TITLE: Indefinite summation of polynomials
QUESTION [7 upvotes]: 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.

REPLY [5 votes]: 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.