Indefinite summation of polynomials I've been experimenting with the summation of polynomials. My line of attack is to treat the subject the way I would for calculus, but not using limits.
By way of a very simple example, suppose I wish to add the all numbers between $10$ and $20$ inclusive, and find a polynomial which I can plug the numbers into to get my answer.  I suspect its some form of polynomial with degree $2$. So I do a integer 'differentiation':
$$
\mathrm{diff}\left(x^{2}\right)=x^{2}-\left(x-1\right)^{2}=2x-1
$$
I can see from this that I nearly have my answer, so assuming an inverse 'integration' operation and re-arranging:
$$
\frac{1}{2}\mathrm{diff}\left(x^{2}+\mathrm{int}\left(1\right)\right)=x
$$
Now, I know that the 'indefinite integral' of 1 is just x, from 'differentiating' $x-(x-1) = 1$. So ultimately:
$$
\frac{1}{2}\left(x^{2}+x\right)=\mathrm{int}\left(x\right)
$$
So to get my answer I take the 'definite' integral: 
$$
\mathrm{int}\left(x\right):10,20=\frac{1}{2}\left(20^{2}+20\right)-\frac{1}{2}\left(9^{2}+9\right)=165
$$
(the lower bound needs decreasing by one)
My question is, is there a general way I can 'integrate' any polynomial, in this way?
Please excuse my lack of rigour and the odd notation.

As other answers have noted, you are about to discover the calculus of finite differences.
For practical calculations, here a most useful fact: the rule
$\frac{d}{dx} x^n = n x^{n-1}$
corresponds to
$\Delta_1 x(x-1)(x-2)\cdots(x-(n-1)) = n x(x-1)(x-2)\cdots(x-(n-2))$
