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.
 A: Your "diff" is actually called (backward) finite difference.
\begin{align}
\nabla_1 [ P ](x) &= P(x) - P(x-1) \\
&= P(x-1+1) - P(x-1) \\
&= \Delta_1[ P ](x-1)
\end{align}
The inverse the forward finite difference is called indefinite sum. Extracted from Wikipedia, the useful formulae for polynomials is:
\begin{align}
\Delta^{-1}_1 x^n &= \frac{B_{n+1}(x)}{n+1} + C \\
\Delta^{-1}_1 af(x) &= a \Delta^{-1}_1 f(x)
\end{align}
(Bn+1(x) is the Bernoulli polynomial.)
The Δ-1 can be converted back to your "int" by substituting $x \mapsto x + 1$.
A: 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.
A: 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.
A: 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
$\Delta_1 x(x-1)(x-2)\cdots(x-(n-1)) = n x(x-1)(x-2)\cdots(x-(n-2))$
