To put things into perspective I've read some posts with the method of undetermined coefficients in it for differential equations. What I'm talking about is the method to determine a closed formula for a sequence. For those of you who are unfamiliar with it I'll try to explain it.
E.g.:
$$0, 2, 5, 9, 14,\ldots$$ $$2, 3, 4, 5,\ldots$$ $$1, 1, 1,\ldots$$
The $1^{st}$ sequence (it also can be a different sequence of course, this is just an example) is always given in an exemplary exercise (in my case I found it myself, just by the brute-force method). The $2^{nd}$ one is acquired by subtracting the number in the 1^{st}sequence by the following number of that sequence. And one is required to stop this algorithm after it get a constant sequence like in the $3^{rd}$ one.
So now one has performed the algorithm two times. Thus concluding we need a polynomial of degree 2. Now let's take a random polynomial of which we need to determine the coefficients:
$$P(x)= ax{^2} + bx + c$$
We know that $P(0)= 0,P(1)= 2$and $P(2)= 5$ .
So $\begin{cases} c= 0 \\ a + b =2 \\ 4a + 2b =5\\ \end{cases}$ which gives us that $P(x)= \frac 12 x^2 + \frac 32 x $
To rephrase my question from early on:
Why does the method of coefficients work for every the simpler sequences as stated above? Does there exist a proof of it? Please explain that proof to me?
PS.:
1) I know that it doesn't work for sequences whose solutions aren't reals or if there are multiple solutions of such a P(x). Then you need to work with generating functions.
2) I've chosen this sequence on purpose because it's the sequence to determine number of diagonals of a convex polygon. Just for those who wondered about it.