\ \ \ \ \ \ \ H_{M,N+1}^{m+1,n}\left[
\lambda \dfrac{u^{k}}{v^{k}};q\left\vert 
\begin{array}{l}
\left( a_{1},\alpha _{1}\right) ,...,\left( a_{M},\alpha _{M}\right) \\ 
\left( 1+\lambda ,-k\right) ,\left( b_{1},\beta _{1}\right) ,...,\left(
b_{N},\beta _{N}\right)\end{array}\right. \right] .$

<span style="font-variant:small-caps;">Proof</span> We prove Part $\left( \mathtt{i}\right) $ since proof of Part $\left( \mathtt{ii}\right) $ is similar. By considering $\left( 45\right) ,$ we have

$N_{q}\left( x^{\lambda }H_{M,N}^{m,n}\left[ \lambda x^{k};q\left\vert 
\begin{array}{l}
\left( a_{1},\alpha _{1}\right) ,\left( a_{1},\alpha _{1}\right) ,...,\left(
a_{M},\alpha _{M}\right) \\ 
\left( b_{1},\beta _{1}\right) ,b_{2},...,\left( b_{N},\beta _{N}\right)\end{array}\right. \right] \right) \left( u;v\right) =$

$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \dfrac{1}{2\pi i}\dint\nolimits_{C}\frac{\dprod\limits_{j=1}^{m}G\left( q^{b_{j}-\beta _{j}z}\right)
\dprod\limits_{j=1}^{n}G\left( q^{1-a_{j}+\alpha _{j}z}\right) \pi \left(
\lambda \right) ^{z}}{\dprod\limits_{j=m+1}^{N}G\left( q^{1-b_{j}+\beta
_{j}z}\right) \dprod\limits_{j=n+1}^{M}G\left( q^{a_{j}-\alpha _{j}z}\right)
G\left( q^{1-z}\right) \sin \pi z}N_{q}\left( x^{\lambda +kz}\right) \left(
u;v\right) d_{q}z.$$$\tag{49}$$By virtue of Theorem 2, $\left( 49\right) $ gives

$N_{q}\left( x^{\lambda }H_{M,N}^{m,n}\left[ \lambda x^{k};q\left\vert 
\begin{array}{l}
\left( a_{1},\alpha _{1}\right) ,\left( a_{1},\alpha _{1}\right) ,...,\left(
a_{M},\alpha _{M}\right) \\ 
\left( b_{1},\beta _{1}\right) ,b_{2},...,\left( b_{N},\beta _{N}\right)\end{array}\right. \right] \right) \left( u;v\right) =$

$\ \ \ \ \dfrac{1}{2\pi i}\dint\nolimits_{C}\frac{\dprod\limits_{j=1}^{m}G\left( q^{b_{j}-\beta _{j}z}\right) \dprod\limits_{j=1}^{n}G\left(
q^{1-a_{j}+\alpha _{j}z}\right) \pi \left( \lambda \right) ^{z}}{\dprod\limits_{j=m+1}^{N}G\left( q^{1-b_{j}+\beta _{j}z}\right)
\dprod\limits_{j=n+1}^{M}G\left( q^{a_{j}-\alpha _{j}z}\right) G\left(
q^{1-z}\right) \sin \pi z}\dfrac{\left( 1-q\right) ^{\lambda +kz}}{v^{\lambda +kz+1}}\Gamma _{q}\left( \lambda +kz+1\right) d_{q}z.$

The fact that $\left( 1-q\right) ^{\lambda +kz}\Gamma _{q}\left(
\lambda 