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Suppose $$a_1=2,a_2=6,a_3=12,a_4=20,....$$ be a sequence. What is $a_n$?

Tom
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  • Not that you would need to do so for this sequence, but a great place to look up sequences is the OEIS. Check it out! – Valborg Sep 25 '16 at 21:32
  • @Valborg -Please more guide. – Tom Sep 25 '16 at 21:36
  • If you do not have more information on the structure of $a_n$ you cannot conclude. You may guess what could be the right answer, but there are infinitely many "good" formulas which will give you these four initial terms. – Beni Bogosel Sep 25 '16 at 21:46
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    Have a look at : http://oeis.org/ that will in fact give you different solutions, $a_n=n(n+1)$ being the most simple to describe... – Jean Marie Sep 25 '16 at 21:52
  • @ JeanMarie, this shoul be an answer (of course, with additon that there exists infinite many other solutions) – z100 Sep 25 '16 at 22:19
  • This question is not about mathematics, within the scope defined in the help center. – Did Sep 25 '16 at 22:25
  • See this question. While it's not about the same sequence, it discusses the same issues, and some useful methods to get a reasonable answer. – Jean-Claude Arbaut Sep 26 '16 at 00:24

3 Answers3

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Look at the differences formed by two consecutive terms: \begin{align} a_2-a_1&=6-2=4\\ a_3-a_2&=12-6=6\\ a_4-a_3&=20-12=8 \end{align} Each time we take the difference of two consecutive terms the result increases by $2$: \begin{align} a_2-a_1&=2\times 2\\ a_3-a_2&=2\times 3\\ a_4-a_3&=2\times 4\\ \dots&\quad\text{so we see }\\ a_{n}-a_{n-1}&=2\times n \end{align} or \begin{align} a_2&=a_1+2\times 2\\ a_3&=a_2+2\times 3\\ a_4&=a_3+2\times 4\\ \dots&\\ a_{n}&=a_{n-1}+2\times n \end{align} Hence $a_n=a_{n-1}+2\times n=a_{n-2}+2\times(n-1)+2\times n\dots=2\sum_{k=1}^{n}k=2\frac12n(n+1)=n(n+1)$.

Daniel Buck
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One of the simplest ways of attempting to extrapolate a formula for a sequence of numbers, particularly if there are so few terms given, is to try and identify if there is a point beyond which the forward difference operator appears to return a constant sequence. Try that out.

Valborg
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It could be $a_n=\sum_{k=1}^n2k=n(n+1)$.

The sequence is A002378: $2, 6, 12, 20, 30, 42, 56, 72, 90, 110,\dots$.

However, this is only one of the possible answers. Consider the sequence A045619, i.e. the increasing sequence of numbers that are the products of two or more consecutive positive integers. It starts in the same way 2,6,12,20, but the next one is $2\cdot 3 \cdot 4=24$.

More terms are: $2, 6, 12, 20, 24, 30, 42, 56, 60, 72, 90, 110, \dots$.

Robert Z
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    We should refrain from simply providing answers on this site. We are here to teach and help people understand math, not do their homework for them. – Valborg Sep 25 '16 at 21:39
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    You are not useful if you simply solve people's homework for them. No one gains anything by that. – Valborg Sep 25 '16 at 21:56
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    Facetiousness does not become you. This isn't about wisdom, it's about the community guidelines regarding problems that are most likely homework. http://meta.math.stackexchange.com/questions/106/what-is-the-proper-way-to-handle-homework-questions – Valborg Sep 25 '16 at 22:11