pointed out below, $\{n^2+2:n\in \mathbb{N}\}$ does the trick nicely. Bit of an oversight on my part :P

share|improve this question
add comment

2 Answers

up vote 4 down vote accepted

How about $$\{ n^2 + 2 : n\in \mathbb N \}$$

Not what you are looking for?

You are also free to write something like “$\{a_n : n\in \mathbb N\}$ where $a_1 = 3$ and $a_{n+1} = a_n + 2n + 1$.”

share|improve this answer
    
Oh man, I totally overlooked that :P. $n^2 + 2$ is much better. Thanks for pointing that out :) –  ivan Feb 24 '13 at 2:36
    
@ivan : then accept the answer! –  Arjang Feb 26 '13 at 12:34
add comment

Yes, perfect.

If $\Bbb N$ is used in the context with $0$, then in the set you can use for example $\Bbb N^+$ or $\Bbb Z^+$ or most clearly $\Bbb N\setminus\{0\}$. Or, you can also write $a_0=3$ instead of $a_1=3$ in that case..

share|improve this answer
    
Good point. Thanks! –  ivan Feb 24 '13 at 2:41
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.