# Set builder of this set $0$, $1$, $3$, $6$, $10$, $15$

I have tried to create the set builder of this infinite set:

0, 1, 3, 6, 10, 15, 21, 28,...


I have notice that

$$n = (n - 1) + (N + 1)$$

where n is the current number, n - 1 is the previous number and N + 1 is a natural number which represents the number of numbers before n.

For example, suppose $$n=3$$

3 = 1 + 2


The problem is that I do not know how to really represent N-1, because, if we do not know how many numbers that are before n, we cannot find N+1 and therefore n.

## 4 Answers

Notice that we start with $z = 0$, and that the zth element of the sequence $a_z$ is defined by $a_z = a_{z-1} + z$. This is basically the set of triangular numbers.

With an induction proof, we also can prove that the correct representation of the zth element is $a_z = \frac{z(z+1)}{2}$.

To express this as a set, I would denote it as $\{\frac{z(z+1)}{2}|z\in \mathbb{N}\}$

• @cell: added that – Maroon Sep 30 '14 at 19:11
• I'm busy at the moment and can't find notation that includes 0, but it shouldn't be too difficult to sort out – Maroon Sep 30 '14 at 19:17

Read about Triangular numbers. The answer is:

$$\left\{{n+1\choose2} :\ n \in \mathbb{N}\right\}$$

Firstly, it's a sequence, not a set. secondly, this might work?: $$a_0=0, a_n=a_{n-1}+n$$

Which yields: $$a_n=\sum_{k=0}^{n}k=\frac{n^2+n}{2}$$ So, like @agha pointed out, as a set it is: $$\{\frac{n^2 + n}{2} \space | \space n \in \mathbb{N} \cup \{0\} \}$$

• Yes of course. But notice that in order to build it, the "subscript" was used, i.e it's a function of "position" (a natural number) – user76568 Sep 30 '14 at 18:23

If you look at double the sequence, we see $0 \times 1, 1 \times 2, 2 \times 3,\ldots$