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I have this sequence: $7,8,10,13...$. Since this is neither an arithmetic sequence or geometric I was not sure how to go about solving. My initial thoughts are that it goes up by $+1$ then $+2$, then $+3$ etc. In which the common difference of those is $+1$. But I am having trouble formulating this into an $n^{th}$ term sequence. Can anyone give me a hand? Thank you :)

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  • $\begingroup$ Something similar here. $\endgroup$ – 雨が好きな人 May 22 at 18:09
  • $\begingroup$ The sequence of triangular numbers, $T(n)=\sum\limits_{i=1}^n i$ which begins $0,1,3,6,10,\dots$ has closed form $T(n)=\frac{n(n-1)}{2}$. Your sequence is just the sequence of triangular numbers plus seven. $\endgroup$ – JMoravitz May 22 at 18:12
  • $\begingroup$ Look up , or figure out, how to sum $1 + 2 + 3 + . . . + n$. Then you should he able to adapt it to your sequence $\endgroup$ – badjohn May 22 at 18:14
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    $\begingroup$ I gave the answer (proved in the comments to auscrypt's answer) earlier and only because he argued that no answer is more valid over the other without realising that both the solutions are same, you accepted his answer! So unfair! $\endgroup$ – Brozovic May 22 at 18:25
  • $\begingroup$ @Brozovic The accepted answer is that one which is most helpful for the asker. No reason to complain. Besides many people (including me) prefer closed form solutions to recurrence relations. In this case one need not compute all previous terms to obtain a certain one. $\endgroup$ – user May 23 at 15:30
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Your formula would be $a_n = 7+\frac{n(n-1)}{2}$; this is just triangular numbers but with a constant added. In general, questions which ask to continue the sequence are usually arbitrary; there are an infinitude of ways to do so, none of which is more valid than another.

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  • $\begingroup$ $$7+n (n-1)/2.$$ $\endgroup$ – user May 22 at 18:16
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    $\begingroup$ $a_n = 7 + \frac{(n-1)(n-2)}{2} = 7 + (1 + 2 + \dots + (n-1))=(7+1)+(2+\dots+(n-1))= a_2 +(2+\dots+(n-1))= \dots = a_{n-1} + (n-1)$ $\endgroup$ – Brozovic May 22 at 18:17
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$a_1 = 7$ and $a_n = a_{n-1} +(n-1)$ , $\forall n \ge 2$

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