1-q\right) _{j}}\Gamma _{q}\left( x\right) ,  \tag{29}$$the previous equation consequently gives$$\begin{aligned}
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( q^{\alpha };q\right) _{n}\text{ }\Gamma
_{q}\left( \alpha \right) \\
&=&\frac{\Gamma _{q}\left( \alpha \right) \left( 1-q\right) ^{\alpha
-1}u^{\alpha -1}}{v^{\alpha }}\sum\limits_{n\geq 0}A_{n}\frac{u^{n}}{v^{n}}\left( q^{\alpha };q\right) _{n}\frac{u^{n}}{v^{n}} \\
&=&\frac{\Gamma _{q}\left( \alpha \right) \left( 1-q\right) ^{\alpha
-1}u^{\alpha -1}}{v^{\alpha }}\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] .\end{aligned}$$This completes the proof of the theorem.

An inspection of the previous two main theorems leads to the following list of inclusions:

On setting $k=m=1,$ Theorem $2\left( \mathtt{ii}\right) \left( 
\text{for }\alpha =1\right) $ yields$$N_{q}\left( E_{q}\left( ax\right) \right) \left( u;v\right) =\frac{1}{v}\text{ }_{1}\Phi _{1}\left[ \left. 
\begin{array}{c}
q \\ 
0\end{array}\right\vert q,\frac{au}{v}\right] .  \tag{30}$$Similarly, by inserting $\alpha =1$, Theorem $2\left( \mathtt{i}\right) $ instantly shows$$\left( N_{q}\left( 1\right) \right) \left( u;v\right) =\frac{1}{v}.  \tag{31}$$Moreover, by setting $\alpha =n+1$ in the first part of the theorem$,$ it yields $$N_{q}\left( x^{n}\right) \left( u;v\right) =\frac{\left( 1-q\right) ^{n}}{v^{n+1}}u^{n}\left( \left[ n\right] _{q}\right) !\text{ }.  \tag{32}$$Therefore, $\left( 31\right) $ and $\left( 32\right) $ when designated reveal$$N_{q}\left( 1+x^{2}+...+x^{n}\right) \left( u;v\right) =\sum\limits_{k\geq 0}\frac{\left( 1-q\right) ^{k}}{v^{k+1}}u^{k}\left( \left[ k\right]
_{q}\right) !\text{ }.$$We finally establish the main third theorem of this section.

<span style="font-variant:small-caps;">Theorem 3.</span> *Let* $f\left( x\right) $* be given as* 