What are some "natural" interpolations of the sequence $\small 0,1,1+2a,1+2a+3a^2,1+2a+3a^2+4a^3,\ldots $? (This is a spin-off of a recent question here)
In fiddling with the answer to that question I came to the set of sequences    
$\qquad \small \begin{array} {llll}
A(1)=1,A(2)=1+2a,A(3)=1+2a+3a^2,A(4)=1+2a+3a^2+4a^3, \ldots \\
B(1)=1,B(2)=1+3a,B(3)=1+3a+6a^2,B(4)=1+3a+6a^2+10a^3, \ldots \\
C(1)=1,C(2)=1+4a,C(3)=1+4a+10a^2,C(4)=1+4a+10a^2+20a^3, \ldots \\
\ldots \\
 \end{array} $
with some indeterminate a .         
We had the discussion often here in MSE, that interpolation to fractional indexes, say A(1.5)=?? is arbitrary, considering, that an initial solution composed with any 1 -periodic function satisfies the condition. But here the embedding in a set of sequences, which are constructed from binomial-coefficients might suggest some "natural" interpolation, such as for      
$\qquad \small K(1)=1, K(2)=1+a, K(3)=1+a+a^2, \ldots $     
the interpolation $\small K(r) = {a^r-1 \over a-1}$ seems the most "natural" which even can smoothly be defined for a=1. This observation made me to refer to "q-analogues" $\small [r]_a $ in my answer in the initiating MSE-question, but it's not obvious how to interpolate the shown sequences of higher orders A , B , C (I think they're not related to the "q-binomial-analogues" , for instance ).   
Q: So what would be some "natural" interpolation to fractional indexes for the sequences A, B, C, and possibly in general for sequences generated in the obvious generalized manner? 

Agreeing mostly with Henning's ansatz I got now the general form as
$$ A_m(n) = {1 \over (1-a)^m} - \sum_{k=0}^{m-1} \binom{n+m}{k}{a^{n+m-k} \over (1-a)^{m-k} } $$
I do not yet see, whether some examples of fractional indexes agree with the solutions of all three given answers so far, for instance: given a=2.0 what is A(1.5), B(4/3), C(7/5)? With my programmed version I get now 
$\qquad \small  A(1.5)\sim 9.48528137424  $
$\qquad \small  B(4/3) \sim 11.8791929545  $
$\qquad \small  C(7/5) \sim 18.4386165488  $
(No interpolation for fractional m yet)

[update 2] the derivative-versions of Sivaram and Michael arrive at the same values so I think, all versions can be translated into each other and mutually support each other to express a "natural" interpolation.
[update 3] I had an index-error in my computation call. Corrected the numerical results.
 A: You have $K(r) = (a^r-1)/(a-1)$.
And ${d \over da} K(r) = 1 + 2a + \cdots + (r-1)a^{r-2} = A(r-1)$.
So it seems natural to define $A(r) = {d \over da} K(r+1)$, where $K(r)$ is defined as in your interpolation.
Similarly 
$$ {d^2 \over da^2} K(r) = 2 B(r-2) $$
and
$$ {d^3 \over da^3} K(r) = 6 C(r-3) $$
which we can solve for $B(r), C(r)$ to get
$$ B(r) = {1 \over 2} {d^2 \over da^2} K(r+2), C(r) = {1 \over 6} {d^3 \over da^3} K(r+3). $$
So I would define
$$ A_n(r) = {1 \over n!} {d^n \over da^n} K(r+n) $$
for any integer $n$ and real number $a$; this seems to be reasonably well-behaved. Here $A_1, A_2, A_3$ are your $A, B, C$.
(While I was writing this answer, Sivaram Ambikasaran's answer agreed. I am not sure if this is the same generalization or not.)
Also, does this extend to fractional $n$? I am not familiar enough with the fractional calculus to tell.
A: Here is an attack leading to a closed forms for the $A$s and $B$s that can be evaluated for fractional $n$:
First, let $ A_n = \sum_{k=0}^n (k+1)a^k $.
For $n>0$ we can write
$$ A_n = a A_{n-1} + 1 + a + \cdots + a^n = a A_{n-1} + \frac{a^{n+1}-1}{a-1} $$
We can then use a base case of $A_0=1=\frac{a-1}{a-1}$ to unfold the entire recurrence into
\begin{align}
A_n &= \frac{1}{a-1}\Bigl( (a^{n+1}-1) + a(a^n-1) + a^2(a^{n-1}-1) + \cdots + a^n(a-1) \Bigr)
\\&= \frac{1}{a-1}\Bigl( (n+1)a^{n+1} - (1+a+\cdots+a^n) \Bigr)
\\&= \frac{1}{a-1}\Bigl( (n+1)a^{n+1} - \frac{a^{n+1}-1}{a-1} \Bigr)
\end{align}
We can push this into the $B$s by noting that
$$ B_n=A_n + a A_{n-1} + a^2 A_{n-2} + \cdots + a^n A_0 $$
so
\begin{align}
B_n &= \frac{1}{a-1}\left(\sum_{k=0}^n k+1\right)a^{n+1}
- \frac{1}{(a-1)^2} \Bigl( (a^{n+1}-1)+a(a^{n}-1)+\cdots+a^n(a-1) \Bigr)
\\&= \frac{1}{a-1} \frac{(n+1)(n+2)}{2} a^{n+1}
- \frac{1}{(a-1)^2} \Bigl( (n+1)a^{n+1} - (1+a+\cdots+a^n) \Bigr)
\\&= \frac{1}{a-1} \frac{(n+1)(n+2)}{2} a^{n+1}
- \frac{1}{(a-1)^2} (n+1)a^{n+1}
+ \frac{1}{(a-1)^3} (a^{n+1}-1)
\end{align}
A pattern is emerging ... I will leave the derivation for $C_n$ to the reader, but the only difficulty should be in keeping the algebra straight.
A: One "natural" interpolation is along the following lines.
$$A(n) = \sum_{k=1}^{n} ka^{k-1} = \frac{d K_n(a)}{da}$$
$$B(n) = \sum_{k=1}^{n} \frac{k(k+1)}{2}a^{k-1} = \frac1{2!} \frac{d^2 K_{n+1}(a)}{da^2}$$
$$C(n) = \sum_{k=1}^{n} \frac{k(k+1)(k+2)}{6}a^{k-1} = \frac1{3!} \frac{d^3 K_{n+2}(a)}{da^3}$$
A generic "natural" interpolation would be $\displaystyle A_m(n) = \sum_{k=1}^{n} \binom{k+m}{m+1} a^{k-1}  = \frac1{(m+1)!} \frac{d^{m+1} K_{n+m}(a)}{da^{m+1}}$.
In our case, $A_0(n) = A(n)$, $A_1(n) = B(n)$, $A_2(n) = C(n)$ and so on...
A: I'll build from Michael's work (thanks for doing the heavy lifting!) and start with
$$A_n(r)=\frac1{n!}\frac{\mathrm d^n}{\mathrm da^n} \frac{a^{r+n}-1}{a-1}$$
Let's switch back to the series representation, and swap summation and differentiation:
$$A_n(r)=\frac1{n!}\sum_{k=0}^{r+n-1} \frac{\mathrm d^na^k}{\mathrm d a^n}$$
and make a suitable replacement:
$$A_n(r)=\frac1{n!}\sum_{k=0}^{r+n-1} \frac{k!a^{k-n}}{(k-n)!}=\frac1{n!}\sum_{k=-n}^{r-1} \frac{(k+n)!a^k}{k!}=\frac1{n!}\sum_{k=0}^{r-1} \frac{(k+n)!a^k}{k!}$$
where a reindexing and removal of extraneous zero terms was done in the last two steps.
The result can then be expressed as:
$$A_n(r)=\frac{\mathrm{I}_{1-a}(n+1,r)}{(1-a)^{n+1}}=\frac{1-\mathrm{I}_a(r,n+1)}{(1-a)^{n+1}}$$
where $\mathrm{I}_z(r,n)$ is a regularized incomplete beta function.
In terms of the Gaussian hypergeometric function, we have
$$\begin{align*}
A_n(r)&=\frac{\mathrm{I}_{1-a}(n+1,r)}{(1-a)^{n+1}}\\
&=\frac{a^r}{(n+1)\mathrm{B}(n+1,r)}{}_2 F_1\left({{n+r+1, 1}\atop{n+2}}\mid1-a\right)\\
&=\binom{n+r}{n+1} {}_2 F_1\left({{1-r,n+1}\atop{n+2}}\mid1-a\right)
\end{align*}$$
where the third one is derived from the second one through the Pfaff transformation. In particular, the third one gives the representation
$$A_n(r)=r\binom{n+r}{r}\sum_{k=0}^{r-1}\binom{r-1}{k}\frac{(a-1)^k}{n+k+1}$$
which is valid for arbitrary $n$ and nonnegative integer $r$.
Other hypergeometric representations can be derived.
