I derived a new formula related to arithmetic sequences, I think!

Your method, while interesting, is not new: this is simply an example of a recurrence relation. An easier way to solve this is to first note that two subsequent subtractions yields a constant (see link in comments above).

Now we set up equations in quadratic form:
$$a(1)^2+b(1)+c = 7 \implies a+b+c = 7\\ a(2)^2+b(2)+c = 14 \implies 4a+2b+c=14\\ a(3)^2+b(3)+c=28 \implies 9a+3b+c = 28$$
Solving this, we get the quadratic $\frac{7}{2}x^2 -\frac{7}{2}x + 7$ which includes all of your points (specifically integers greater than or equal to $1$)

You can take this farther: Let $A^{(1)}_n=A^{(0)}_{n+1}-A^{(0)}_n$ and $A^{(2)}=A^{(1)}_{n+1}-A^{(1)}_n$ and $A^{(3)}_n=A^{(2)}_{n+1}-A^{(2)}_n$ and so on. Then $A^{(j)}_n$ is a non-zero constant if and only if $A^{(0)}_n$ is a polynomial function of $n,$ with degree $j.$ You can prove this by induction on $j.$