# Formula for a sequence

I have this sequence: $-2, 1, 6, 13, 22, 33, ...$

where each term is the previous plus an odd number. The odd numbers are increasing from 3. I am asked to find an explicit formula with respect to $n$ which can give me the $n$-th number in the sequence. The sequence starts at $n = 1$.

I tried to find a pattern, without success. I then tried to write the sequence as a recursive formula: $a_1 = -2$
$a_{n + 1} = a_n + 2n + 1$

and then I got stuck.

Thanks,
rubik

• André's answer is exactly what you need, but I think this question touches on something interesting about sequences. If you have a sequence with terms $a_n$, then if the first difference $a_n-a_{n-1}=k$ for some constant k, then you have an arithmetic progression. If the second difference, $(a_n-a_{n-1})-(a_{n-1}-a_{n-2})$, is constant then you have a quadratic progression (your $a_n=n^2-3$, for example). More generally, if the $n^{th}$ difference is a constant, then $a_n=P(n)$ for some polynomial P of degree n. This is a consequence of the fact that $\frac{d^n}{dx^n}x^n=n!$, which is constant – preferred_anon May 7 '12 at 20:03
• Conway & Guy "The Book of Numbers" has a number of proofs without words in Chapter 2, and some observations on how to extend difference methods in Chapter 3 (these are both mentioned in answers below) plus lots of other material. I'm sure there are other sources, and there is plenty online too. – Mark Bennet May 7 '12 at 20:16
• @DanLitt: Thank you! That's really a useful thing. So if I need to find a formula for a sequence I can try the differences method. – rubik May 8 '12 at 12:51

Hint: Add $3$ to every number in your sequence.
Remark: A related result is that the sum of the first $n$ odd positive integers is equal to $n^2$. This follows easily from the fact that $n^2-(n-1)^2=2n-1$. There is also an attractive proof without words.