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.

You seem to be reaching for the
calculus of finite differences, once a well-known topic but rather
unfashionable these days. The answer to your question is yes: given a polynomial
$f(x)$ there is a polynomial $g(x)$ (of degree one greater than $f$) such that
$$f(x)=g(x)-g(x-1).$$
This polynomial $g$ (like the integral of $f$) is unique save for
its constant term.
Once one has $g$ then of course
$$f(a)+f(a+1)+\cdots+f(b)=g(b)-g(a-1).$$
When $f(x)=x^n$ is a monomial, the coefficients
of $g$ involve the endlessly fascinating Bernoulli numbers.
