Closed form solutions for a family of hypergeometric sums. In Closed form solution of a hypergeometric sum. we provided a family of hypergeometric sums that have a closed form solution. Now we would like to investigate another family and come up with corresponding closed form solutions. Let $n \ge 0$ and $m \ge 0$ be integers. We consider a following sum:
\begin{equation}
{\mathfrak S}^{(m)}_n(x) := \sum\limits_{i=0}^\infty \binom{m \cdot i}{i+n} \cdot x^i
\end{equation}
Now, by using the duplication formula for the factorial, meaning by using the identity:
\begin{equation}
(2 i)! = 4^i i! \frac{(i-1/2)!}{(-1/2)!}
\end{equation}
for $i \ge 0$ and then by using known binomial identities we computed the sum in question for $m=2$. We have:
\begin{eqnarray}
{\mathfrak S}^{(2)}_n(x) = 
\frac{\sqrt{\pi} n!}{(n-1/2)!} \left[ 
\sum\limits_{p=0}^n \sum\limits_{q=0}^n \binom{n}{q} \binom{n-p+q-1}{q-1} \binom{-1/2}{p} (-1)^{n+q} \frac{(4 x)^{p-q}}{(1-4 x)^{p+1/2}}\right.\\
-
\sum\limits_{p=0}^{n-1} \binom{n-1/2}{p} (-1)^p \binom{2 n-p-1}{n} \frac{1}{(4 x)^{n-p}} 
\left.\right]
\end{eqnarray}
valid for $x \in (0,1/4)$. 
Now the obvious question is of course what is the result for arbitrary values of $n$ and $m$?
 A: Let us denote the sum in question as:
\begin{equation}
{\mathfrak S}^{(m)}_j(w) := \sum\limits_{i=0}^\infty \binom{m\cdot i}{i+j} \cdot w^i
\end{equation}
We provide an answer for arbitrary $m\ge 2$ and $n=0$.
In https://www.researchgate.net/profile/Larry_Glasser/publication/252738212_Hypergeometric_functions_and_the_trinomial_equation/links/5454c36a0cf2bccc490c41ba.pdf the author proves an interesting result.
Let $|w| < (m-1)^{m-1}/m^m$. Out of the solutions of the  following transcendental equation:
\begin{equation}
1-x+w \cdot x^m=0
\end{equation}
we choose the one the one that is closest to unity.
Then we have:
\begin{equation}
x=1+ \sum\limits_{i=1}^\infty \binom{m \cdot i}{i} \cdot \frac{w^i}{((m-1)i+1)}
\end{equation}
Now by substituting for $\tilde{w}:=w^{1/(m-1)}$ then by multiplying both sides of the above by $\tilde{w}$ and differentiating with respect to $\tilde{w}$ we readily get the following identity:
\begin{equation}
\sum\limits_{i=0}^\infty \binom{m \cdot i}{i} \cdot w^i = \frac{x\cdot\left(1-w \cdot x^{m-1} \right)}{1-m \cdot w \cdot x^{m-1}}=\frac{x}{(1-m) x+m}
\end{equation}
Now, let us take arbitary value of $j\ge 0$. Then clearly we have:
\begin{eqnarray}
\binom{m \cdot i}{i+j}&=& \binom{m \cdot i}{i} \cdot \frac{((m-1) \cdot i)_{(j)}}{(i+1)^{(j)}}\\
&=&\binom{m \cdot i}{i} \cdot\left[(m-1)^j + \sum\limits_{l=1}^j \binom{l \cdot m+j-l-1}{j} j \binom{j-1}{l-1} \cdot \frac{(-1)^{j-l+1}}{(i+l)}\right]
\end{eqnarray}
where in the second line we decomposed the ratio of Pochhammer factors into partial fractions with respect to the variable $i$.
Now since $1/(i+l) = \int\limits_0^1 \theta^{i+l-1} d\theta$ we have:
\begin{eqnarray}
{\mathfrak S}^{(m)}_j(w)&=& (m-1)^j \cdot \frac{x}{(1-m) x+m} +\sum\limits_{l=1}^j \binom{l \cdot m+j-l-1}{j} j \binom{j-1}{l-1} \cdot (-1)^{j-l+1}\cdot \\
&&\int\limits_1^x \left( \frac{\xi-1}{w \xi^m}\right)^{l-1} \cdot \frac{1}{w \xi^{m}} d\xi\\
&=& (m-1)^j \cdot \frac{x}{(1-m) x+m} +\sum\limits_{l=1}^j \binom{l \cdot m+j-l-1}{j} j \binom{j-1}{l-1} \cdot \frac{(-1)^{j-l+1}}{w^l} \\
&&\int\limits_0^{x-1} \frac{u^{l-1}}{(1+u)^{l m}} du\\
\end{eqnarray}
Therefore the final result reads:
\begin{eqnarray}
{\mathfrak S}^{(m)}_j(w)
&=& \frac{x}{((1-m)x+m)(x-1)^j}+\sum\limits_{l=1}^j (-1)^{j-l+1} \binom{l \cdot m+j-l-1}{j-l} \cdot \frac{1}{w^{l}}
\end{eqnarray}
