<s>[INST] Yet another family of hypergeometric sums that has a closed form solution. Let $m \ge 2$ and $j\ge 0$ be integers. Now, let $0 < z < \frac{(m-1)^{m-1}}{m^m}$ be a real number. Consider a following sum:
\begin{equation}
{\mathfrak S}^{(m,j)}(z) := \sum\limits_{i=0}^\infty \binom{m \cdot i + j}{i}\cdot z^i
\end{equation}
By using both my answer to Closed form solutions for a family of hypergeometric sums. and by generalizing the approach from my answer to About the identity $\sum\limits_{i=0}^{\infty}\binom{2i+j}{i}z^i=\frac{B_2(z)^j}{\sqrt{1-4z}}$  I have derived the following results:
\begin{eqnarray}
{\mathfrak S}^{(m,0)}(z) &=& \frac{x \left(1-z x^{m-1}\right)}{1-m z x^{m-1}}\\
{\mathfrak S}^{(m,1)}(z) &=& \frac{m}{m-1} \cdot\frac{ x \left(1-z x^{m-1}\right)}{ \left(1-m z x^{m-1}\right)}-\frac{1}{m-1}\cdot \left(\frac{x-1}{x z}\right)^{\frac{1}{m-1}}\\
{\mathfrak S}^{(m,2)}(z) &=& \frac{m^2}{(m-1)^2}\cdot \frac{x \left(1-z x^{m-1}\right)}{ \left(1-m z x^{m-1}\right)}
-\frac{(m-2) }{(m-1)^2}\cdot \left(\frac{x-1}{x z}\right)^{\frac{1}{m-1}}+\\
&&
-\frac{1}{(m-1)^2}\cdot \left(\frac{x-1}{x z}\right)^{\frac{2}{m-1}} \cdot \frac{((m-1) x+2) }{ x}\\
{\mathfrak S}^{(m,3)}(z) &=& \frac{m^3}{(m-1)^3}\cdot \frac{x \left(1-z x^{m-1}\right)}{ \left(1-m z x^{m-1}\right)}-
\frac{(m-2)(2 m-3)}{2(m-1)^3}\cdot \left(\frac{x-1}{x z}\right)^{\frac{1}{m-1}}+\\
&&-\frac{m-3}{(m-1)^3}\cdot \left(\frac{x-1}{x z}\right)^{\frac{2}{m-1}}\cdot \frac{((m-1) x+2)}{x}+\\
&&-\frac{1}{2(m-1)^3}\cdot \left(\frac{x-1}{x z}\right)^{\frac{3}{m-1}}\cdot
\frac{2(m-1)^2 x^2+6(m-1)x+3(m+2)}{x^2}
\end{eqnarray}
Here $x:=x(z)$ is obtained in the following way. Out of the solutions of the equation:
\begin{equation}
1-x+z \cdot x^m=0
\end{equation}
we choose the one that is the closest to unity.
Now the question is how does the result look like for arbitrary $j \ge 2$.
 [/INST] Hint: The series ${\mathfrak S}^{(m,j)}(z)$ is strongly related with the generalized binomial series $B_m(z)$
\begin{align*}
B_m(z)=\sum_{i=0}^\infty\binom{mi+1}{i}\frac{1}{mi+1}z^i
\end{align*}
 defined in (5.58)  in Concrete Mathematics
by R.L. Graham, D.E. Knuth and O. Patashnik.

The series $B_m(z)$ satisfies the identities (see (5.60), (5.61)):
  \begin{align*}
B_m(z)^j&=\sum_{i=0}^\infty \binom{mi+j}{i}\frac{j}{mi+j}z^i\\
\frac{B_m(z)^j}{1-m+mB_m(z)^{-1}}&=\sum_{i=0}^\infty\binom{mi+j}{i}z^i
\end{align*}
  so that the following holds
  \begin{align*}
\color{blue}{{\mathfrak S}^{(m,j)}(z)=\frac{B_m(z)^j}{1-m+mB_m(z)^{-1}}}
\end{align*}

</s>