Solve for equation algebraically Is it possible to write the following function as $H(x)=$'some expresssion` ?
$$D(x) = H(x) + H(x-1)$$
Edit:
Hey everyone, thanks for all the great responses, and just to clarify H(x) and D(x) are always going to be polynomials, I wasn't sure if that made too big of a difference so I didn't mention it before.
Edit 2: 
I'm sincerely sorry if I was too general, I've only really used polynomial equations in my education thus far, so I didn't realize they might not seam just regular to everyone else. Once again I'm very sorry If I didn't give everyone an opportunity to answer the question because of my vagueness. I sincerely appreciate all of your answers and time.

For $x$ an integer, up to a constant, $H(x)$ is the alternating sum of $D(x)$ from 0 up to $x$ (no reason to start at $-\infty$ like Arthur, although this might sometimes be preferable) since
$$((-1)^x D(0)- (-1)^x D(1)+\dots + D(x))+ ((-1)^{x-1}D(0) - (-1)^{x-1}D(1) + \dots + D(x-1)) = D(x) $$
and for $H(x)=0$ the equation gives a recurrence that fixes $D$ given $D(0)$.
Therefore, your problem includes the calculations of all finite sums  and you cannot expect a reasonable answer without additional assumptions on your functions.
For example, if your $D$ is a polynomial, then you know that the answer is also a polynomial that can easily be found by an interpolation formula. Clearly, the polynomial equation will then hold for all $x$.
For a hypergeometric expression, the Gosper algorithm can be adapted.
For an integrable function, the Euler-Maclaurin formula can sometimes be of use.
But there is no general answer. Sums do not always have simple answers and even if they do, there is no simple way to find them.
