first type*$.$* Then, we have*$$N_{q}\left( x^{\alpha -1}f\left( x\right) \right) \left( u;v\right) =\dfrac{\left( 1-q\right) ^{\alpha -1}}{v^{\alpha }}u^{\alpha -1}\Gamma _{q}\left(
\alpha \right) .$$

$\left( \mathtt{ii}\right) $ *Let* $a\in 
\mathbb{R}
$ *and* $f\left( x\right) =_{m-k}\Phi _{m-1}\left[ \left. 
\begin{array}{c}
a_{1},a_{2},...,a_{m-k} \\ 
b_{1},b_{2},...,b_{m-1}\end{array}\right\vert q,ax\right] .$ *Then, we have*$$\begin{aligned}
\left( N_{q}x^{\alpha -1}f\left( x\right) \right) \left( u;v\right) &=&\dfrac{\Gamma _{q}\left( \alpha \right) \left( 1-q\right) ^{\alpha
-1}u^{\alpha -1}}{v^{\alpha }}  \notag \\
&&\text{ }_{m-k+1}\Phi _{m}\left[ \left. 
\begin{array}{c}
a_{1},a_{2},...,a_{m-k}q^{\alpha } \\ 
b_{1},b_{2},...,b_{m-1},0\end{array}\right\vert q,\frac{au}{v}\right] .  \TCItag{27}\end{aligned}$$<span style="font-variant:small-caps;">Proof of</span> $\left( \mathtt{i}\right) $ By assuming $A_{0}=1$ and $A_{n}=0,$ $\forall n\geq 1,$ it follows from $\left( 23\right) $ that $f\left( x\right) =1.$

Hence, the first part obviously follows.

<span style="font-variant:small-caps;">Proof of</span> $\left( \mathtt{ii}\right) $ Appealing to the $q$-analogue $\left( 14\right) ,f\left( x\right) $ can fairly be written as$$f\left( x\right) =\sum\limits_{n\geq 0}\frac{\left(
a_{1},a_{2},...,a_{m-k};q\right) _{n}}{\left(
b_{1},b_{2},...,b_{m-1};q\right) _{n}}\left( \left( -1\right) ^{n}q^{\frac{n\left( n-1\right) }{2}}\right) \frac{q^{n}}{\left( q;q\right) _{n}}x^{n}.$$

On setting$$A_{n}=\frac{\left( a_{1},a_{2},...,a_{m-k};q\right) _{n}}{\left(
b_{1},b_{2},...,b_{m-1};q\right) _{n}}\left( \left( -1\right) ^{n}q^{\frac{n\left( n-1\right) }{2}}\right) ^{k}a^{n},  \tag{28}$$we get the power series representation of type $\left( 23\right) .$ Hence, Theorem 1 reveals$$N_{q}\left( x^{\alpha -1}f\left( x\right) \right) \left( u;v\right) =\frac{\left( 1-q\right) ^{\alpha -1}}{v^{\alpha }}u^{\alpha -1}\sum\limits_{n\geq
0}A_{n}\frac{u^{n}}{v^{n}}\left( 1-q\right) ^{n}\Gamma _{q}\left( \alpha
+n\right) .$$Therefore, on using the identity of the gamma function $\left[ 1\right] $$$\Gamma _{q}\left( x+j\right) =\frac{\left( q^{x};q\right) _{j}}{\left(
