1
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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 $\{\frac{z(z+1)}{2}|z\in \mathbb{N}\}$

$\endgroup$
2
  • $\begingroup$ @cell: added that $\endgroup$ – Maroon Sep 30 '14 at 19:11
  • $\begingroup$ I'm busy at the moment and can't find notation that includes 0, but it shouldn't be too difficult to sort out $\endgroup$ – Maroon Sep 30 '14 at 19:17
1
$\begingroup$

Read about Triangular numbers. The answer is:

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

$\endgroup$
1
$\begingroup$

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: $$\{\frac{n^2 + n}{2} \space | \space n \in \mathbb{N} \cup \{0\} \}$$

$\endgroup$
1
  • $\begingroup$ 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) $\endgroup$ – user76568 Sep 30 '14 at 18:23
0
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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