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
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$\begingroup$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 $a_{n+1} = a_n + 2n + 1$.”
$\endgroup$ 1 $\begingroup$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..
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