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Good evening to everybody.

I was reading a chapter today on a book related to recursive sequences, and I saw as an example the following simple sequence : $$a_{n+1}=a_n + n $$ I was thinking if we could find an explicit formula for this one, something like $a_n=C^n+R $ for example , or even if we can find a recursive formula (that is a formula involving only the values $ a_{n+1} , a_n , a_{n+2} $ and not the number $n$). Is this possible?

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By looking at a few terms explicitly you will find the following structure:
$$ a_{n} = a_{1} + \sum_{k=1}^{n-1}k $$ You can prove this easily by induction. If you want, you can further simplify this term using the formula for the sum of first $m$ positive integers, i.e., $$ \sum_{j=1}^{m}j=\frac{m^{2}+m}{2} $$

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Yes, you can. $$a_2=a_1+1.$$ $$a_3=a_2+2=a_1+2+1.$$ $$a_4=a_3+3=a_2+2+1=a_1+3+2+1.$$ Find out a few more terms and conjecture a closed form for $a_n$.

Prove it using induction and you should be through.

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Yes,thank you all for your answers. In fact, one can see that $ a_n=a_1+(1+2+...+n-1)=a_1+\frac{n(n-1)}{2}=\frac{n^2-n+2}{2} $ . And from this, one can get also a recursive formula: $a_n=\frac{n^2-n+2}{2} $ , hence, $ n^2-n-2a_n+2=0 $ , and using the discriminant , we get that $ n=\frac{1+\sqrt{8a_n-7}}{2}$ and putting this to $a_{n+1}=\frac{n^2+n+2}{2} $ we get that $ a_{n+1}=\frac{(\frac{1+\sqrt{8a_n-7}}{2})(\frac{3+\sqrt{8a_n-7}}{2})+2}{2} $ .

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