I found this series:
$2;10;32;88$
My teacher has told me that there is no methodology or equation to find the n-term ( Im be in high school yet ).
It really is that? If so, at least there are techniques? It takes a lot of time to do the test-error, thinking practically at random.
Things like this:
$-2;4;-6;8$ can be solved more easy, because is a it is a more concordant pattern.
= $2n(-1)^n$ ( With $a_1$ as first index)
On
Your teacher is right. There is no general method for finding the $n$th term of a sequence when you know just the first few. There are infinitely many continuations, and it may well happen that more than one of them is interesting, or correct in some particular context.
You might have fun browsing at The Online Encyclopedia of Integer Sequences: https://oeis.org/ . Your's isn't there. Where did it come from? If you discovered it in some mathematical exploration perhaps you should tell OEIS about it.
You like to fit these values $$(1,2);(2,10);(3,32);(4,88)$$ into a model.
While there are many different ways to find such a model, it is easy to find a polynomial to fit the data.
Check the following model:
$$a_n=(44/3)(n-1)(n-2)(n-3)-16(n-1)(n-2)(n-4)+5(n-1)(n-3)(n-4)-(1/3) (n-2)(n-3)(n-4)$$
The simplified polynomial takes the form $$ a_n=\frac {1}{3}(10n^3-39n^2+71n-36)$$ Your data fits perfectly in this model.
Note that the answer is not unique because there are many curves passing through your data.