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.

For any particular polynomial there is an easier way to do indefinite summation than using the Bernoulli numbers, going off of Greg Graviton's answer.  Here we'll use the forward difference $\Delta f(x) = f(x+1) - f(x)$.  Then 
$\displaystyle \Delta {x \choose n} = {x \choose n-1}.$
This implies that we can perform a "Taylor expansion" on any polynomial to write it in the form $f(x) = \sum a_n {x \choose n}$ by evaluating the finite differences $\Delta^n f(0)$ at zero.  For any particular polynomial $f$ it is easy to write these finite differences down by constructing a table.  In general, the formula is
$\displaystyle a_n = \Delta^n f(0) = \sum_{k=0}^{n} (-1)^{n-k} {n \choose k} f(k)$
as one can readily prove by writing $\Delta = S - I$ where $S$ is the shift operator $S f(x) = f(x+1)$ and $I$ is the identity operator $I f(x) = f(x)$.  Then the indefinite sum of $f$ is just $\sum a_n {x \choose n+1}$.  This is the easiest way I know how to do such computations by hand, and it also leads to a fairly easy method for polynomial interpolation given the values of a polynomial at consecutive integers.
