# Is there exact formula or recursive relation for the sequence $a_{n+1}=a_n + n$? [closed]

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?

$$a_{n} = a_{1} + \sum_{k=1}^{n-1}k$$ 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., $$\sum_{j=1}^{m}j=\frac{m^{2}+m}{2}$$
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 $$a_n$$.
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 $$a_{n+1}=\frac{(\frac{1+\sqrt{8a_n-7}}{2})(\frac{3+\sqrt{8a_n-7}}{2})+2}{2}$$ .