# Expansion of summation of power series raised to a power

I need to simplify the expression written below to get the $x$ term in its simplest form: $$\ E=\left(\sum_{k=1}^b \sum_{n=0}^\infty Z_n(a,k) x^\frac{n+k}{2} \right)^t ,\$$ where $$\ Z_n(a,k)=\frac{\Gamma[a-k]}{{(1+k-a)_n}n!} \left(\frac{ab}{gb+\omega}\right)^{k+n} . \$$ $(.)_n$ denotes the Pochhammer symbol, and $a,g,\omega$ are positive constants. $b$ and $t$ are natural numbers. The infinite summation (power series) can be solved by using the relations given in this Link. However, a finite summation ( from $k=1$ to $b$ ) is involved in the above expression. I thought of using multinomial theorem, but I got confused.

I will greatly appreciate any hint or help. Thanks.

The expression $E$ can be rewritten as: $$E=\left(\sum_{n=0}^\infty Z_n(a,1)x^\frac{n+1}{2}+\dots+\sum_{n=0}^\infty Z_n(a,b)x^\frac{n+b}{2} \right)^t \tag{1}$$ Now, using Multinomial theorem, $(1)$ can be simplified as: $$E= \sum_{r_1+ r_2+\dots+r_b =t} \dbinom{t}{r_1, r_2 ,\dots , r_b} \left(\sum_{n=0}^\infty Z_n(a,1)x^\frac{n+1}{2} \right)^{r_1} \dots \left( \sum_{n=0}^\infty Z_n(a,b)x^\frac{n+b}{2} \right)^{r_b} \tag{2}$$
From $(0.314)$ in Gradshteyn and Ryzhik book, $(2)$ can be written as: $$E=\sum_{r_1+ r_2+\dots+r_b =t} \dbinom{t}{r_1, r_2 ,\dots , r_b} \left(\sum_{n=0}^\infty Z_n^\left({r_1}\right)(a,1)x^\frac{n+r_1}{2} \right) \dots \left( \sum_{n=0}^\infty Z_n^\left({r_b}\right)(a,b)x^\frac{n+br_b}{2} \right) \tag{3}$$
where $z^\left(r\right)$ denotes $z$ is convolved $(r-1)$ times with itself. Finally, $(3)$ is simplifed with the help of $(0.316)$ in Gradshteyn and Rydzhik book as:
$$E=\sum_{r_1+ r_2+\dots+r_b =t} \dbinom{t}{r_1, r_2 ,\dots , r_b} \sum_{n=0}^\infty C_n \left(Z_n^\left({r_1}\right)(a,1), \dots , Z_n^\left({r_b}\right)(a,b), \right)x^\frac{n+r_1+\dots+{br_b}}{2} \tag{4}$$
$C_n(Z_1,Z_2)$ indicates the convolution of $Z_1$ and $Z_2$, and can be calculated as explained in this Link.