Partial sums of falling factorials I want to know if there exists some way, approximate or exact, to do a partial sum of falling factorials of the kind:
$$\sum_{k=i}^{n}(a+k)_{h}$$ where all are constants (here $(r)_s:=r(r-1)\cdots (r-s+1)$ represent a falling factorial). 
And I'm interested too in some partial sum like this
$$\sum_{k=i}^{n}(a+k)_{h}r^k$$
In particular I want a closed form to this formula:
$$\sum_{m=0}^{3}\left(\sum_{k=1}^{19-m}(19-k)_m\right )^{-1} \left(-(19)_m+\sum_{k=0}^{19-m}(19-k)_m q^k\right )$$
Possibly there is not a closed form but I don't know. I started to read about hypergeometric series but this topic is completely new to me so I don't have a clear way to approach to my question by now.
I will appreciate any help. If you can show me via some link or bibliography is fine too. Thank you in advance.

UPDATE
Ok, I was reading the book of Graham that @ncmathsadist said to me and I have a partial answer. The question is close to some general topics on discrete maths (that I unfortunately forget).
The point is that an analogue to $\int_{a}^{b}x^n dx=\frac{x^{n+1}}{n+1}\Big|_{a}^{b}\ $ on difference calculus is
$$\sum\nolimits_{a}^{b}(k)_n\delta k =\frac{(k)_{n+1}}{n+1}\bigg|_{a}^{b}$$
For the second case I can use an analogue to integration by parts that is named summation by parts:
$$\sum f(k)\Delta g(k) \delta k=f(k)g(k)-\sum \Delta f(k) g(k+1)\delta k$$
But I dont get any closed form, so I assumed these formulas haven't closed forms.
 A: I like Knuth's notation for falling factorials better:
$$
\alpha^{\underline{h}} 
  = \alpha \cdot (\alpha - 1) \cdot \dotsm \cdot (\alpha - h + 1)
$$
First note that:
$$
\Delta \alpha^{\underline{h}}
  = (\alpha + 1)^{\underline{h}} -\alpha^{\underline{h}}
  = (\alpha + 1) \cdot \alpha^{\underline{h - 1}}
      - \alpha^{\underline{h - 1}} \cdot (\alpha - h + 1)
  = h \alpha^{\underline{h - 1}}
$$
This suggests:
$$
\sum_{0 \le k \le n} (\alpha + k)^{\underline{h}}
  = \frac{(\alpha + n + 1)^{\underline{h + 1}}}{h + 1}
      - \frac{\alpha^{\underline{h + 1}}}{h + 1}
$$
This we prove by induction.
Base: $n = 0$ gives:
$\begin{align}
\sum_{0 \le k \le 0} (\alpha + k)^{\underline{h}}
  &= \alpha^{\underline{h}} \\
\frac{(\alpha + 1)^{\underline{h + 1}}}{h + 1}
  - \frac{\alpha^{\underline{h + 1}}}{h + 1}
  &= \frac{1}{h + 1}
       \Delta \alpha^{\underline{h + 1}} \\
  &= \alpha^{\underline{h}}
\end{align}$
This checks out.
Induction: Assume it is true for $n$, look at $n + 1$:
$\begin{align}
\sum_{0 \le k \le n + 1} (\alpha + k)^{\underline{h}}
  &= \sum_{0 \le k \le n} (\alpha + k)^{\underline{h}}
       + (\alpha + n + 1)^{\underline{h}} \\
  &= \frac{(\alpha + n + 1)^{\underline{h + 1}}}{h + 1}
       - \frac{\alpha^{\underline{h + 1}}}{h + 1}
       + (\alpha + n + 1)^{\underline{h}} \\
  &= \frac{(\alpha + n + 1)^{\underline{h}} \cdot (\alpha + n + 1 - h)}
          {h + 1}
       + \frac{(\alpha + n + 1)^{\underline{h}} \cdot (h + 1)}
              {h + 1}
       - \frac{\alpha^{\underline{h + 1}}}{h + 1} \\
  &= \frac{(\alpha + n + 1)^{\underline{h}} \cdot (\alpha + n + 2)}
          {h + 1}
       - \frac{\alpha^{\underline{h + 1}}}{h + 1} \\
  &= \frac{(\alpha + n + 2)^{\underline{h + 1}}}{h + 1}
       - \frac{\alpha^{\underline{h + 1}}}{h + 1}
\end{align}$
This is exactly as claimed.
For your second sum, note that:
$$
(\alpha + k)^{\underline{h}} x^{\alpha + k - h}
  = \frac{\mathrm{d}^h}{\mathrm{d} x^h} x^{\alpha + k}
$$
Thus the sum is essentially:
$\begin{align}
\sum\limits_{k=0}^n (\alpha + k)^{\underline{h}} r^k
  &= r^{h - \alpha} \sum\limits_{k=0}^n (\alpha + k)^{\underline{h}} r^{\alpha + k - h} \\
  &= r^{h - \alpha} 
       \left.
         \frac{\mathrm{d}^h}{\mathrm{d} x^h} \sum\limits_{k=0}^n x^{\alpha + k}
       \right|_{x = r} \\
  &= r^{h - \alpha} 
       \left.
         \frac{\mathrm{d}^h}{\mathrm{d} x^h} \left(x^\alpha \sum\limits_{k=0}^n x^k\right)
       \right|_{x = r} \\
  &=\left.r^{h-\alpha} \frac{d^h}{d x^h} \left( \frac{x^\alpha - x^{\alpha+n+1}}{1- x} \right)\right|_{x=r} \\
  &=r^{h-\alpha} \sum\limits_{l=0}^h \binom{h}{l} \left(\alpha^{\underline{h-l}} r^{\alpha-(h-l)} - (\alpha+n+1)^{\underline{h-l}} r^{\alpha+n+1-(h-l)}\right) \cdot \frac{l!}{(1-r)^{l+1}} \\
 &= \frac{1}{(1-r)^{h+1}} \sum\limits_{l=0}^h \binom{h}{l} \left(\alpha^{\underline{h-l}} r^{l} - (\alpha+n+1)^{\underline{h-l}} r^{n+1+l}\right) \cdot l! (1-r)^{h-l} \\
 &= \frac{h!}{(1-r)^{h+1}} \sum\limits_{l=0}^h \frac{1}{(h-l)!} \left(\alpha^{\underline{h-l}} r^{l} - (\alpha+n+1)^{\underline{h-l}} r^{n+1+l}\right) \cdot  (1-r)^{h-l} \\
 &= \frac{h!}{(1-r)^{h+1}} \sum\limits_{l=0}^h  \left(\binom{\alpha}{h-l} r^{l} - \binom{\alpha+n+1}{h-l} r^{n+1+l}\right) \cdot  (1-r)^{h-l} \\
 &= \frac{h!}{(1-r)} \left\{
\binom{\alpha}{h} \cdot 
F_{2,1} \left[\begin{array}{ll}1 & -h \\ \alpha-h+1 \end{array};\frac{r}{r-1}\right]
- r^{n+1} \binom{\alpha+n+1}{h} \cdot 
F_{2,1} \left[\begin{array}{ll}1 & -h \\ \alpha+n-h+2 \end{array};\frac{r}{r-1}\right]
\right\}
\end{align}$
This is quite ugly. The remaining sum is geometric, and can be expressed as a fraction. Leibnitz' formula for multiple derivatives of a product reduce that somewhat, but it is still a mess. If $h$ is  small integer, perhaps a CAS gives somewhat manageable.
Note: It is a matter of taste whether an expression is ugly or not. In my opinion the final result has a closed form if we use hypergeometric functions.
