Find $\sum_{i=1}^{\infty}i(2y^i-y^{2i})^b$ for $0I am trying to find a closed form expression for $\sum_{i=1}^{\infty}i(2y^i-y^{2i})^b$ where $0<y<1$ and b is a positive integer. I was thinking binomial expansion but didn't get anywhere with it.
Any help appreciated.
 A: Using the  binomial theorem: 
$$
\begin{align}
  & \quad \quad \quad =\sum\nolimits_{i=1}^{\infty }{\left[ i\sum\nolimits_{k=0}^{b}{\left( \begin{matrix}
  b \\ 
  k \\ 
\end{matrix} \right){{2}^{k}}{{y}^{ik}}{{\left( -{{y}^{2i}} \right)}^{b-k}}} \right]} \\ 
 & \quad \quad \quad =\sum\nolimits_{i=1}^{\infty }{\left[ i\sum\nolimits_{k=0}^{b}{{{\left( -1 \right)}^{b-k}}{{2}^{k}}\left( \begin{matrix}
  b \\ 
  k \\ 
\end{matrix} \right){{y}^{2ib-ik}}} \right]} \\ 
 & \quad \quad \quad =\sum\nolimits_{i=1}^{\infty }{\left[ \sum\nolimits_{k=0}^{b}{c\left( a,k \right)i{{y}^{2ib-ik}}} \right]}\quad \quad where\quad c\left( a,k \right)={{\left( -1 \right)}^{b-k}}{{2}^{k}}\left( \begin{matrix}
  b \\ 
  k \\ 
\end{matrix} \right) \\ 
 & \quad \quad \quad =\sum\nolimits_{k=0}^{b}{\left[ c\left( a,k \right)\sum\nolimits_{i=1}^{\infty }{i{{y}^{2ib-ik}}} \right]} \\ 
 & \quad \quad \quad =\sum\nolimits_{k=0}^{b}{\left[ c\left( a,k \right)\frac{{{y}^{2b+k}}}{{{\left( {{y}^{k}}-{{y}^{2b}} \right)}^{2}}} \right]} \\ 
\end{align}
$$
in the last line i used the identity(for a positive integer$ \alpha$):
$$\sum\nolimits_{i=1}^{\infty }{i{{y}^{i\alpha }}}=\frac{{{y}^{\alpha }}}{{{\left( {{y}^{\alpha }}-1 \right)}^{2}}}$$
