In the quest for closed form solutions for a large class of hypergeometric sums. Let $j_1\ge 0$ and $j_2\ge 0$ and $m\ge 2$ be integers. Now, let $0 < z < (m-1)^{m-1}/m^m$ be a real number. Consider a following sum:
\begin{equation}
{\mathfrak S}^{(m,j_1)}_{j_2}(z):= \sum\limits_{i=0}^\infty \binom{m\cdot i+j_1}{i+j_2} \cdot z^i
\end{equation}
Now with considerable help of other users, in Yet another family of hypergeometric sums that has a closed form solution. and in Closed form solutions for a family of hypergeometric sums. we have derived the following results:
\begin{eqnarray}
{\mathfrak S}^{(m,j)}_{0}(z) &=&\frac{x^{j+1}}{(1-m) \cdot x+m}\\
{\mathfrak S}^{(m,j)}_{1}(z) &=&\frac{x^{j+1}}{((1-m) \cdot x+m)\cdot(x-1)}-\frac{1}{z}\\
{\mathfrak S}^{(m,j)}_{2}(z) &=&\frac{x^{j+1}}{((1-m) \cdot x+m)\cdot(x-1)^2}-\frac{1}{z^2}+\frac{(m-j)}{z}\\
\vdots\\
{\mathfrak S}^{(m,0)}_{j}(z) &=& \frac{x}{((1-m) x+m) \cdot (x-1)^j}+\sum\limits_{l=1}^j \binom{l \cdot m+j-l-1}{j-l} \cdot \frac{(-1)^{j-l+1}}{z^l}
\end{eqnarray}
where $x:=x(z)$ is computed in the following way. 
Out of the solutions of the polynomial equation:
\begin{equation}
1-x+z \cdot x^m=0
\end{equation}
choose the one that is the closest to unity.
Now, the question is to provide the answer for arbitrary values of $j_1$ and $j_2$.
 A: The final result is actually quite simple and reads:
\begin{eqnarray}
{\mathfrak S}^{(m,j)}_{j_1}(z) = \frac{x^{j+1}}{((1-m) x+m) \cdot (x-1)^{j_1}} + \sum\limits_{l=1}^{j_1}\binom{l\cdot m-j+j_1-l-1}{j_1-l} \frac{(-1)^{j_1-l+1}}{z^l}
\end{eqnarray}
We prove it by induction in $j_1$. Let us assume the formula above is valid for some value $j_1 \ge 0$. Then we compute:
\begin{eqnarray}
{\mathfrak S}^{(m,j)}_{j_1+1}(z)=\sum\limits_{i=0}^\infty \binom{m\cdot i+j}{i+j_1} \cdot \underbrace{\frac{(m-1)\cdot i+j-j_1}{i+j_1+1}}_{(m-1)+\frac{(j-(j_1+1)\cdot m+1)}{i+j_1+1}}\cdot z^i
\end{eqnarray}
Therefore we have:
\begin{eqnarray}
&&\frac{{\mathfrak S}^{(m,j)}_{j_1+1}(z)-(m-1)\cdot {\mathfrak S}^{(m,j)}_{j_1}(z)}{(j-(j_1+1)\cdot m+1)}=\\
&&  \int\limits_0^1 \theta^{j_1} \cdot {\mathfrak S}^{(m,j)}_{j_1}(\theta z)d\theta=\\
&&\int\limits_1^x \left(\frac{\xi-1}{z \xi^m} \right)^{j_1} \left(\frac{(1-m) \xi+m}{z \cdot \xi^{m+1}}\right)\cdot \\
&&\left[\frac{\xi^{j+1}}{((1-m) \xi+m) \cdot (\xi-1)^{j_1}} + \sum\limits_{l=1}^{j_1}\binom{l\cdot m-j+j_1-l-1}{j_1-l} (-1)^{j_1-l+1} \left(\frac{\xi^m}{\xi-1}\right)^l\right] d\xi=\\
&&\frac{1}{z^{j_1+1}}\int\limits_1^x \\
&&\left[\xi^{j-(j_1+1)\cdot m} + \sum\limits_{l=1}^{j_1}\binom{l\cdot m-j+j_1-l-1}{j_1-l} (-1)^{j_1-l+1} \xi^{m \cdot l-(j_1+1)\cdot m-1}  \cdot(\xi-1)^{j_1-l} ((1-m)\xi+m)\right] d\xi=\\
&&\frac{1}{z^{j_1+1}} \cdot\\
&&\left[
\frac{x^{j-(j_1+1) \cdot m+1}-1}{j-(j_1+1) \cdot m+1}+
\sum\limits_{l=1}^{j_1}\binom{l\cdot m-j+j_1-l-1}{j_1-l} (-1)^{j_1-l+1} 
\frac{(-1+x)^{j_1-l+1}\cdot x^{-(j_1-l+1) \cdot m}}{j_1-l+1}
\right]=\\
&&\frac{1}{z^{j_1+1}} \cdot\\
&&\left[
\frac{x^{j+1}\cdot \left(\frac{z}{x-1}\right)^{j_1+1}-1}{j-(j_1+1) \cdot m+1}+
\sum\limits_{l=1}^{j_1}\binom{l\cdot m-j+j_1-l-1}{j_1-l} (-1)^{j_1-l+1} 
\frac{z^{j_1-l+1}}{j_1-l+1}
\right]=\\
&&\left[
\frac{\frac{x^{j+1}}{(x-1)^{j_1+1}}-\frac{1}{z^{j_1+1}}}{j-(j_1+1) \cdot m+1}+
\sum\limits_{l=1}^{j_1}\binom{l\cdot m-j+j_1-l-1}{j_1-l} (-1)^{j_1-l+1} 
\frac{z^{-l}}{j_1-l+1}
\right]
\end{eqnarray}
Therefore we have shown that :
\begin{eqnarray}
{\mathfrak S}^{(m,j)}_{j_1+1}(z)-(m-1)\cdot {\mathfrak S}^{(m,j)}_{j_1}(z)=
\frac{x^{j+1}}{(x-1)^{j_1+1}}-\frac{1}{z^{j_1+1}}+
\sum\limits_{l=1}^{j_1}\binom{l\cdot m-j+j_1-l-1}{j_1-l} \frac{(-1)^{j_1-l+1}}{z^l} 
\frac{(j-(j_1+1)\cdot m+1)}{j_1-l+1}
\end{eqnarray}
which is a tautology as you can check plugging that expression to Mathematica and settting $j_1=1,2,3,\cdots$. This completes the proof.
