I have the sequence: ($a_1=-1 ,a_{n+1}=a_n+2n , n\in N$)

I'm trying to find an explicit equation for this sequence. I am not familiar to how to find closed form of a recursive sequence so I tried to write first terms of the sequence:

$$\begin{array}{rcc} n:&1&2&3&4&5&6\\ a_n:&-1&1&5&11&19&29 \end{array}$$

And form here I guessed there is a relation between $n^2$ and $a_n$. and after trying some equations I got: $$a_n=n^2-n-1$$

But as I said earlier, I don't know how to write explicit equation of a recursive sequence. So how can I find this equation mathematically? and how can I learn some basic technics to convert such a sequence to explicit one?(looking for some resources)

  • $\begingroup$ You can see this for solving recurrence relations. $\endgroup$
    – cosmo5
    Oct 21, 2020 at 17:17

2 Answers 2


You have for all $k \geq 1$ $$a_{k+1}-a_k = 2k$$

so if you sum these equalities for $k=1, ..., n-1$, you get $$\sum_{k=1}^{n-1} (a_{k+1}-a_k) = \sum_{k=1}^{n-1} 2k$$ i.e. $$a_n-a_1 = n(n-1)$$ i.e. finally because $a_1=-1$, $$a_n=n^2-n-1$$

  • $\begingroup$ Is this relevant to telescoping series? $\endgroup$
    – Aligator
    Oct 21, 2020 at 17:02
  • $\begingroup$ Yes, the way from step 2 to step 3 is a telecoping trick. $\endgroup$ Oct 21, 2020 at 17:03
  • $\begingroup$ Do we usually use this method for some simple recursive sequences or there are other tricks ? $\endgroup$
    – Aligator
    Oct 21, 2020 at 17:05
  • $\begingroup$ It really depends. There is no general method to express a recursive sequence directly in terms of $n$. It happens very frequently that it is purely impossible. $\endgroup$ Oct 21, 2020 at 17:08
  • 1
    $\begingroup$ @Aryadeva The sum $1+2+... +n$ is equal to $n(n+1)/2$. You can see en.wikipedia.org/wiki/1_%2B_2_%2B_3_%2B_4_%2B_%E2%8B%AF for example. Or you can try to prove it by induction. Tell me if you have some difficulties proving this famous identity. $\endgroup$ Oct 21, 2020 at 17:22

Observe the finite differences of the first 6 values of the series.

enter image description here

After $2$ iterations of successive finite differences, the differences become constant. Therefore, the polynomial that generates the sequence has degree 2.

Let $a(n) = An^2 + Bn + C$.

Use $a(0), a(1)$ and $a(2)$ for $n = 1, 2, 3$ respectively and obtain a system of equations for variables $A, B, C$ and solve them. (You could use any three values of $n$).

Here's the worked solution for the example on WolframALpha: This gives $A = 1, B = -1, C = -1$.

So, $a(n) = n^2 - n - 1$

  • $\begingroup$ Ok, the differences are $2n$. so differences are $\{2,4,6,8,\cdots\}$ and finding differences again we have $\{2,2,2,\cdots\}$. doing that again we have all zeros. But I don't understand how we conclude the polynomial that generates the sequence has degree $2$ from this. $\endgroup$
    – Aligator
    Oct 21, 2020 at 17:34
  • $\begingroup$ For the values generated by a polynomial of degree $n$, after $n$ successive iterations, the difference becomes constant. See: Newton's Interpolation formula of finite differences. $\endgroup$
    – vvg
    Oct 21, 2020 at 17:48

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