Extension of the q-Pfaff-Saalschütz Theorem by Two Integer Parameters

Abstract

We investigate a class of terminating ${}_3{\varphi }_2$-series that comes from the balanced series perturbed by two extra integer parameters. By making use of the linearization method, a general summation formula is established that extends the well-known q-Pfaff-Saalschütz theorem. Five closed formulae are exemplified as applications.

1. Introduction and Outline

Denote by $\mathbb{N}$ the set of natural numbers with ${\mathbb{N}}_0=\left\{0\right\}\cup \mathbb{N}$. For an indeterminate x, the rising and falling q-shifted factorials of order $n\in {\mathbb{N}}_0$ are defined by

$$\begin{array}{l}{(x;q)}_0=1\mathrm{and}{(x;q)}_n=(1-x)(1-xq)\cdots (1-q^{n-1}x)\\ {\langle x;q\rangle }_0=1\mathrm{and}{\langle x;q\rangle }_n=(1-x)(1-x/q)\cdots (1-q^{1-n}x)\end{array}\}\mathrm{for}n\in \mathbb{N}$$

with the quotient form for the former being abbreviated compactly to

$${\left[\begin{array}{c}a,b,\dots ,c\\ \alpha ,\beta ,\dots ,\gamma \end{array}|q\right]}_n=\frac{{(a;q)}_n{(b;q)}_n\cdots {(c;q)}_n}{{(\alpha ;q)}_n{(\beta ;q)}_n\cdots {(\gamma ;q)}_n}.$$

Following Gasper and Rahman ([1], §1.2), the basic hypergeometric series reads as

$${}_{1+p}{\varphi }_p\left[\begin{array}{cccc}{a}_0,& a_1,& \dots ,& a_p\\ & b_1,& \dots ,& b_p\end{array}|q;z\right]=\sum _{n=0}^{\infty }{\left[\begin{array}{cccc}{a}_0,& a_1,& \cdots ,& a_p\\ q,& b_1,& \cdots ,& b_p\end{array}|q\right]}_n{z}^n.$$

This series is terminating when one of the numerator parameters has the form $q^{-m}$ with $m\in {\mathbb{N}}_0$. It is said to be balanced when the quotient between the product of its denominator parameters and the product of its numerator parameters is equal to q. If that quotient results in $q^m$, where m is a small positive integer greater than 1, then the series will be said to be “quasi-balanced”.

There are numerous q-series identities. Among them, the q-Saalschütz summation theorem (cf. [1], Equation (II.12)) is fundamental:

$${}_3{\varphi }_2\left[\begin{array}{c}{q}^{-n},a,b\\ c,q^{1-n}ab/c\end{array}|q;q\right]={\left[\begin{array}{c}c/a,c/b\\ c,c/\left(ab\right)\end{array}|q\right]}_n.$$

(1)

For $\lambda ,\mu \in {\mathbb{N}}_0$ subject to $\lambda \ge \mu$, we shall investigate by means of the linearization method (cf. [2,3,4]), the following quasi-balanced series

$${\mathsf{\Omega }}_n(\lambda ,\mu |a,b,c)={}_3{\varphi }_2\left[\begin{array}{c}{q}^{-n},a,b\\ c,q^{1-n+\lambda }ab/c\end{array}|q;q^{1+\mu }\right].$$

(2)

The main theorem will be proved in the next section which expresses the series ${\mathsf{\Omega }}_n(\lambda ,\mu |a,b,c)$ explicitly as a linear combination of ${\mathsf{\Omega }}_n(0,0|a,b,c)$ with the number of terms being independent of n. Then the paper will end in Section 3, where five explicit formulae are presented as examples.

2. Main Theorem and Proof

Recently, the linearization method has successfully been employed in [2,3,4], to extend several important known theorems of hypergoemetric series and q-series by integer parameters. In this section, we are going first to establish the linearization lemma below and then apply it to reduce the series ${\mathsf{\Omega }}_n(\lambda ,\mu |a,b,c)$ in terms of ${\mathsf{\Omega }}_n(0,0|a,b,c)$, which has the closed expression as in (1).

Lemma 1.

For a variable y, non-negative integers $\{\lambda ,\mu \}$ with $\lambda \ge \mu$ and two indeterminates $\{A,B\}$, the following linear relation holds:

$$y^{\mu }=\sum _{i=0}^{\lambda }{(Ay;q)}_{\lambda -i}{\langle By;q\rangle }_i{W}_i(\lambda ,\mu ).$$

(3)

The connection coefficients $W_i(\lambda ,\mu )$ are independent of y and given explicitly by the expressions

$$\begin{array}{cc}\hfill W_i(\lambda ,\mu )& =\frac{{(-1)}^i{B}^{-\mu }}{{(A/B;q)}_{\lambda }}\left[{\displaystyle \genfrac{}{}{0.0pt}{}{\lambda }{i}}\right]{\left[{\displaystyle \genfrac{}{}{0.0pt}{}{\lambda }{\mu }}\right]}^{-1}\sum _{j=0}^{\mu }{q}^{\left(\genfrac{}{}{0pt}{}{i}{2}\right)+(i-j)(\mu -j)}\left[{\displaystyle \genfrac{}{}{0.0pt}{}{i}{j}}\right]\left[{\displaystyle \genfrac{}{}{0.0pt}{}{\lambda -i}{\mu -j}}\right]{\left(\frac{A}{B}\right)}^{i-j}\hfill \end{array}$$

(4)

$$\begin{array}{cc}& ={(-1)}^i\left[{\displaystyle \genfrac{}{}{0.0pt}{}{\lambda }{i}}\right]{\left[{\displaystyle \genfrac{}{}{0.0pt}{}{\lambda }{\mu }}\right]}^{-1}\frac{A^{i-\mu }{q}^{\left(\genfrac{}{}{0pt}{}{i+\mu }{2}\right)-{\mu }^2}}{B^{\mu +i}{(A/B;q)}_{\lambda }}\left[x^{\mu }\right]{(-qAx;q)}_{\lambda -i}{\langle -Bx;q\rangle }_i,\hfill \end{array}$$

(5)

where $\left[x^n\right]P\left(x\right)$ stands for the coefficient of $x^n$ in the polynomial $P\left(x\right)$ and the Gaussian binomial coefficient is defined by

$$\left[{\displaystyle \genfrac{}{}{0.0pt}{}{\lambda }{\mu }}\right]=\frac{{\langle q^{\lambda };q\rangle }_{\mu }}{{(q;q)}_{\mu }}=\frac{{(q;q)}_{\lambda }}{{(q;q)}_{\mu }{(q;q)}_{\lambda -\mu }}.$$

Proof. 

It suffices to substitute (4) or (5) into (3) and then verify that the resulting sum equals $y^{\mu }$ for $\mu \le \lambda$. First, denote by $\mathrm{S}$ the resulting double sum after having replaced $W_i(\lambda ,\mu )$ in (3) by (4). In view of the q-binomial product

$$\left[{\displaystyle \genfrac{}{}{0.0pt}{}{\lambda }{i}}\right]{\left[{\displaystyle \genfrac{}{}{0.0pt}{}{\lambda }{\mu }}\right]}^{-1}\left[{\displaystyle \genfrac{}{}{0.0pt}{}{i}{j}}\right]\left[{\displaystyle \genfrac{}{}{0.0pt}{}{\lambda -i}{\mu -j}}\right]=\left[{\displaystyle \genfrac{}{}{0.0pt}{}{\mu }{j}}\right]\left[{\displaystyle \genfrac{}{}{0.0pt}{}{\lambda -\mu }{i-j}}\right],$$

we can manipulate, by interchanging the summation order, $\mathrm{S}$ as follows:

$$\begin{array}{cc}\hfill \mathrm{S}=& \sum _{i=0}^{\lambda }\frac{{(-1)}^i{B}^{-\mu }}{{(A/B;q)}_{\lambda }}\left[{\displaystyle \genfrac{}{}{0.0pt}{}{\lambda }{i}}\right]{\left[{\displaystyle \genfrac{}{}{0.0pt}{}{\lambda }{\mu }}\right]}^{-1}{(Ay;q)}_{\lambda -i}{\langle By;q\rangle }_i\hfill \\ & \times \sum _{j=0}^{\mu }{\left(\frac{A}{B}\right)}^{i-j}\left[{\displaystyle \genfrac{}{}{0.0pt}{}{i}{j}}\right]\left[{\displaystyle \genfrac{}{}{0.0pt}{}{\lambda -i}{\mu -j}}\right]q^{\left(\genfrac{}{}{0pt}{}{i}{2}\right)+(i-j)(\mu -j)}\hfill \\ \hfill =& \sum _{j=0}^{\mu }{\left(\frac{B}{A}\right)}^j\left[{\displaystyle \genfrac{}{}{0.0pt}{}{\mu }{j}}\right]\frac{B^{-\mu }}{{(A/B;q)}_{\lambda }}\mathrm{S}\left(j\right),\hfill \end{array}$$

(6)

where $\mathrm{S}\left(j\right)$ is given by

$$\mathrm{S}\left(j\right)=\sum _{i=0}^{\lambda }{\left(-\frac{A}{B}\right)}^i\left[{\displaystyle \genfrac{}{}{0.0pt}{}{\lambda -\mu }{i-j}}\right]{(Ay;q)}_{\lambda -i}{\langle By;q\rangle }_i{q}^{\left(\genfrac{}{}{0pt}{}{i}{2}\right)+(i-j)(\mu -j)}.$$

Observe that for the above sum $\mathrm{S}\left(j\right)$, the summation index i runs, in fact, from j to $\lambda$ since $\left[{\displaystyle \genfrac{}{}{0.0pt}{}{\lambda -\mu }{i-j}}\right]$ vanishes when $i<j$. We can rewrite it, by letting $i=k+j$, as

$$\begin{array}{cc}\hfill \mathrm{S}\left(j\right)& =\sum _{k=0}^{\lambda -j}{\left(-\frac{A}{B}\right)}^{k+j}\left[{\displaystyle \genfrac{}{}{0.0pt}{}{\lambda -\mu }{k}}\right]{(Ay;q)}_{\lambda -j-k}{\langle By;q\rangle }_{j+k}{q}^{\left(\genfrac{}{}{0pt}{}{k+j}{2}\right)+k(\mu -j)}\hfill \\ & =q^{\left(\genfrac{}{}{0pt}{}{j}{2}\right)}{\left(-\frac{A}{B}\right)}^j{\langle By;q\rangle }_j{(Ay;q)}_{\lambda -j}\sum _{k=0}^{\lambda -j}\frac{{(q^{\mu -\lambda };q)}_k{(q^j/\left(By\right);q)}_k}{{(q;q)}_k{(q^{1+j-\lambda }/\left(Ay\right);q)}_k}{q}^k,\hfill \end{array}$$

where the three equalities below have been utilized:

$$\begin{array}{cc}\hfill \left[{\displaystyle \genfrac{}{}{0.0pt}{}{\lambda -\mu }{k}}\right]& ={(-1)}^k\frac{{(q^{\mu -\lambda };q)}_k}{{(q;q)}_k}{q}^{k(\lambda -\mu )-\left(\genfrac{}{}{0pt}{}{k}{2}\right)},\hfill \\ \hfill {\langle By;q\rangle }_{j+k}& ={(-By)}^k{\langle By;q\rangle }_j{(q^j/\left(By\right);q)}_k{q}^{-kj-\left(\genfrac{}{}{0pt}{}{k}{2}\right)},\hfill \\ \hfill {(Ay;q)}_{\lambda -j-k}& ={(-Ay)}^{-k}\frac{{(Ay;q)}_{\lambda -j}}{{(q^{1+j-\lambda }/\left(Ay\right);q)}_k}{q}^{k(1+j-\lambda )+\left(\genfrac{}{}{0pt}{}{k}{2}\right)}.\hfill \end{array}$$

Evaluating the last sum with respect to k by the q-Chu-Vandermonde theorem (cf. [1], Equation (II.6))

$${}_2{\varphi }_1\left[\begin{array}{c}{q}^{\mu -\lambda },q^j/\left(By\right)\\ q^{1+j-\lambda }/\left(Ay\right)\end{array}|q;q\right]=\frac{{(q^{1-\lambda }B/A;q)}_{\lambda -\mu }}{{(q^{1+j-\lambda }/\left(Ay\right);q)}_{\lambda -\mu }}{(q^j/\left(By\right))}^{\lambda -\mu }=\frac{{(q^{\mu }A/B;q)}_{\lambda -\mu }}{{(q^{\mu -j}Ay;q)}_{\lambda -\mu }},$$

we find that

$$\begin{array}{cc}\hfill \mathrm{S}\left(j\right)& =q^{\left(\genfrac{}{}{0pt}{}{j}{2}\right)}{\left(-\frac{A}{B}\right)}^j{\langle By;q\rangle }_j{(Ay;q)}_{\lambda -j}\frac{{(q^{\mu }A/B;q)}_{\lambda -\mu }}{{(q^{\mu -j}Ay;q)}_{\lambda -\mu }}\hfill \\ & =q^{\left(\genfrac{}{}{0pt}{}{j}{2}\right)}{\left(-\frac{A}{B}\right)}^j{(q^{\mu }A/B;q)}_{\lambda -\mu }{\langle By;q\rangle }_j{(Ay;q)}_{\mu -j}.\hfill \end{array}$$

Substituting this into (6) leads us to the expression

$$\mathrm{S}=\frac{{(q^{\mu }A/B;q)}_{\lambda -\mu }}{B^{\mu }{(A/B;q)}_{\lambda }}\sum _{j=0}^{\mu }{(-1)}^j\left[{\displaystyle \genfrac{}{}{0.0pt}{}{\mu }{j}}\right]q^{\left(\genfrac{}{}{0pt}{}{j}{2}\right)}{\langle By;q\rangle }_j{(Ay;q)}_{\mu -j}.$$

Taking into account that

$$\begin{array}{cc}\hfill \left[{\displaystyle \genfrac{}{}{0.0pt}{}{\mu }{j}}\right]& ={(-1)}^j\frac{{(q^{-\mu };q)}_j}{{(q;q)}_j}{q}^{j\mu -\left(\genfrac{}{}{0pt}{}{j}{2}\right)},\hfill \\ \hfill {\langle By;q\rangle }_j& ={(-By)}^j{(1/\left(By\right);q)}_j{q}^{-\left(\genfrac{}{}{0pt}{}{j}{2}\right)},\hfill \\ \hfill {(Ay;q)}_{\mu -j}& ={(-Ay)}^{-j}\frac{{(Ay;q)}_{\mu }}{{(q^{1-\mu }/\left(Ay\right);q)}_j}{q}^{\left(\genfrac{}{}{0pt}{}{j}{2}\right)-j\mu +j};\hfill \end{array}$$

we can reduce further

$$\mathrm{S}=\frac{{(Ay;q)}_{\mu }}{B^{\mu }{(A/B;q)}_{\mu }}\sum _{j=0}^{\mu }{\left(\frac{qB}{A}\right)}^j\frac{{(q^{-\mu };q)}_j{(1/\left(By\right);q)}_j}{{(q;q)}_j{(q^{1-\mu }/\left(Ay\right);q)}_j}.$$

According to another q-Chu-Vandermonde theorem (cf. [1], Equation (II.7)), the last sum admits the closed form

$${}_2{\varphi }_1\left[\begin{array}{c}{q}^{-\mu },1/\left(By\right)\\ q^{1-\mu }/\left(Ay\right)\end{array}|q;\frac{qB}{A}\right]=\frac{{(q^{1-\mu }B/A;q)}_{\mu }}{{(q^{1-\mu }/\left(Ay\right);q)}_{\mu }}=\frac{{(A/B;q)}_{\mu }}{{(Ay;q)}_{\mu }}{\left(By\right)}^{\mu }.$$

This confirms finally that

$$\mathrm{S}=y^{\mu }\mathrm{for}\mu \le \lambda .$$

In order to show the equivalent expression (5), consider the q-binomial expansions ([1], Equation (II.4))

$$\begin{array}{cc}\hfill {\langle -Bx;q\rangle }_i& =\sum _{j=0}^i{\left(Bx\right)}^j\left[{\displaystyle \genfrac{}{}{0.0pt}{}{i}{j}}\right]q^{\left(\genfrac{}{}{0pt}{}{j+1}{2}\right)-ij},\hfill \\ \hfill {(-qAx;q)}_{\lambda -i}& =\sum _{k=0}^{\lambda }{\left(Ax\right)}^k\left[{\displaystyle \genfrac{}{}{0.0pt}{}{\lambda -i}{k}}\right]q^{\left(\genfrac{}{}{0pt}{}{k+1}{2}\right)}.\hfill \end{array}$$

Then we can extract the coefficient

$$\left[x^{\mu -j}\right]{\langle -Bx;q\rangle }_i{(-qAx;q)}_{\lambda -i}=\sum _{j=0}^i{A}^{\mu -j}{B}^j\left[{\displaystyle \genfrac{}{}{0.0pt}{}{i}{j}}\right]\left[{\displaystyle \genfrac{}{}{0.0pt}{}{\lambda -i}{\mu -j}}\right]q^{\left(\genfrac{}{}{0pt}{}{\mu +1}{2}\right)+j(j-i-\mu )}.$$

Substituting this into (5) and then making some simplifications, we find that the resulting expression coincides with (4). This completes the proof of Lemma 1. □

Now, specify the parameters in Lemma 1 by

$$y\to q^k,A\to a\mathrm{and}B\to q^{\lambda -n}ab/c.$$

Then the linear relation corresponding to (3) becomes

$$q^{k\mu }=\sum _{i=0}^{\lambda }{(q^{k}a;q)}_{\lambda -i}{\langle q^{\lambda -n+k}ab/c;q\rangle }_i{𝒲}_i(\lambda ,\mu ),$$

(7)

where the connection coefficients ${𝒲}_i(\lambda ,\mu )$ as in (4) are given explicitly by

$$\begin{array}{ll}{𝒲}_i(\lambda ,\mu )& =\frac{{(-1)}^i{(c/\left(ab\right))}^{\mu }}{{(q^{n-\lambda }c/b;q)}_{\lambda }}\left[{\displaystyle \genfrac{}{}{0.0pt}{}{\lambda }{i}}\right]{\left[{\displaystyle \genfrac{}{}{0.0pt}{}{\lambda }{\mu }}\right]}^{-1}\\ & \times {\displaystyle \sum _{j=0}^{\mu }}{\left(\frac{c}{b}\right)}^{i-j}\left[{\displaystyle \genfrac{}{}{0.0pt}{}{i}{j}}\right]\left[{\displaystyle \genfrac{}{}{0.0pt}{}{\lambda -i}{\mu -j}}\right]q^{\left(\genfrac{}{}{0pt}{}{i}{2}\right)+(i-j)(\mu -j)+(\mu +i-j)(n-\lambda )}.\end{array}$$

(8)

By putting the linear relation (7) inside the series ${\mathsf{\Omega }}_n(\lambda ,\mu |a,b,c)$ and then exchanging the summation order, we can proceed with the double series in the following manner:

$$\begin{array}{cc}\hfill {\mathsf{\Omega }}_n(\lambda ,\mu |a,b,c)=& \sum _{k=0}^n{q}^k{\left[\begin{array}{c}{q}^{-n},a,b\\ q,c,q^{1-n+\lambda }ab/c\end{array}|q\right]}_k\hfill \\ \hfill \times & \sum _{i=0}^{\lambda }{(q^{k}a;q)}_{\lambda -i}{\langle q^{\lambda -n+k}ab/c;q\rangle }_i{𝒲}_i(\lambda ,\mu )\hfill \\ \hfill =& \sum _{i=0}^{\lambda }\frac{{(a;q)}_{\lambda -i}{𝒲}_i(\lambda ,\mu )}{{(q^{1-n+\lambda }ab/c;q)}_{-i}}\sum _{k=0}^n{q}^k{\left[\begin{array}{c}{q}^{-n},q^{\lambda -i}a,b\\ q,c,q^{1+\lambda -n-i}ab/c\end{array}|q\right]}_k.\hfill \end{array}$$

Evaluating the rightmost sum with respect to k by (1)

$${\mathsf{\Omega }}_n(0,0|q^{\lambda -i}a,b,c)=\frac{{(q^{i-\lambda }c/a,c/b;q)}_n}{{(c,q^{i-\lambda }c/\left(ab\right);q)}_n}$$

and then simplifying the result, we establish the following theorem.

Theorem 2.

Assuming ${𝒲}_i(\lambda ,\mu )$ as in (8), we have the following reduction formula

$$\begin{array}{cc}\hfill {\mathsf{\Omega }}_n(\lambda ,\mu |a,b,c)={\left[\begin{array}{c}c/a,c/b\\ c,c/\left(ab\right)\end{array}|q\right]}_n& \sum _{i=0}^{\lambda }{\left[\begin{array}{c}a,qa/c,q^{1-n}ab/c\\ qab/c,q^{1-n}a/c\end{array}|q\right]}_{\lambda -i}\hfill \\ & \times {\langle q^{\lambda -n}ab/c;q\rangle }_i{𝒲}_i(\lambda ,\mu ).\hfill \end{array}$$

Particular cases of this theorem have previously been investigated in the literature.

• Al-Salam and Fields [5] (see also [6]) derived symmetric expressions for the case $\lambda = \mu$.
• Srivastava [7] (see also [6]) considered the case $\mu = 0$.
• Verma and Joshi [8] examined both cases $\lambda = \mu$ and $\mu = 0$.

Furthermore, an extension of the q-Saalschütz summation theorem along a different direction into double series can be found in [9].

3. Special Cases as Examples

By assigning $\lambda$ and $\mu$ to small integers in Theorem 2, we can derive concrete formulae. Five examples are highlighted below as applications.

Example 1

(${\mathsf{\Omega }}_n(1,0)$).

$${}_3{\varphi }_2\left[\begin{array}{c}{q}^{-n},a,b\\ c,q^{2-n}ab/c\end{array}|q;q\right]={\left[\begin{array}{c}{q}^{-1}c/a,q^{-1}c/b\\ c,q^{-1}c/\left(ab\right)\end{array}|q\right]}_n\left\{1-\frac{(1-q^n)(1-qab/c)}{(1-qa/c)(1-qb/c)}\right\}.$$

Example 2

(${\mathsf{\Omega }}_n(1,1)$).

$${}_3{\varphi }_2\left[\begin{array}{c}{q}^{-n},a,b\\ c,q^{2-n}ab/c\end{array}|q;q^2\right]={\left[\begin{array}{c}{q}^{-1}c/a,q^{-1}c/b\\ c,q^{-1}c/\left(ab\right)\end{array}|q\right]}_n\left\{q^n-\frac{q(1-q^n)(1-qab/c)}{c(1-qa/c)(1-qb/c)}\right\}.$$

Example 3

(${\mathsf{\Omega }}_n(2,0)$).

$$\begin{array}{cc}\hfill {}_3{\varphi }_2& \left[\begin{array}{c}{q}^{-n},a,b\\ c,q^{3-n}ab/c\end{array}|q;q\right]={\left[\begin{array}{c}{q}^{-1}c/a,q^{-2}c/b\\ c,q^{-2}c/\left(ab\right)\end{array}|q\right]}_n\frac{(1-b)(1-qb)(1-\frac{c}{qab})}{(1-q)(1-q^{n}b)(1-\frac{c}{qa})}\hfill \\ \hfill \times & \{1-\frac{q{b}^3(1-q)(1-q^n)(1-q^{n-1})(1-\frac{q^{2}ab}{c})(1-q^{n-1}c)(1-q^{n-2}c)}{ac(1-b)(1-qb)(1-\frac{qb}{c})(1-\frac{q^{2}b}{c})(1-q^{n-1}b)(1-\frac{q^{n-2}c}{a})}\hfill \\ \hfill -& \frac{(1-\frac{q^{2}ab}{c})(1-q^{n}b)(qc+q^2{b}^2-q^{2}bc-q^{3}b-q^{n}bc+q^{2+n}bc)}{qc(1-qb)(1-\frac{qab}{c})(1-\frac{q^{2}b}{c})(1-q^{n-1}b)}\}.\hfill \end{array}$$

Example 4

(${\mathsf{\Omega }}_n(2,1)$).

$$\begin{array}{c}{}_3{\varphi }_2\left[\begin{array}{c}{q}^{-n},a,b\\ c,q^{3-n}ab/c\end{array}|q;q^2\right]={\left[\begin{array}{c}{q}^{-1}c/a,q^{-2}c/b\\ c,q^{-2}c/\left(ab\right)\end{array}|q\right]}_n\frac{(1-b)(1-q^{-1}/b)}{(1-qa/c)(1-q^{n}b)}\hfill \\ \times \{q^n+\frac{q^{2}b(1-q^{n}b)(1-q^{2}ab/c)(1-q^{n-2}c/b)}{c^2(1-b)(1-qb)(1-qb/c)(1-q^{2}b/c)}(c-qb-bc+q^{n}bc)\hfill \\ \hspace{1em}\hspace{1em}\hspace{1em}+\frac{q^2{b}^2(1-q^n)(1-q^{n-1})(1-q^{n-1}c)(1-q^{n-2}c)(1-q^{2}ab/c)}{ac(1-b)(1-qb)(1-qb/c)(1-q^{2}b/c)(1-q^{n-2}c/a)}\}.\end{array}$$

Example 5

(${\mathsf{\Omega }}_n(2,2)$).

$$\begin{array}{cc}\hfill {}_3{\varphi }_2\left[\begin{array}{c}{q}^{-n},a,b\\ c,q^{3-n}ab/c\end{array}|q;q^3\right]=& {\left[\begin{array}{c}{q}^{-1}c/a,c/b\\ c,q^{-2}c/\left(ab\right)\end{array}|q\right]}_n\hfill \\ \hfill \times & \{1-\frac{(1-q^n)(1-b)(1-qb)}{q^{2}b(1-qa/c)(1-q^{n-1}c/b)(1-q^{n-2}c/b)}\hfill \\ \hfill -& \frac{c(1-q^n)(1-q^{n-1}c)(1-q^{n-2}c)}{q^{2}a{b}^2(1-q^{n-2}c/a)(1-q^{n-1}c/b)(1-q^{n-2}c/b)}\}.\hfill \end{array}$$

Concluding Comments

In the course of proving Theorem 2, the linear representation formula in Lemma 1 plays a crucial role. The authors believe that this lemma should be useful also for examining other related q-series. The interested reader is enthusiastically encouraged to make further exploration.