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 😛
Best Answer
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 $a_{n+1} = a_n + 2n + 1$.”