1-q\right) ^{\alpha -1}\Gamma
_{q}\left( \alpha \right) }{K\left( \frac{u}{v};\alpha \right) } \\
&&\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }_{m-k+1}\Phi _{m-1}\left[
\left. 
\begin{array}{l}
a_{1},a_{2},...,a_{m-k},q^{\alpha } \\ 
b_{1},b_{2},...,b_{m-1}\end{array}\right\vert q,\frac{au}{vq^{\alpha }}\right] .\end{aligned}$$<span style="font-variant:small-caps;">Proof</span> Let the hypothesis of the theorem be satisfied. A charity of $\left( 14\right) $ gives$$f\left( x\right) =\sum\limits_{0}^{\infty }\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}\frac{a^{n}}{\left( q;q\right) _{n}}x^{n}.$$On setting $A_{n}=\dfrac{\left( a_{1},a_{2},...,a_{m-k};q\right) _{n}}{\left( b_{1},b_{2},...,b_{m-1},q\right) _{n}}\left( \left( -1\right) ^{n}q^{\frac{n\left( n-1\right) }{2}}\right) ^{k}\dfrac{a^{n}}{\left( q;q\right)
_{n}}$ and using Theorem 9 it implies$$_{q}N\left( x^{\alpha -1}f\left( x\right) \right) \left( u;v\right) =\left( 
\frac{u}{v}\right) ^{\alpha }\left( 1-q\right) ^{\alpha -1}\Gamma _{q}\left(
\alpha \right) \sum\limits_{0}^{\infty }A_{n}\frac{\left( q^{\alpha
};q\right) _{n}}{K\left( \frac{u}{v};\alpha +n\right) }\left( \frac{u}{v}\right) ^{n}.$$By using the fact that $K\left( A,\alpha \right) =q^{\alpha -1}K\left(
A,\alpha -1\right) ,$ the preceding equation gives$$\begin{aligned}
_{q}N\left( x^{\alpha -1}f\left( x\right) \right) \left( u;v\right)
&=&\left( \frac{u}{v}\right) ^{\alpha }\frac{\left( 1-q\right) ^{\alpha
-1}\Gamma _{q}\left( \alpha \right) }{K\left( \frac{u}{v};\alpha \right) }\sum\limits_{0}^{\infty }A_{n}\left( q^{\alpha };q\right) _{n} \\
&&\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ }\left( \left( -1\right) ^{n}q^{\frac{n\left( n-1\right) }{2}}\right) ^{-1}\left( \frac{-1}{q^{\alpha }}\right)
^{n}\left( \frac{u}{v}\right) ^{n} \\
&=&\frac{\left( \frac{u}{v}\right) ^{\alpha }\left( 1-q\right) ^{\alpha
-1}\Gamma _{q}\left( \alpha \right) }{K\left( \frac{u}{v};\alpha \right) 