\magnification=\magstep1
\input texmacros
\TagsOnRight

\def\sumn{\sum^\infty_{n=0}}
\def\sumk{\sum^\infty_{k=0}}
\def\sumj{\sum^\infty_{j=0}}
\def\sumnk{\sum^n_{k=0}}
\def\G{\Gamma}
\def\D{\Delta}
\def\cb{\left(\frac cb\right)}
\def\sa{\sqrt a}
\def\gfrac#1{\Gamma\left(\frac{#1}3\right)}
\def\pt{\frac\pi3}
\def\spt{\sin\pt}
\def\cpt{\cos\pt}
\def\pa{\pi\alpha}
\def\pb{\pi\beta}
\def\spa{\sin\pa}
\def\cpa{\cos\pa}
\def\spb{\sin\pb}
\def\cpb{\cos\pb}
\def\ab{(\alpha+\beta)}
\def\amb{(\alpha-\beta)}
\def\i{\infty}
\def\q{q^2}

\title
An Indefinite Bibasic Summation Formula and Some Quadratic, \\
Cubic, and Quartic Summation and Transformation Formulas \endtitle

\author George Gasper \footnote{ 
Department of Mathematics, Northwestern University, Evanston, Illinois 60208. 
The research of this author was supported in part by the National Science 
Foundation under grant DMS-8601901.} 
 and Mizan Rahman \footnote{
Department of Mathematics and Statistics, Carleton University, Ottawa, 
Ontario, Canada K1S 5B6.  The research of this author was supported in part by 
the National Science and Engineering Research Council under grant A6197.}
\endauthor

\centerline{ (Published in Canad. J. Math. 42 (1990), 1--27)}

{\narrower\ms\ni
Abstract.   An indefinite bibasic summation formula containing four 
arbitrary parameters is derived and used to to derive a bilateral bibasic 
summation formula and some rather general quadratic, cubic, and quartic 
summation and transformation formulas.   In particular, a cubic 
transformation formula with four parameters is derived which contains as a 
special case a $q$-extension of Gosper's previously unproven strange 
evaluation of a  ${}_7F_6$ series. \ms}

\subheading{1. Introduction} Let
$$(a;q)_n = \cases
1,  & n=0, \\
(1-a)(1-aq)\cdots(1-aq^{n-1}), & n=1,2,\dots, \\
[(1-aq^{-1})(1-aq^{-2}) \cdots(1-aq^n)]^{-1}, & n=-1,-2,\dots, \endcases$$
denote the $q$-shifted factorial,
$$(a;q)_\infty=\prod^\infty_{k=0} (1-aq^k), \qquad |q|<1,$$
and set
$$\align
(a_1,a_2,\dots,a_m;q)_n &= (a_1;q)_n(a_2;q)_n\cdots (a_m;q)_n, \\
(a_1,a_2,\dots,a_m;q)_\infty &= (a_1;q)_\infty(a_2;q)_\infty\cdots 
(a_m;q)_\infty. \endalign$$
Recently, Gasper [7] showed that some bibasic summation formulas derived 
by Carlitz [5], Al-Salam and Verma [1], and Wm.~Gosper could be extended to 
the indefinite bibasic summation formula
$$\align
&{} \sum^n_{k=0} \frac{(1-ap^kq^k)(1-bp^kq^{-k})}{(1-a)(1-b)} \,
\frac{(a,b;p)_k(c,a/bc;q)_k}{(q,aq/b;q)_k (ap/c,bcp;p)_k}\, q^k \tag 1.1 \\
&= \frac{(ap,bp;p)_n(cq,aq/bc;q)_n}{(q,aq/b;q)_n (ap/c,bcp;p)_n}, 
\qquad n=0,1,2,\dots, \endalign$$
\ni where  $p$ and $q$ are independent bases and $a$, $b$, $c$  are arbitrary 
parameters.   He also showed that this formula could be used to derive a 
bibasic extension of Euler's transformation formula [7, (1.6)], a bibasic 
Lagrange inversion formula, a bibasic extension of Verma's  $q$-analogue [13, 
p\. 349] of the Fields and Wimp expansion formula [6, (1.3)], and some 
quadratic, cubic, and quartic summation formulas containing at least two 
parameters.

In particular, he derived the cubic summation formula
$$\align
&{} \sumk \frac{1-aq^{4k}}{1-a}\, \frac{(a,b;q)_k (q/b,q^2/b;q^2)_k 
(c,a^2b/c;q^3)_k}{(q^3,aq^3/b;q^3)_k (abq,ab;q^2)_k (aq/c,cq/ab;q)_k} \, q^k
\tag 1.2 \\
&+ \frac{ab^2}c \, \frac{(aq,bq,1/b;q)_\infty (c,a^2bq^3/c^2; 
q^3)_\infty}{(ab,cq/ab,aq/c;q)_\infty (ab^2/c,aq^3/b;q^3)_\infty} \\
&\cdot {}_2\phi_1 \left[\matrix
ab^2/c,a^2b/c \\ a^2bq^3/c^2 \endmatrix ; q^3,q^3 \right] \\
&= \frac{(aq,ab/c;q)_\infty (ab^2,aq^3/bc;q^3)_\infty}{(ab,aq/c;q)_\infty 
(ab^2/c,aq^3/b;q^3)_\infty}, \qquad |q|<1, \endalign$$
and its hypergeometric limit case
$$\align
&{}_7F_6 \left[\matrix
3a,1+3a/4,3b,(1-3b)/2, 1-3b/2,c,2a+b-c \\
3a/4,1+a-b,(1+3a+3b)/2,3(a+b)/2, 1+3a-3c,1+3c-3a-3b \endmatrix ;1 \right] \\
&= \frac{\Gamma(3a+3b)\Gamma (1+3a-3c) \Gamma(a+2b-c)\Gamma(1+a-b)}{\Gamma 
(3a+1)\Gamma(3a+3b-3c) \Gamma(a+2b)\Gamma(1+a-b-c)} \\
&\cdot \left\{ 1+\frac{\sin3\pi b \sin\pi c}{\sin3\pi(a+b-c) \sin\pi(a+2b)}
\right\}. 
\tag 1.3 \endalign$$

The ${}_2\phi_1$ function in (1.2) is a special case of the ${}_r\phi_s$ basic 
hypergeometric series defined by
$${}_r\phi_s \left[\matrix a_1,\dots,a_r \\ b_1,\dots,b_s\endmatrix ;q,z\right]
= \sumn \frac{(a_1,\dots,a_r;q)_n}{(q,b_1,\dots,b_s;q)_n} \,
z^n [(-1)^n q^{\binom n2}]^{1+s-r}
\tag 1.4 $$
with $\binom n2=n(n-1)/2$. In (1.4) and all formulas containing infinite sums 
or products it is assumed that $|q|<1$. An ${}_rF_s$ hypergeometric series is 
defined by
$${}_rF_s \left[\matrix
a_1,\dots,a_r \\ b_1,\dots,b_s \endmatrix ;z\right] = \sumn \frac{(a_1)_n 
\cdots (a_r)_n}{n!(b_1)_n\cdots(b_s)_n}\, z^n
\tag 1.5 $$
\ni where the shifted factorial  $(a)_n$ is defined by
$$(a)_n = \prod^{n-1}_{k=0} (a+k).$$

In their paper [9] on strange evaluations of hypergeometric series, 
Gessel and Stanton were able to prove all of the ``mysterious-looking 
evaluations'' with more than one parameter that were stated by Wm.~Gosper 
in a letter to R. Askey except for the evaluation
$$\align
&{}_7F_6 \left[\matrix
a,a+1/2,b,1-b,c, (2a+1)/3-c,a/2+1 \\
1/2,(2a-b+3)/3, (2a+b+2)/3,3c, 2a+1-3c,a/2 \endmatrix ;1 \right] 
\tag 1.6 \\
&= \frac2{\sqrt3}\, \frac{\G\left( c+\frac13\right) \G\left(c+\frac23\right)
\G\left( \frac{2a-b+3}3\right)\G\left( \frac{2a+b+2}3\right)}{\G\left( 
\frac{2a+2}3\right) \G\left( \frac{2a+3}3\right) \G\left(\frac{3c+b+1}3\right) 
\G\left(\frac{3c+2-b}3\right)} \\
&\cdot \frac{\G\left(\frac{2+2a-3c}3\right)\G\left( \frac{3+2a-3c}3\right)\sin 
\frac\pi3\,(b+1)}{\G\left(\frac{2+2a+b-3c}3\right) \G\left(\frac{3+2a-b-3c}3 
\right)}. \endalign$$
Since the case $b=2a$ of (1.6) is a special case of (1.3), this raised the 
question of whether (1.6) and a $q$-extension of it could be derived by 
extending the method in [7] and it led the authors to look for extensions of 
(1.1) and (1.2).

In this paper we show that (1.1) can be extended to
$$\align
&{} \sum^n_{k=-m} \frac{(1-adp^kq^k)(1-bp^k/dq^k)}{(1-ad)(1-b/d)}\, 
\frac{(a,b;p)_k(c,ad^2/bc;q)_k}{(dq,adq/b;q)_k (adp/c,bcp/d;p)_k}\, q^k 
\tag 1.7 \\
&= \frac{(1-a)(1-b)(1-c)(1-ad^2/bc)}{d(1-ad)(1-b/d)(1-c/d)(1-ad/bc)} \\
&\cdot \left\{ \frac{(ap,bp;p)_n (cq,ad^2q/bc;q)_n}{(dq,adq/b;q)_n 
(adp/c,bcp/d;p)_n} - \frac{(c/ad, d/bc;p)_{m+1} (1/d,b/ad;q)_{m+1}}{(1/c, 
bc/ad^2;q)_{m+1} (1/a,1/b;p)_{m+1}} \right\}, \endalign$$
where $n,m=0,\pm1,\pm2,\dots$, and then use this formula to derive the cubic 
summation formula
$$\align
&{} \sumk \frac{1-a^2q^{4k}}{1-a^2} \, \frac{(b,q/b;q)_k(a,aq;q^2)_k 
(c^3,a^2q/c^3;q^3)_k}{(a^2q^3/b,a^2bq^2;q^3)_k (q^2,q;q^2)_k 
(a^2q/c^3,c^3;q)_k}\, q^k \tag 1.8 \\
&- \frac{(b,q/b,a^2q;q)_\infty 
(c^3,q^3,a^2q/c^3,c^6q^2/a^2;q^3)_\infty}{(q,c^3, a^2q/c^3;q)_\infty(bc^3/a^2, 
a^2bq^2,a^2q^3/b,c^3q/a^2b;q^3)_\infty} \, {}_2\phi_1 \left[\matrix
bc^3/a^2,c^3q/a^2b \\ c^6q^2/a^2 \endmatrix ; q^3,q^3 \right] \\
&= \frac{(a^2q^2,a^2q^3,c^3/a^2,c^3q/a^2,bc^3q, 
bq,q^2/b,c^3q^2/b;q^3)_\infty}{(q,q^2,c^3q,c^3q^2,a^2q^3/b, 
a^2bq^2,bc^3/a^2,c^3q/a^2b;q^3)_\infty}, \qquad |q|<1, \endalign$$
which contains Gosper's sum (1.6) as a limit case.   It  should be noted that 
in (1.7) and elsewhere we employ the standard convention of defining 
$$\sum^n_{k=m} a_k =\cases
a_m+a_{m+1}+\cdots+a_n, & m\le n, \\
0, & m=n+1, \\
-(a_{n+1}+a_{n+2}+\cdots+a_{m-1}), & m\ge n+2, \endcases
\tag 1.9 $$
for  $n,m = 0,\pm1,\pm2,\dots$.

The bibasic sum (1.7) will also be used to derive a common 
generalization of (1.2) and (1.8) and some rather general quadratic, cubic, 
and quartic summation and transformation formulas.   

\subheading{2. Derivation of (1.7)} Let
$$s_n=\frac{(ap,bp;p)_n (cq,ad^2q/bc;q)_n}{(dq,adq/b;q)_n (adp/c,bcp/d;p)_n}
\tag 2.1 $$
for $n=0,\pm1,\pm2,\dots$, and define the difference operator $\Delta$ by
$$\Delta s_n=s_n-s_{n-1}.$$
Then
$$\sum^n_{k=-m} \Delta s_k=s_n-s_{-m-1} 
\tag 2.2 $$
and
$$\align
\D s_k &= s_k-s_{k-1} \tag 2.3 \\
&= \frac{(ap,bp;p)_{k-1} (cq,ad^2q/bc;q)_{k-1}}{(dq,adq/b;q)_k (adp/c,bcp/d; 
p)_k} \\
&\cdot \big\{ (1-ap^k)(1-bp^k)(1-cq^k)(1-ad^2q^k/bc) \\
&\quad - (1-dq^k)(1-adq^k/b) (1-adp^k/c)(1-bcp^k/d)\big\} \\
&= \frac{d(1-c/d)(1-ad/bc) (1-adp^kq^k)(1-bp^k/dq^k)}{(1-a) (1-b)(1-c)(1-
ad^2/bc)} \, \frac{(a,b;p)_k(c,ad^2/bc;q)_k q^k}{(dq,adq/b;q)_k 
(adp/c,bcp/d;p)_k} \endalign$$
for $k=0,\pm1,\pm2,\dots$. Hence
$$\align
&{} \sum^n_{k=-m} \frac{(1-adp^kq^k)(1-bp^k/dq^k)}{(1-ad)(1-b/d)}\,
\frac{(a,b;p)_k(c,ad^2/bc;q)_k}{(dq,adq/b;q)_k (adp/c,bcp/d;p)_k}\,q^k
\tag 2.4 \\
&= \frac{(1-a)(1-b)(1-c) (1-ad^2/bc)}{d(1-ad)(1-b/d)(1-c/d)(1-ad/bc)}\, 
\{s_n-s_{-m-1}\} \\
&= \frac{(1-a)(11-b)(1-c)(1-ad^2/bc)}{d(1-ad)(1-b/d)(1-c/d)(1-ad/bc)} \\
&\cdot \left\{ 
\frac{(ap,bp;p)_n(cq,ad^2q/bc;q)_n}{(dq,adq/b;q)_n(adp/c,bcp/d;p)_n} - 
\frac{(c/ad,d/bc;p)_{m+1}(1/d,b/ad;q)_{m+1}}{(1/c,bc/ad^2;q)_{m+1}(1/a, 
1/b;p)_{m+1}} \right\} \endalign$$ 
for  $n,m=0,\pm1,\pm2,\dots$, which completes the proof of (1.7).    
In (2.4) we used the fact that 
$$(a;q)_{-n} = \frac{(-q/a)^n q^{\binom n2}}{(q/a;q)_n}. 
\tag 2.5 $$

Observe that (1.1) is the case $d=1$, $m=0$ of (1.7) and that, just as in 
(1.1), the series in (1.7) is partially of well-poised type in the sense that   
$a(dq) = b(adq/b) = adq$ and $c(adp/c) = (ad^2/bc)(bcp/d)=adp$. It is these 
observations and the simplification that occurred on the right side of (2.3) 
which motivated the choice of $s_n$ in (2.1).

If  $|p|<1$ and $|q|<1$, then by letting  $n$ or $m$ tend to infinity we find 
that (1.7) also holds with $n$ or $m$ replaced by $\infty$.  In particular, 
this gives the following evaluation of a bilateral bibasic series
$$\align
&{} \sum^\infty_{k=-\infty} \frac{(1-adp^kq^k)(1-bp^k/dq^k)}{(1-ad)(1-b/d)} \, 
\frac{(a,b;p)_k(c,ad^2/bc;q)_k}{(dq,adq/b;q)_k (adp/c,bcp/d;p)_k}\, q^k
\tag 2.6 \\
&= \frac{(1-a)(1-b)(1-c)(1-ad^2/bc)}{d(1-ad)(1-b/d)(1-c/d)(1-ad/bc)} \\
&\cdot \left\{ \frac{(ap,bp,p)_\infty(cq,ad^2q/bc;q)_\infty}{(dq,
adq/b;q)_\infty (adp/c,bcp/d;p)_\infty} - \frac{(c/ad,d/bc;p)_\infty 
(1/d,b/ad;q)_\infty}{(1/c,bc/ad^2;q)_\infty (1/a,1/b;p)_\infty} \right\},
\endalign$$
where  $|p|<1$ and $|q|<1$.

In the following sections we shall use the $m=0$ case of (1.7) in the form
$$\align
&{} \sumnk \frac{(1-adp^kq^k)(1-bp^k/dq^k)}{(1-ad)(1-b/d)} \, 
\frac{(a,b;p)_k(c,ad^2/bc;q)_k}{(dq,adq/b;q)_k(adp/c,bcp/d;p)_k}\, q^k 
\tag 2.7 \\
&= \frac{(1-a)(1-b)(1-c)(1-ad^2/bc)}{d(1-ad)(1-b/d)(1-c/d)(1-ad/bc)}\, 
\frac{(ap,bp;p)_n (cq,ad^2q/bc;q)_n}{(dq,adq/b;q)_n(adp/c,bcp/d;p)_n} \\
&- \frac{(1-d)(1-ad/b)(1-ad/c)(1-bc/d)}{d(1-ad)(1-b/d) (1-c/d)(1-ad/bc)}.
\endalign$$
There is no loss in generality because, by setting $k=j-m$ in (1.7), it is 
easily seen that (1.7) is equivalent to (2.7) with  $n$, $a$, $b$, $c$, $d$
replaced by $n+m$, $ap^{-m}$, $bp^{-m}$, $cq^{-m}$, $dq^{-m}$, respectively.    
We shall also use the special case  $c=q^{-n}$, $n=0,1,2,\dots$, of (2.7) in 
the form
$$\align
&{} \sumnk \frac{(1-adp^kq^k)(1-bp^k/dq^k)}{(1-ad)(1-b/d)} \, 
\frac{(a,b;p)_k(q^{-n},ad^2q^n/b;q)_k}{(dq,adq/b;q)_k(adpq^n,bp/dq^n;p)_k}\, 
q^k \tag 2.8 \\
&= \frac{(1-d)(1-ad/b)(1-adq^n)(1-dq^n/b)}{(1-ad)(1-d/b)(1-dq^n)(1-adq^k/b)}, 
\qquad n=0,1,2,\dots. \endalign$$





\subheading{3.  $q$-Extensions of Gosper's sum (1.6)}
In view of our previous observation that the case $b=2a$ of (1.6) is a 
special case of the sum (1.3), it is natural to try to use the bibasic sum 
(2.8) to extend the proof in [7] of the $q$-analogue (1.2) of (1.3)  to derive 
a summation formula which contains both (1.3) and (1.6) as limit cases.   
Therefore, let us set $q=p^3$ in (2.8), replace $d$ by $c$, and then change  
$p$ to $q$ to obtain the sum 
$$\align
&{} \sumnk \frac{(1-acq^{4k})(1-b/cq^{2k})}{(1-ac)(1-b/c)} \, \frac{(a,b;q)_k 
(q^{-3n},ac^2q^{3n}/b;q^3)_k}{(cq^3,acq^3/b;q^3)0_k(acq^{3n+1},b/cq^{3n-
1};q)_k}\, q^{3k} \tag 3.1 \\
&= \frac{(1-c)(1-ac/b)(1-acq^{3n}) (1-cq^{3n}/b)}{(1-ac) (1-c/b)(1-cq^{3n}) 
(1-acq^{3n}/b)}, \qquad n=0,1,2,\dots, \endalign$$
which reduces to [7, (5.18)] when  $c = 1$.

Multiply both sides of (3.1) by
$$\frac{(ac^2/b;q^3)_n(c/b;q)_{3n}}{(q^3;q^3)_n(acq;q)_{3n}}\, A_n$$
and sum over  $n$  to get
$$\align
&{} \sumn \frac{(ac^2/b;q^3)_n(cq/b;q)_{3n}}{(q^3;q^3)_n(ac;q)_{3n}}\, 
\frac{(1-c)(1-ac/b)}{(1-cq^{3n})(1-acq^{3n}/b)}\, A_n
\tag 3.2 \\
&= \sumn\sumnk \frac{(1-acq^{4k})(1-b/cq^{2k})}{(1-ac)(1-b/c)}\, 
\frac{(a,b;q)_k}{(cq^3,acq^3/b;q^3)_k} \\
&\cdot \frac{(ac^2/b;q^3)_{n+k}(c/b;q)_{3n-k}}{(q^3;q^3)_{n-
k}(acq;q)_{3n+k}}\, q^{k(k+1)} \left(\frac cb\right)^k A_n \\
&= \sumk \frac{(1-acq^{4k})(1-b/cq^{2k})}{(1-ac)(1-b/c)}\, 
\frac{(a,b;q)_k(ac^2/b;q^3)_{2k}(c/b;q)_{2k}}{(cq^3,acq^3/b;q^3)_k 
(acq;q)_{4k}}\, q^{k(k+1)} \left(\frac cb\right)^k \\
&\cdot \sumj \frac{(ac^2q^{6k}/b;q^3)_j (cq^{2k}/b;q)_{3j}}{(q^3; 
q^3)_j(acq^{4k+1};q)_{3j}}\, A_{j+k}. 
\endalign$$
Now choose
$$A_n=\frac{1-ac^2q^{6n}/b}{1-ac^2/b}\, \frac{(d,e;q^3)_n}{(ac^2q^3/bd, 
ac^2q^3/be;q^3)_n}\, z^n$$
to obtain the expression
$$\align
&{}_{10}W_9 (ac^2/b;c,d,e,ac/b,cq/b,cq^2/b,cq^3/b;q^3,z) 
\tag 3.3 \\
&= \sumk \frac{(1-acq^{4k})(1-b/cq^{2k})}{(1-ac)(1-b/c)}\, 
\frac{(a,b;q)_k (ac^2q^3/b;q^3)_{2k}(c/b;q)_{2k}}{(cq^3,acq^3/b;q^3)_k 
(acq;q)_{4k}} \\
&\cdot \frac{(d,e;q^3)_k}{(ac^2q^3/bd,ac^2q^3/bc;q^3)_k}\, q^{k(k+1)} 
\left(\frac cb\right)^k z^k \\
&\cdot {}_8W_7 (ac^2q^{6k}/b; dq^{3k}, eq^{3k},cq^{2k}/b,cq^{2k+1}/b, 
cq^{2k+2}/b; q^3,z), \endalign$$
where we let
$$\align
&{}_{r+1}W_r (a;b_1,b_2,\dots,b_{r-2};q,z) \tag 3.4 \\
&= {}_{r+1}\phi_r \left[\matrix
a,q\sqrt a,-q\sqrt a,b_1,b_2,\dots,b_{r-2} \\
\sqrt a,-\sqrt a,aq/b_1,aq/b_2,\dots,aq/b_{r-2} \endmatrix ;q,z\right].
\endalign$$

Observe that if  $e=a^2bcq^{3n}$, $z=q^3$, and $d=q^{-3n}$, $n=0,1,2,\dots$, 
then the ${}_8W_7$ in (3.3) can be summed by Jackson's sum [12,(3.3.1.1)] 
$$\align
&{}_8W_7 (a;b,c,d,e,q^{-n};q,q) \tag 3.5 \\
&= \frac{(aq,aq/bc,aq/cd,aq/bd;q)_n}{(aq/b,aq/c,aq/d,aq/bcd;q)_n}, \qquad 
n=0,1,2,\dots, \endalign$$
where $a^2q^{n+1}=bcde$, giving the formula
$$\align
&{} \sumnk \frac{1-acq^{4k}}{1-ac}\, \frac{(a,b;q)_k 
(cq/b;q)_{2k}(a^2bcq^{3n}, q^{-3n};q^3)_k}{(cq^3,acq^3/b;q^3)_k 
(ab;q)_{2k}(1/abq^{3n-1},acq^{3n+1};q)_k}\, q^k 
\tag 3.6 \\
&= \frac{(acq;q)_{3n}(ab^2/c;q^3)_n}{(ab;q)_{3n}(ac^2q^3/b;q^3)_n}\, 
{}_{10}W_9 (ac^2/b;c,ac/b,cq/b,cq^2/b,cq^3/b,a^2bcq^{3n},q^{-3n};q^3,q^3)
\endalign$$
for $n=0,1,2,\dots$.

If  $b=q/a$, then the above ${}_{10}W_9$ reduces to a ${}_8W_7$ that can also 
be summed by (3.5), giving the cubic summation formula
$$\align
&{} \sumnk \frac{1-acq^{4k}}{1-ac}\, \frac{(a,q/a;q)_k (ac;q)_{2k}(acq^{3n+1},
q^{-3n};q^3)_k}{(cq^3,a^2cq^2;q^3)_k (q;q)_{2k} (q^{-3n},acq^{3n+1};q)_k}\, q^k 
\tag 3.7 \\
&= \frac{(aq,q^2/a,acq^2,acq^3;q^3)_n}{(q,q^2,a^2cq^2,cq^3;q^3)_n}, \qquad
n=0,1,2,\dots, \endalign$$
which is a $q$-extension of the terminating case of Gosper's sum (1.6).

To derive a nonterminating extension of (3.7), we first set $e=a^2bc/d$ 
and $z=q^3$ in (3.3) to obtain
$$\align
&{}_{10}W_9 (ac^2/b; c,d,ac/b,a^2bc/d,cq/b,cq^2/b,cq^3/b ;q^3,q^3)
\tag 3.8 \\
&= \sumk \frac{(1-acq^{4k})(1-b/cq^{2k})}{(1-ac)(1-b/c)}\, \frac{(a,b;q)_k 
(ac^2q^3/b;q^3)_{2k}(c/b;q)_{2k}}{(cq^3,acq^3/b;q^3)_k (acq;q)_{4k}}\,
q^{k(k+4)}\cb^k \\
&\cdot \frac{(d,a^2bc/d;q^3)_k}{(ac^2q^3/bd,cdq^3/ab^2;q^3)_k}\, {}_8W_7 
(ac^2q^{6k}/b; dq^{3k},a^2bcq^{3k}/d, cq^{2k}/b, cq^{2k+1}/b, 
cq^{2k+2}/b;q^3,q^3). \endalign$$
Next we apply Bailey's nonterminating extension [12, (IV.15). p. 248] of (3.5)
$$\align
&{}_8W_7 (a;b,c,d,e,f;q,q) \tag 3.9 \\
&= \frac{(aq,b/a,aq/de,aq/ce,aq/cd,aq/cf,aq/df,aq/ef;q)_\infty}{(aq/c, 
aq/d,aq/e,aq/f,bc/a,bd/a,be/a,bf/a;q)_\infty} \\
&- \frac{(bq/c,bq/d,bq/e,bq/f,aq,c,d,e,f,b/a;q)_\infty}{(b^2q/a,bc/a,bd/a,be/a,
bf/a,aq/c,aq/d,aq/e,aq/f,a/b;q)_\infty} \\
&\cdot {}_8W_7 (b^2/a;b,bc/a,bd/a,be/a,bf/a;q,q), \endalign$$

\ni where  $a^2q=bcdef$, to the ${}_8W_7$ in (3.8) to obtain 
$$\align
&{} {}_8W_7 (ac^2q^{6k}/b; dq^{3k},a^2bcq^{3k}/d,cq^{2k}/b,cq^{2k+1}/b, 
cq^{2k+2}/b;q^3,q^3) \tag 3.10 \\
&= \frac{(ac^2q^{6k+3}/b, bd/ac^2q^{3k},dq^{k+1}/ab, dq^{k+2}/ab,dq^{k+3}/ab, 
abq^{2k},abq^{2k+1},abq^{2k+2};q^3)_\infty}{(cdq^{3k+3}/ab^2,acq^{4k+1}, 
acq^{4k+2}, acq^{4k+3}, ab^2/c,dq^{-k}/ac, dq^{1-k}/ac,dq^{2-
k}/ac;q^3)_\infty} \\
&= \frac{bdq^{-3k}}{ac^2}\, \frac{(ac^2q^{6k+3}/b, a^2bcq^{3k}/d, cq^{2k}/b, 
cq^{2k+1}/b, cq^{2k+2}/b, bdq^{3-3k}/ac^2, d^2q^3/a^2bc;q^3)_\infty}{(ac^2 
q^{3k+3}/bd, cdq^{3k+3}/ab^2,acq^{4k+1},acq^{4k+2}, acq^{4k+3},ab^2/c, 
bd^2q^3/ac^2;q^3)_\infty} \\
&\cdot \frac{(bdq^{k+1}/c,bdq^{k+2}/c, bdq^{k+3}/c;q^3)_\infty}{(dq^{-k}/ac, 
dq^{1-k}/ac,dq^{2-k}/ac;q^3)_\infty}\\
&\cdot {}_8W_7 (bd^2/ac^2;dq^{3k}, 
ab^2/c,dq^{-k}/ac,dq^{1-k}/ac,dq^{2-k}/ac;q^3,q^3) \\
&= \frac{(ac^2q^3/b,bd/ac^2;q^3)_\infty 
(dq/ab,ab;q)_\infty}{(ab^2/c,cdq^3/ab^2;q^3)_\infty(acq,d/ac;q)_\infty}\, 
\frac{(cdq^3/ab^2,ac^2q^3/bd;q^3)_k(acq;q)_{4k} q^{-k(k+1)}}{(ac^2q^3/b;  
q^3)_{2k}(ab;q)_{2k}(dq/ab,acq/d;q)_k}\, \left(\frac bc\right)^k \\
&+\frac{bd}{ac^2}\, \frac{(ac^2q^3/b, a^2bc/d,d^2q^3/a^2bc, 
bdq^3/ac^2;q^3)_\infty (c/d,bdq/c;q)_\infty}{(ac^2q^3/bd,cdq^3/ab^2,ab^2/c, 
bd^2q^3/ac^2;q^3)_\infty (acq,d/ac;q)_\infty} \\
&\cdot \frac{(cdq^3/ab^2,ac^2 q/bd, ac^2/bd;q^3)_k}{(ac^2q^3/b;q^3)_{2k}
(a^2bc/d;q^3)_k} \, \frac{(acq;q)_{4k} q^{-k(k+1)}(b/c)^k}{(acq/d,bdq/c;q)_k
 (c/b;q)_{2k}} \\
&\cdot {}_8W_7 (bd^2/ac^2;dq^{3k},ab^2/c,dq^{-k}/ac,dq^{1-k}/ac,dq^{2-k}/ac; 
q^3,q^3). \endalign$$

\ni Using (3.10) in (3.8), we find that
$$\align
&{}_{10}W_9 (ac^2/b;c,d,ac/b,a^2bc/d,cq/b,cq^2/b, cq^3/b;q^3,q^3) 
\tag 3.11 \\
&+ \frac{(ac^2q^3/b,a^2bc/d, d^2q^3/a^2bc, bd/ac^2;q^3)_\infty 
(c/b,bdq/c;q)_\infty}{(ac^2/bd,cdq^3/ab^2,ab^2/c, bd^2q^3/ac^2;q^3)_\infty 
(acq,d/ac;q)_\infty} \\
&\cdot \sumk \frac{(1-acq^{4k}) (1-b/cq^{2k})}{(1-ac)(1-b/c)}\, 
\frac{(a,b;q)_k(d,ac^2/bd; q^3)_k}{(cq^3,acq^3/b;q^3)_k (acq/d,bdq/c;q)_k}\, 
q^{3k} \\
&\cdot {}_8W_7 (bd^2/ac;dq^{3k},ab^2/c,dq^{-k}/ac,dq^{1-k}/ac,dq^{1-
k}/ac;q^3,q^3) \\
&= \frac{(ac^2q^3/b, bd/ac^2;q^3)_\infty 
(ab,dq/ab;q)_\infty}{(ab^2/c,cdq^3/ab^2;q^3)_\infty (acq,d/ac;q)_\infty} \\
&\cdot \sumk \frac{1-acq^{4k}}{1-ac}\, \frac{(a,b;q)_k(qc/b;q)_{2k} 
(d,a^2bc/d;q^3)_k}{(cq^3,acq^3/b;q^3)_k (ab;q)_{2k}(acq/d,dq/ab;q)_k}\, q^k.
\endalign$$
Now observe that the sum over $k$ on the left side of (3.11) is
$$\align
&{} \sumk \frac{(1-acq^{4k})(1-b/cq^{2k})}{(1-ac)(1-b/c)}\, 
\frac{(a,b;q)_k}{(cq^3,acq^3/b;q^3)_k} \tag 3.12 \\
&\cdot \sumj \frac{(bd^2/ac^2,ab^2/c;q^3)_j (1-bd^2q^{6j}/ac^2) 
(d;q^3)_{k+j}(ac^2/bd;q^3)_{k-j}}{(q^3, d^2q^3/a^2bc;q^3)_j (1-bd^2/ac^2) 
(acq/d;q)_{k-3j} (bdq/c;q)_{k+3j}}\, q^{3j^2+3k} 
\left(\frac{d^2}{a^2bc}\right)^j \\
&= \sumj \frac{(bd^2/ac^2,ab^2/c,d;q^3)_j (1-bd^2q^{6j}/ac^2) 
(d/ac;q)_{3j}}{(q^3, d^2q^3/a^2bc, bdq^3/ac^2;q^3)_j (1-
bd^2/ac^2)(bdq/c;q)_{3j}} \, q^{3j} \\
&\cdot \sumk \frac{(1-acq^{4k})(1-b/cq^{2k})}{(1-ac)(1-b/c)}\, 
\frac{(a,b;q)_k(dq^{3j},ac^2q^{-3j}/bd;q^3)_k}{(cq^3,acq^3/b;q^3)_k (acq^{1-
3j}/d,bdq^{1+3j}/c;q)_k}\, q^{3k} \endalign$$


\ni and that, by (2.7),
$$\align
&{} \sumk \frac{(1-acq^{4k})(1-b/cq^{2k})}{(1-ac)(1-b/c)}\, 
\frac{(a,b;q)_k(dq^{3j},ac^2q^{-3j}/bd;q^3)_k}{(cq^3,acq^3/b;q^3)_k
(acq^{1-3j}/d, bdq^{1+3j}/c;q)_k}\, q^{3k} \tag 3.13 \\
&= \frac{(1-c)(1-ac/b)(1-bdq^{3j}/c) (1-ac/dq^{3j})}{c(1-ac)(1-b/c) 
(1-dq^{3j}/c)(1-ac/bdq^{3j})}\, \frac{(a,b;q)_\infty 
(dq^{3j},ac^2/bdq^{3j};q^3)_\infty}{(c,ac/b;q^3)_\infty 
(ac/dq^{3j},bdq^{3j}/c;q)_\infty} \\
&- \frac{(1-c)(1-ac/b)(1-bdq^{3j}) (1-ac/dq^{3j})}{c(1-ac)(1-b/c)(1-
dq^{3j}/c)(1-ac/bdq^{3j})}. \endalign$$
Hence the right side of (3.12) is
$$\align
&- \frac{bd(1-c)(1-ac/b)}{ac^2(1-ac)(1-b/c)} \sumj \frac{(bd^2/ac^2,  
ab^2/c;q^3)_j (1-bd^2q^{6j}/ac^2)}{(q^3,d^2q^3/a^2bc;q^3)_j(1-bd^2/ac^2)} 
\tag 3.14 \\
&\cdot \frac{(d^2q^3/a^2bc)^jq^{3j^2}}{(1-dq^{3j}/c)(1-bdq^{3j}/ac)}\, 
\frac{(a,b;q)_\infty(d,ac^2/bd;q^3)_\infty}{(c,ac/b;q^3)_\infty(acq/d,bdq/c; 
q)_\infty} \\
&- \frac{b(1-c)(1-ac/b)(1-bd/c)(1-d/ac)}{c(1-ac)(1-b/c)(1-d/c)(1-bd/ac)} \\
&\cdot {}_{10}W_9 (bd^2/ac^2; ab^2/c,d/c,bd/ac,d,dq/ac, dq^2/ac,dq^3/ac; 
q^3,q^3) \endalign$$
and it follows from (3.11) that
$$\align
&{}_{10}W_9 (ac^2/b;d,c,a^2bc/d,ac/b,cq/b,cq^2/b, cq^3/b;q^3,q^3)
\tag 3.15 \\
&+ \frac{(1-c)(1-ac/b)}{(1-d/c)(1-bd/ac)} \, \frac{(cq/b,bd/c;q)_\infty 
(ac^2q^3/b,a^2bc/d,d^2q^3/a^2bc, bd/ac^2;q^3)_\infty}{(ac,dq/ac;q)_\infty 
(ac^2/bd,cdq^3/ab^2,ab^2/c, bd^2q^3/ac^2;q^3)_\infty} \\
&\cdot {}_{10}W_9 (bd^2/ac^2;d,bd/ac, ab^2/c,d/c,dq/ac,dq^2/ac,dq^3/ac;q^3,
 q^3) \\
&+ \frac{d(1-c)(1-ac/b)}{ac(1-d/c)(1-bd/ac)} \, \frac{(a,b,cq/b;q)_\infty 
(d,ac^2q^3/b,a^2bc/d,d^2q^3/a^2bc, 
bd/ac^2;q^3)_\infty}{(ac,d/ac,acq/d;q)_\infty 
(c,ac/b,cdq^3/ab^2,ab^2/c,bd^2q^3/ac^2;q^3)_\infty} \\
&\cdot \sumj \frac{(bd^2/ac^2;q^3)_j(1-bd^2q^{6j}/ac^2) 
(ab^2/c,d/c,bd/ac;q^3)_j}{(q^3;q^3)_j (1-bd^2/ac^2) 
(d^2q^3/a^2bc,bdq^3/ac,dq^3/c;q^3)_j}\, q^{3j^2} \left( 
\frac{d^2q^3}{a^2bc}\right)^j \\
&= \frac{(ab,dq/ab;q)_\infty 
(ac^2q^3/b,bd/ac^2;q^3)_\infty}{(acq,d/ac;q)_\infty (ab^2/c,cdq^3/ab^2; 
q^3)_\infty} \\
&\cdot \sumk \frac{1-acq^{4k}}{1-ac}\, \frac{(a,b;q)_k (cq/b;q)_{2k} 
(d,a^2bc/d;q^3)_k}{(cq^3,acq^3/b;q^3)_k (ab;q)_{2k} (acq/d,dq/ab;q)_k}\, q^k. 
\endalign$$
Using the transformation formula [7, (5.21)]
$$\align
&{} \sumj \frac{(a,q\sa,-q\sa,d,e,f;q)_j}{(q,\sa,-\sa,aq/d,aq/e,aq/f;q)_j}\,
\left(\frac{a^2q^2}{def}\right)^j q^{j(j-1)} \tag 3.16 \\
&= \frac{(aq,aq/ef;q)_\infty}{(aq/e,aq/f;q)_\infty}\, {}_2\phi_1 
\left[\matrix e,f \\ aq/d \endmatrix ;q,\frac{aq}{ef}\right] \endalign$$
and Heine's transformation formula [8, (1.4.5)]
$${}_2\phi_1 \left[\matrix a,b \\ c \endmatrix ;q,z\right] = 
\frac{(c/b,bz;q)_\infty}{(c,z;q)_\infty}\, {}_2\phi_1 \left[\matrix abz/c,b \\ 
bz \endmatrix ;q,\frac cb\right],
\tag 3.17 $$
we find that the sum over  $j$ in (3.15) equals
$$\align
&{} \frac{(bd^2q^3/ac^2,dq^3/a^2b;q^3)_\infty}{(d^2q^3/a^2bc, 
bdq^3/ac;q^3)_\infty}\, {}_2\phi_1 \left[\matrix ab^2/c,d/c \\ dq^3/c 
\endmatrix ;q,\frac{dq^3}{a^2b}\right] \tag 3.18 \\
&= \frac{(q^3,bd^2q^3/ac^2;q^3)_\infty}{(dq^3/c,bdq^3/ac;q^3)_\infty}\, 
{}_2\phi_1 \left[\matrix d/c,bd/ac \\ d^2q^3/a^2bc \endmatrix ;q^3,q^3 
\right]. \endalign$$
Thus, we have the cubic transformation formula
$$\align
&{}_{10}W_9 (ac^2/b;d,c,a^2bc/d,ac/b,cq/b,cq^2/b,cq^3/b;q^3,q^3) 
\tag 3.19 \\
&+ \frac{(1-c)(1-ac/b)}{(1-d/c)(1-bd/ac)}\, \frac{(cq/b,bd/c;q)_\infty 
(ac^2q^3/b, a^2bc/d, d^2q^3/a^2bc, bd/ac^2;q^3)_\infty}{(ac,dq/ac;q)_\infty 
(ac^2/bd, cdq^3/ab^2,ab^2/c, bd^2q^3/ac^2;q^3)_\infty} \\
&\cdot {}_{10}W_9 (bd^2/ac^2; 
d,bd/ac,ab^2/c,d/c,dq/ac,dq^2/ac,dq^3/ac;q^3,q^3) \\
&- \frac{(a,b,cq/b;q)_\infty (q^3,d,ac^2q^3/b, a^2bc/d, d^2q^3/a^2bc, 
bd/ac^2;q^3)_\infty}{(ac,dq/ac,ac/d;q)_\infty (cq^3,acq^3/b, d/c,bd/ac,
cdq^3/ab^2,ab^2/c;q^3)_\infty} \\
&\cdot {}_2\phi_1 \left[\matrix d/c,bd/ac \\ d^2q^3/a^2bc \endmatrix 
;q,q\right] \\
&= \frac{(ab,dq/ab;q)_\infty (bd/ac^2,ac^2q^3/b;q^3)_\infty}{(acq,d/ac
;q)_\infty (ab^2/c,cdq^3/ab^2; q^3)_\infty} \\
&\cdot \sumk \frac{1-acq^{4k}}{1-ac}\, \frac{(a,b;q)_k(cq/b;q)_{2k} 
(d,a^2bc/d;q^3)_k}{(cq^3,acq^3/b;q^3)_k (ab;q)_{2k} (acq/d,dq/ab;q)_k}\,
q^k. \endalign$$

Formula (1.2) follows from (3.19) by setting  
$c=1$ and  then replacing  $d$ by  $a^2b/c$.  To derive the cubic summation 
formula (1.8) which contains Gosper's sum (1.6) as a limit case,  set  
$b=q/a$ in (3.19) to get
$$\align
&{}_8W_7 (a^2c^2/q;d,c,acq/d, a^2c/q,acq;q^3,q^3) \tag 3.20 \\
&+ \frac{(1-c)(1-a^2c/q)}{(1-d/c)(1-dq/a^2c)}\, \frac{(a^2c^2q^2,acq/d, 
d^2q^2/ac, dq/a^2c^2;q^3)_\infty}{(a^2c^2/dq,acdq,q^2/ac, 
d^2q^4/a^2c^2;q^3)_\infty} \\
&\cdot {}_8W_7 (qd^2/a^2c^2;d,dq/a^2c, q^2/ac,d/c,dq^2/ac;q^3,q^3) \\
&- \frac{(a,q/a;q)_\infty (q^3,d,a^2c^2q^2,acq/d, 
d^2q^2/ac,dq/a^2c^2;q^3)_\infty}{(ac/d,dq/ac;q)_\infty 
(cq^3,a^2cq^2,d/c,dq/a^2c,acdq,q^2/ac;q^3)_\infty} \\
&\cdot {}_2\phi_1 \left[\matrix d/c,dq/a^2c \\ d^2q^2/ac \endmatrix ;q^3,q^3 
\right] \\
&= \frac{(q,d;q)_\infty (dq/a^2c^2,a^2c^2q^2;q^3)_\infty}{(acq,d/ac;q)_\infty 
(q^2/ac,acdq;q^3)_\infty} \\
&\cdot \sumk \frac{1-acq^{4k}}{1-ac}\, \frac{(a,q/a;q)_k (ac;q)_{2k} 
(d,acq/d;q^3)_k}{(cq^3,a^2cq^2;q^3)_k(q;q)_{2k} (acq/d,d;q)_k}\, q^k 
\endalign$$
and then apply (3.9) to the above sum of two ${}_8W_7$ series to get
$$\align
&{} \sumk \frac{1-acq^{4k}}{1-ac}\, \frac{(a,q/a;q)_k (ac;q)_{2k} 
(d,acq/d;q^3)_k}{(cq^3,a^2cq^2;q^3)_k(q;q)_{2k} (acq/d,d;q)_k}\, q^k 
\tag 3.21 \\
&= 
\frac{(acq^2,acq^3,d/ac,dq/ac,adq,aq,q^2/a,dq^2/a;q^3)_\infty}{(q,q^2,dq,dq^2,
a^2cq^2,cq^3, dq/a^2c,d/c;q^3)_\infty} \\
&+ \frac{d(a,q/a,acq;q)_\infty (q^3,d,acq/d,d^2q^2/ac;q^3)_\infty}{ac(q,d,
acq/d; q)_\infty (cq^3,a^2cq^2,d/c,dq/a^2c;q^3)_\infty} \\
&\cdot {}_2\phi_1 \left[\matrix d/c,dq/a^2c \\ d^2q^2/ac \endmatrix ;q^3,q^3 
\right]. \endalign$$

\ni
Formula (1.8) follows from (3.21) by replacing  $a$, $c$, $d$  by  $b$, 
$a^2/b$,  $c^3$, respectively, and using the identity 
$(a;q)_{2k}=(a,aq;q^2)_k$.

To verify that Gosper's formula (1.6) is a limit case of (3.21), replace  
$a$, $b$, $c$ in (3.21) by  $q^a$, $q^b$, $q^c$, respectively, take the limit
$q\uparrow 1$ and use the limits [2], [8, Chapter 1]
$$\lim_{q\uparrow 1} \frac{(q^x;q)_n}{(1-q)^n} = (x)_n, \quad
\lim_{q\uparrow1} \frac{(q;q)_\infty(1-q)^{1-x}}{(q^x;q)_\infty}=\G(x),
\tag 3.22 $$
Gauss' formula [12, (III.3), p\. 243] and the identity $\G(x)\G(1-x)=\pi/\sin 
\pi x$ to find that
$$\align
&{}_7F_8 \left[\matrix a,a+1/2,b,1-b,c, (2a+1)/3-c, a/2+1 \\
1/2,(2a-b+3)/3, (2a+b+2)/3,3c,2a+1-3c,a/2 \endmatrix ;1 \right] 
\tag 3.23 \\
&= \frac{\gfrac1 \gfrac2 \G\left( c+\frac13\right) \G\left( c+\frac23\right) 
\gfrac{2a-b+3} \gfrac{2a+b+2} \gfrac{3c+b-2a} \gfrac{3c+1-2a-b}}{\gfrac{2a+2} 
\gfrac{2a+3} \gfrac{3c-2a} \gfrac{3c+1-2a} \gfrac{b+1+3c} \gfrac{b+1} 
\gfrac{2-b} \gfrac{3c+2-b}} \\
&+ \frac{\G(3c) \G(1+2a-3c) \gfrac{3c-2a+b} \gfrac{2a+b+2} \gfrac{2a-b+3} 
\gfrac{3c+1-2a-b} \gfrac{1+2a}}{\G(b)\G(1-b) \G(1+2a)\G(c) \gfrac{2a+1-3c} 
\gfrac{3c+2-b}\gfrac{3c+1+b}} \\
&= \frac{\gfrac1\gfrac2 \G\left( c+\frac13\right) \G\left( c+\frac23\right) 
\gfrac{2a-b+3} \gfrac{2a+b+2} \gfrac{3c-2a+b} \gfrac{3c-2a-b+1}}{\gfrac{2a+2} 
\gfrac{2a+3} \gfrac{b+1} \gfrac{2-b} \gfrac{3c+b+1} \gfrac{3c+2-b} \gfrac{3c-
2a} \gfrac{3c-2a+1}}\, B \endalign$$
with
$$\align
B &= 1+\frac{\gfrac{3c-2a}\gfrac{3c-2a+1}\gfrac{3+2a-2c} \gfrac{2+2a-
3c}}{\gfrac b\G\left(1-\frac b3\right) \gfrac{1-b}\gfrac{2+b}} 
\tag 3.24 \\
&= 1+\frac{\sin\frac{\pi b}3\sin \frac\pi3\,(1-b)}{\sin\frac\pi3\,(3c-2a) 
\sin\pt\,(3c-2a+1)}. \endalign$$
Now observe that by setting $\alpha=b/3$, and $\beta=(3c-2a)/3$,  
and using trigonometric identities we have
$$\align
&{} \sin\pi\beta\sin\left(\pt+\pi\beta\right) + \sin\pi\alpha 
 \sin\left(\pt-\pi\alpha\right) \tag 3.25 \\
&= \sin\pi\ab \spt(1+\beta-\alpha) \\
&= \spt (3c-2a+b) \spt(1+3c-2a-b). \endalign$$
Therefore
$$\align
B &= \frac{\spt(3c-2a+b)\spt (1+3c-2a-b)}{\spt(3c-2a)\spt(1+3c-2a)} 
\tag 3.26 \\
&= \frac{\gfrac{3c-2a} \gfrac{3+2a-3c} \gfrac{1+3c-2a} \gfrac{2+2a-
3c}}{\gfrac{3c-2a+b} \gfrac{3+2a-b-3c} \gfrac{1+3c-2a-b} \gfrac{2+2a+b-3c}}
\endalign$$
and thus
$$\align
&{}_7F_6 \left[\matrix a,a+\frac12,b,1-b,c,(2a+1)/3-c, a/2+1 \\
1/2,(2a-b+3)/3,(2a+b+2)/3, 3c,3a+1-3c,a/2 \endmatrix ;1 \right]
\tag 3.27 \\
&= \frac{\gfrac1\gfrac2 \G\left(c+\frac13\right)\G\left(c+\frac23\right) 
\gfrac{2a-b+3} \gfrac{2a+b+2} \gfrac{2+2a-3c} \gfrac{3+2a-3c}}{\gfrac{b+1} 
\gfrac{2-b} \gfrac{2a+2}\gfrac{2a+3} \gfrac{3c+b+1} \gfrac{3c+2-b} 
\gfrac{2+2a+b-3c} \gfrac{3+2a-b-3c}} \\
&= \frac2{\sqrt3}\, \frac{\G\left(c+\frac13\right)\G\left(c+\frac23\right) 
\gfrac{2a-b+3} \gfrac{2a+b+2} \gfrac{2+2a-3c} \gfrac{3+2a-3c} 
\spt(b+1)}{\gfrac{2a+2} \gfrac{2a+3} \gfrac{3c+b+1} \gfrac{3c+2-b} 
\gfrac{2+2a+b-3c} \gfrac{3+2a-b-3c}}, \endalign$$
which gives (1.6). 

\subheading{4.  Some consequences of formula (3.19)}    
When  $d=a^2bcq^{3n}$and $n=0,1,2,\dots$, reduces to the transformation formula
$$\align
&{} \sumnk \frac{1-acq^{4k}}{1-ac}\, \frac{(a,b;q)_k(cq/b;q)_{2k} 
(a^2bcq^{3n},q^{-3n};q^3)_k}{(cq^3,acq^3/b;q^3)_k (ab;q)_{2k} (q^{1-3n}/ab, 
acq^{3n+1};q)_k}\, q^k \tag 4.1 \\
&= \frac{(acq;q)_{3n} (ab^2/c;q^3)_n}{(ab;q)_{3n}(ac^2q^3/b;q^3)_n} \\
&\cdot {}_{10}W_9 (ac^2/b;c,ac/b,cq/b,cq^2/b,cq^3/b,a^2 bcq^{3n}, q^{-3n}; 
q^3,q^3). \endalign$$
If we use Bailey's [3, 8.5(1)] transformation
$$\align
&{}_{10}W_9 (a;c,d,e,f,g,h,q^{-n};q,q) \tag 4.2 \\
&= \frac{(aq,aq/fq,aq/fh,aq/gh;q)_n}{(aq/f,aq/g,aq/h,aq/fgh;q)_n} \\
&\cdot {}_{10}W_9 (\sigma;\sigma c/a,\sigma d/a,\sigma e/a, f,g,h,q^{-n}; 
q,q), \endalign$$
where  $\sigma=a^2q/cde$, $a^3q^2=cdefghq^{-n}$ and  $n=0,1,2,\dots$, 
to show that the  ${}_{10}W_9$ in (4.1) equals
$$\align
&{} \frac{(ac^2q^3/b,ab/q^2,q^{1-3n}/ab,q^{-3n}/ab;q^3)_n}{(ac,acq,cq^{3-
3n}/ab^2,q^{-3n-2}/ac;q^3)_n} \\
&\cdot {}_{10}W_9 (acq^2;q^3,aq^2,bq^2,cq^2/b,cq^3/b, a^2bcq^{3n}, q^{-3n}; 
q^3,q^3), \endalign$$
then we find that (4.1) can be written in the equivalent form
$$\align
&{} \sumnk \frac{1-acq^{4k}}{1-ac}\, 
\frac{(a,b;q)_k(cq/b;q)_{2k}(a^2bcq^{3n},q^{-3n};q^3)_k}{(cq^3,acq^3/b;q^3)_k
(ab;q)_{2k}(q^{1-3n}/ab, acq^{3n+1};q)_k}\,  q^k \tag 4.3 \\
&= \frac{(1-acq^2)(1-ab/q^2) (1-abq^{3n}) (1-acq^{3n})}{(1-acq^{3n+2}) (1-
abq^{3n-2})(1-ab)(1-ac)} \\
&\cdot \sumnk \frac{1-acq^{6k+2}}{1-acq^2}\, \frac{(aq^2,bq^2,cq^2/b,cq^3/b,
a^2bcq^{3n},q^{-3n};q^3)_k}{(cq^3,acq^3/b,abq^3,abq^2,q^{5-3n}/ab, 
acq^{3n+5};q^3)_k}\, q^{3k}, \endalign$$
where  $n=0,1,2,\dots$.   Notice the rather striking similarities and 
differences between the two series in (4.3).    The other equivalent forms of 
(4.1) which follow by applying (4.2) are not as interesting and so will be 
omitted.

When  $b=q^2/a$, the first two terms on the left side of (3.19) give  
$$\align
&{}_8W_7 (a^2c^2/q^2;d,c,acq^2/d, a^2c/q^2,ac/q;q^3,q^3)
\tag 4.4 \\
&+ \frac{(1-c)(1-a^2c/q^2)}{(1-d/c))(1-dq^2/a^2c)}\, 
\frac{(ac/q,dq^2/ac;q)_\infty (a^2c^2q, acq^2/d, d^2q/ac,dq^2/a^2c^2; 
q^3)_\infty}{(ac,dq/ac;q)_\infty 
(a^2c^2/dq^2,acd/q,q^4/ac,d^2q^5/a^2c^2;q^3)_\infty} \\
&\cdot {}_8W_7 (d^2q^2/a^2c^2; d,dq^2/a^2c,q^4/ac,d/c,dq/ac;q^3,q^3) \\
&= \frac{(a^2c^2q, dq^2/a^2c^2,ad/q,q^3,aq^2,dq/a,d,q^4/a;q^3)_\infty}{(a^2 
cq,acd/q,cq^3, acq^2,dq^2/a^2c, q^4/ac,d/c,dq/ac;q^3)_\infty} \endalign$$
by (3.9).   Hence, we have the summation formula
$$\align
&{} \sumk \frac{1-acq^{4k}}{1-ac}\, \frac{(a,q^2/a;q)_k (ac/q;q)_{2k} 
(d,acq^2/d;q^3)_k}{(cq^3,a^2cq;q^3)_k(q^2;q)_{2k}(acq/d, d/q;q^3)_k}\, q^k 
\tag 4.5 \\
&+ \frac{(a,aq,aq^2,q^2/a, q^3/a,q^4/a,ac/q,acq, 
acq^3,d/ac,d^2q/ac;q^3)_\infty}{(q^2,q^4,d/q,dq,ac/d, acq/d, dq^3/ac, 
dq^2/a^2c, cq^3, a^2cq, d/c;q^3)_\infty} \\
&\cdot {}_2\phi_1 \left[\matrix d/c,dq^2/a^2c \\ d^2q/ac \endmatrix ;q^3,q^3 
\right] \\
&= \frac{(aq^2,q^4/a,ad/q,dq/a,d/ac,dq^2/ac,acq,acq^3;q^3)_\infty}{(q^2,q^4, 
d/q,dq,d/c,cq^3,a^2cq,dq^2/a^2c;q^3)_\infty}. \endalign$$
Replace  $a$, $c$, $d$  by  $b$, $a^2/b$, $c^3$, respectively to get
$$\align
&{} \sumk \frac{1-a^2q^{4k}}{1-a^2}\, \frac{(b,q^2/b;q)_k (a^2/q;q)_{2k}(c^3, 
a^2q^2/c^3;q^3)_k}{(a^2q^3/b, a^2bq;q^3)_k (q^2;q)_{2k} (a^2q/c^3,c^3/q;q)_k}
\, q^k \tag 4.6 \\
&= \frac{(bq^2,q^4/b,bc^3/q,c^3q/b, c^3/a^2,c^3q^2/a^2, 
a^2q,a^2q^3;q^3)_\infty}{(q^2,q^4,c^3/q, c^3q,bc^3/a^2,a^2q^3/b, 
a^2bq,c^3q^2/a^2b; q^3)_\infty} \\
&- \frac{(b,bq,bq^2,q^2/b, q^3/b, q^4/b, qa^2/q,a^2q, a^2q^3,c^3/a^2, 
c^6q/a^2;q^3)_\infty}{(q^2,q^4,c^3/q, c^3q,a^2/c^3,a^2q/c^3, c^3q^3/a^2, 
c^3q^2/a^2b, a^2q^3/b, a^2bq,bc^3/a^2; q^3)_\infty} \\
&\cdot {}_2\phi_1 \left[\matrix bc^3/a^2,c^3q^2/a^2b \\ c^6q/a^2 \endmatrix
;q^3,q^3 \right]. \endalign$$
As in the derivation of (3.27), it can be shown that (4.6) has the  
$q\uparrow 1$ limit case 
$$\align
&{}_7F_6 \left[\matrix a-1/2,a,b,2-b,c,(2a+2-3c)/3,a/2+1 \\ 3/2,(2a-b+3)/3, 
(2a+b+1)/3, 3c-1,2a+1-3c,a/2 \endmatrix ;1 \right]
\tag 4.7 \\
&= \frac{\gfrac2\gfrac4 \G\left(c-\frac13\right) \G\left(c+\frac13\right)
\gfrac{2a-b+3} \gfrac{2a+b+1} \gfrac{2a-3c+3} \gfrac{2a-3c+1}}{\gfrac{b+2} 
\gfrac{4-b} \gfrac{b+3c-1} \gfrac{3c-b+1} \gfrac{2a+1} \gfrac{2a+3} 
\gfrac{2a-3c-b+3} \gfrac{2a-3c+b+1}}. \endalign$$

It should be noted that the evaluation
$$\align
&{}_7F_6 \left[\matrix 1, (d-b+2)/2, (a+d+4)/4, a,b,c/3,(a+2d-b-c)/3 \\
(d-b)/2, (a+d)/4,(d+3)/3, (a+d+3-b)/3, a+d+1-c,b+c+1-d \endmatrix ;1\right]
\tag 4.8 \\
&= \frac{abc(a+2a-b-c)}{(a+d)(b-d)(c-d)(a+d-b-c)}\, \frac{\G(a+d+1-c)\G(b+c+1 
-d)\gfrac{d+3} \gfrac{a+d+3-b}}{\gfrac{c+3}\gfrac{a+2d+3-b-c} \G(a+1)\G(b+1)} 
\\
&+ \frac{d(a+d-b)(a+d-c)(b+c-d)}{(a+d)(b-d)(d-c)(a+d-b-c)} \endalign$$
and the corresponding truncated series formula are limit cases of the $q=p^3$
case of (2.7).

By reversing the order of summation of the series on the left side of 
(4.1) this series is converted to a series which is truncated at the 
$(n+1)^{\text{th}}$ term.   Another transformation 
formula for such a series in terms of a ${}_{10}W_9$ series can be derived 
by proceeding as follows.   Use (2.7) and (3.5) to obtain
$$\align
&{}\sumnk \frac{(1-aq^{4k})(1-bq^{-2k})}{(1-b)}\, \frac{(a,b;q)_k 
(c,a/bc;q^3)_k}{(q^3, aq^3/b;q^3)_k (aq/c,bcq;q)_k} \, q^{3k} \tag 4.9 \\
&= \frac{(aq,bq;q)_n (cq^3,aq^3/bc;q^3)_n}{(aq/c, bcq;q)_n(q^3,aq^3/b; 
q^3)_n}, \qquad n=0,1,2,\dots, \endalign$$
and
$$\align
&{}_8W_7 (a/b;aq^k,aq^{k+1},aq^{k+2},1/ab^2, q^{-3k};q^3,q^3) \tag 4.10 \\
&= \frac{(aq^3/b,q^{-2k}/ab, q^{1-2k}/ab, q^{2-2k}/ab;q^3)_k}{(q^{-3k}/a^2b,
b^{-1}q^{1-k},b^{-1}q^{2-k}, b^{-1}q^{3-k};q^3)_k} \\
&= \frac{(aq^3/b;q^3)_k (1/ab;q)_k (abq;q)_{2k}}{(a^2bq^3;q^3)_k 
(b;q)_k(q/b;q)_{2k}}, \qquad k=0,1,2,\dots. \endalign$$
Hence
$$\align
&{} \sumnk \frac{(1-aq^{4k})(1-bq^{-2k})}{(1-a)(1-b)}\, \frac{(a,1/ab;q)_k 
(abq;q)_{2k}(c,a/bc;q^3)_k}{(q^3,a^2bq^3;q^3)_k (q/b;q)_{2k} (aq/c,bcq;q)_k}\, 
q^{3k} \tag 4.11 \\
&= \sumnk \frac{(1-aq^{4k})(1-bq^{-2k})}{(1-a)(1-b)}\, 
\frac{(c,a/bc;q^3)_k}{(aq/c,bcq;q)_k}\, q^{3k} \\
&\cdot \sum^k_{j=0} \frac{(a/b;q^3)_j (1-aq^{3j}/b) (a;q)_{k+3j} (bq^{-
3j};q)_k (1/ab^2;q^3)_j(-1)^j q^{3j(j+1)/2}}{(q^3;q^3)_{k-j} (q^3;q^3)_j 
(aq^3/b;q^3)_{k+j} (q/b;q)_{3j} (a^2bq^3;q^3)_j (1-a/b)} \\
&= \sum^n_{j=0} \frac{(a/b;q^3)_j (1-aq^{3j}/b) (1/ab^2;q^3)_j (a;q)_{4j} 
(bq^{-3j};q)_j (c,a/bc;q^3)_j}{(q^3;q^3)_j (1-a/b) (a^2bq^3;q^3)_j  
(aq^3/b;q^3)_{2j} (q/b;q)_{3j} (aq/c,bcq;q)_j} \\
&\cdot \frac{(1-aq^{4j})(1-bq^{-2j})}{(1-a)(1-b)}\, (-1)^j q^{3j(j+1)/2+3j} \\
&\cdot \sum^{n-j}_{m=0} \frac{(1-aq^{4j+4m})(1-bq^{-2j-2m})}{(1-aq^{4j}) (1-
bq^{-2j})}\, \frac{(aq^{4j},bq^{-2j};q)_m (cq^{3j},aq^{3j}/bc;q^3)_m}{(q^3, 
aq^{6j+3}/b;q^3)_m (aq^{j+1}/c, bcq^{j+1};q)_m}\, q^{3m} \\
&= \frac{(aq,bq;q)_n (cq^3,aq^3/bc;q^3)_n}{(aq/c,bcq;q)_n (q^3,aq^3/b;q^3)_n} 
\\
&\cdot \sum^n_{j=0} \frac{(a/b,1/ab^2,aq^{n+1},aq^{n+2}, aq^{n+3}, q^{-
3n};q^3)_j}{(q^3,a^2bq^3,q^{-n}/b,q^{1-n}/b, q^{2-n}/b, aq^{3n+3}/b; q^3)_j} 
\\                                                                        
&\cdot \frac{(1-aq^{3j}/b)(1-c)(1-a/bc)}{(1-a/b)(1-cq^{3j})(1-aq^{3j}/bc)}\, 
q^{3j} \endalign$$
by (2.7),  which yields the transformation formula
$$\align
&{} \sumnk \frac{(1-aq^{4k})(1-bq^{-2k})}{(1-a)(1-b)}\, \frac{(a,1/ab;q)_k 
(abq;q)_{2k} (c,a/bc;q^3)_k}{(q^3,a^2bq^3; q^3)_k (q/b;q)_{2k} (aq/c,bcq;q)_k}
\, q^{3k} \tag 4.12 \\
&= \frac{(aq,bq;q)_n (cq^3,aq^3/bc;q^3)_n}{(aq/c,bcq;q)_n(q^3,aq^3/b;q^3)_n} 
\\
&\cdot {}_{10}W_9 (a/b;1/ab^2,c,a/bc,aq^{n+1},aq^{n+2}, aq^{n+3},q^{-3n}; 
q^3,q^3), \endalign$$
where $n=0,1,2,\dots$.

\subheading{5.  A quadratic transformation formula with five parameters}
In [7,(5.15)] and [10,(4.7)] special cases of (2.8) were used to derive the 
quadratic summation formula
$$\align
&{} \sumk \frac{1-aq^{3k}}{1-a}\, \frac{(a,b,q/b;q)_k 
(c,d,a^2q/cd;q^2)_k}{(q^2,aq^2/b,abq;q^2)_k (aq/c,aq/d,cd/a;q)_k}\, q^k
\tag 5.1 \\
&- \frac{aq(aq,b,q/b;q)_\infty (c,d,a^2q^3/cd^2,a^2q^3/c^2d; q^2)_\infty}{cd 
(aq/c,aq/d,cd/a;q)_\infty (abq,abq/cd,aq^2/b,aq^2/bcd;q^2)_\infty} \\
&\cdot {}_3\phi_2 \left[\matrix abq/cd, a^2q/cd,aq^2/bcd \\ a^2q^3/cd^2, 
a^2q^3/c^2d \endmatrix ;q^2,q^2 \right] \\
&= \frac{(aq,aq/cd;q)_\infty (abq/c,abq/d, aq^2/bc,aq^2/bd;q^2)_\infty}{(aq/c, 
aq/d;q)_\infty (abq,abq/cd,aq^2/b,aq^2/bcd;q^2)_\infty} \endalign$$
and this formula was used to derive a quadratic summation formula [7, (5.1)] 
stated earlier by Gosper.    Here we show how (2.8) can be used to extend 
(5.1) to a quadratic transformation formula containing five arbitrary 
parameters.

Take  $q=p^2$ in (2.8) and then change  $p$ and  $d$ to  $q$ and $c$ to get
$$\align
&{} \sumnk \frac{(1-acq^{3k})(1-b/cq^k)}{(1-ac)(1-b/c)}\, \frac{(a,b;q)_k (q^{-
2n},ac^2q^{2n}/b;q^2)_k}{(cq^2,acq^2/b;q^2)_k (acq^{2n+1}, bq^{1-2n}/c; q)_k} 
\, q^{2k} \tag 5.2 \\
&= \frac{(1-c)(1-ac/b)(1-acq^{2n})(1-cq^{2n}/b)}{(1-ac)(1-c/b) (1-cq^{2n}) (1-
acq^{2n}/b)}, \qquad n=0,1,2,\dots. \endalign$$
Multiply both sides of (5.2) by
$$\frac{(ac^2/b;q^2)_n(c/b;q)_{2n}}{(q^2;q^2)_n (acq;q)_{2n}}\, C_n$$
and sum over  $n$  to obtain
$$\align
&{} \sumn \frac{(ac^2/b;q^2)_n(cq/b;q)_{2n} (1-c)(1-ac/b)}{(q^2;q^2)_n 
(ac;q)_{2n} (1-cq^{3n}) (1-acq^{2n}/b)}\, C_n \tag 5.3 \\
&= \sumn\sumnk \frac{(1-acq^{3k}) (1-b/cq^k)}{(1-ac)(1-b/c)}\, 
\frac{(a,b;q)_k(ac^2/b;q^2)_{n+k} (c/b;q)_{2n-k}}{(cq^2, acq^2/b;q^2)_k 
(q^2;q^2)_{n-k} (acq;q)_{2n+k}}\, \left(\frac{cq}b\right)^k q^{\binom k2} 
C_n\\
&= \sumk \frac{(1-acq^{3k})(1-b/cq^k)}{(1-ac)(1-
b/c)}\,\frac{(a,b;q)_k(ac^2/b;q^2)_{2k} (c/b;q)_k}{(cq^2,acq^2/b;q^2)_k 
(acq;q)_{3k}}\, \left(\frac{cq}b\right)^k q^{\binom k2} \\
&\cdot \sum^\infty_{m=0} \frac{(ac^2q^{4k};q^2)_m (cq^k/b;q)_{2m}}{(q^2;q^2)_m 
(acq^{3k+1};q)_{2m}}\, C_{k+m}. \endalign$$
Choose
$$C_n=\frac{(1-ac^2q^{4n}/b) (d,e,f;q^2)_n(a^2q^3/def)^n}{(1-ac^2/b) 
(ac^2q^2/bd, ac^2q^2/be, ac^2q^2/bf;q^2)_n}. 
\tag 5.4 $$
Then (5.3) gives
$$\align
&{}_{10}W_9 (ac^2/b; ac/b,c,cq/b,cq^2/b,d,e,f; q^2,a^2c^2q^3/def) 
\tag 5.5 \\
&= \sumk \frac{(1-acq^{3k})(1-b/cq^k)}{(1-ac)(1-b/c)}\, \frac{(a,b,c/b;q)_k 
(ac^2q^2/b;q^2)_{2k}}{(cq^2,acq^2/b;q^2)_k (acq;q)_{3k}} \\
&\cdot \frac{(d,e,f;q^2)_k (a^2c^3q^4/bdef)^k}{(ac^2q^2/bd, ac^2q^2/be, 
ac^2q^2/bf;q^2)_k}\, q^{\binom k2} \\
&\cdot {}_8W_7 (ac^2q^{4k}/b; cq^k/b, cq^{k+1}/b, dq^{2k}, eq^{2k},fq^{2k}; 
q^2,a^2c^2q^3/def). \endalign$$

\def\bslash{\kern-2pt/}

Now assume that
$$a^2c^2q=def. \tag 5.6 $$
Then we can apply (3.9) to the ${}_8W_7$ in (5.5) to get
$$\align
&{}_8W_7 (ac^2q^{4k}/b; cq^k/b,cq^{k+1}/b, dq^{2k},eq^{2k},fq^{2k};q^2,q^2) 
\tag 5.7 \\           
&= \frac{(ac^2q^{4k+2}/b, bf/ac^2q^{2k}, abq^{2k+1}, acq^{k+2}/d, 
acq^{k+2}\bslash e, 
acq^{k+1}\bslash d, acq^{k+1}\bslash e, ac^2q^2\bslash bde;
q^2)_\infty}{(acq^{3k+2}, acq^{3k+1}, 
ac^2q^{2k+2}/bd, ac^2q^{2k+2}/be, bef/ac^2, bdf/ac^2, f/acq^k, f/acq^{k-1}; 
q^2)_\infty} \\
&+ \frac{bfq^{-2k}  (ac^2q^{4k+2}/b, cq^k/b, cq^{k+1}/b, dq^{2k}, eq^{2k}, 
fq^2/e, fq^2/d, bfq^{2-2k}/ac^2;q^2)_\infty}{ac^2 (ac^2q^{2k+2}/bf, 
acq^{3k+2}, acq^{3k+1}, ac^2q^{2k+2}/bd, ac^2q^{2k+2}/be, bef/ac^2, bdf/ac^2; 
q^2)_\infty} \\
&\cdot \frac{(bfq^{k+1}/c,bfq^{k+2}/c;q^3)_\infty}{(f/acq^k,f/acq^{k-
1};q^3)_\infty}\, {}_8W_7 (bf^2/ac^2;fq^{2k},bef/ac^2,bdf/ac^2,f/acq^k, 
f/acq^{k-1};q^2,q^2) \endalign$$
which, combined with (5.5), gives
$$\align
&{}_{10}W_9 (ac^2/b;c,d,e,f,ac/b, cq/b,cq^2/b;q^2,q^2) \tag 5.8 \\
&= \frac{(ac^2q^2/b,ac^2q^2/bde, abq,bf/ac^2;q^2)_\infty 
(acq/d,acq/e;q)_\infty}{(ac^2q^2/bd,ac^2q^2/be,bdf/ac^2,bef/ac^2;q^2)_\infty 
(acq,f/ac;q)_\infty} \\
&\cdot \sumk \frac{(1-acq^{3k})(1-b/cq^k)}{(1-ac)(1-b/c)}\, 
\frac{(a,b,c/b;q)_k(d,e,f;q^2)_k}{(cq^2,acq^2/b,abq;q^2)_k (acq/d, 
acq/e,acq/f;q)_k}\, q^{2k} \\
&+ \frac{bf(ac^2q^2/b, d,e,fq^2/d,fq^2/e, bfq^2/ac^2;q^2)_\infty 
(bfq/c,c/b;q)_\infty}{ac^2(ac^2q^2/bf,ac^2q^2/be,ac^2q^2/bd, 
bdf/ac^2,bef/ac^2,bf^2q^2/ac^2; q^2)_\infty (acq,f/ac;q)_\infty} \\
&\cdot \sumk \frac{(1-acq^{3k}) (1-b/cq^k)}{(1-ac)(1-b/c)}\, 
\frac{(a,b;q)_k(ac^2/bf,f;q^2)_k}{(cq^2,acq^2/b;q^2)_k(bfq/c,acq/f;q)_k}\, 
q^{2k} \\
&\cdot {}_8W_7 (bf^2/ac^2;fq^{2k},bdf/ac^2, bef/ac^2, f/acq^k, f/acq^{k-
1};q^2,q^2). \endalign$$
The last sum over  $k$ is

$$\align
&{} \sumj \frac{(bf^2/ac^2,f,bdf/ac^2,bef/ac^2,f/ac,fq/ac;q^2)_j (1-
bf^2q^{4j}/ac^2)}{(q^2,bfq^2/ac^2,fq^2/d, fq^2/e, bfq^2/c, bfq/c;q^2)_j (1-
bf^2/ac^2)}\, q^{2j} \tag 5.9 \\
&\cdot \sumk \frac{(1-acq^{3k})(1-b/cq^k)}{(1-ac)(1-b/c)}\, \frac{(a,b;q)_k 
(ac^2/bfq^{2j},fq^{2j};q^2)_k}{(cq^2,acq^2/b;q^2)_k (bfq^{2j+1}/c,
acq^{1-2j}/f; q)_k}\, q^{2k} \\
&= \sumj \frac{(bf^2/ac^2,f,bdf/ac^2,bef/ac^2, f/ac,fq/ac; q^2)_j (1-
bf^2q^{4j}/ac^2)}{(q^2,bfq^2/ac^2,fq^2/d, fq^2/e, bfq^2/c, bfq/c;q^2)_j (1-
bf^2/ac^2)}\, q^{2j} \\
&\cdot \frac{(1-c)(1-ac/b)(1-ac/fq^{2j}) (1-bfq^{2j}/c)}{c(1-ac)(1-b/c)(1-
fq^{2j}/c) (1-ac/bfq^{2j})} \\
&\cdot \left\{ \frac{(a,b;q)_\infty (fq^{2j},ac^2/bfq^{2j}; 
q^2)_\infty}{(c,ac/b;q^2)_\infty (ac/fq^{2j},bfq^{2j}/c;q)_\infty} -1 \right\} 
\\
&= \frac{(1-c)(1-ac/b) (1-f/ac)(1-bf/c)}{(1-c/b)(1-ac)(1-f/c)(1-bf/ac)} \\
&\cdot {}_{10}W_9 (bf^2/ac^2; f,bdf/ac^2,bef/ac^2,bf/ac,f/c, fq/ac,fq^2/ac; 
q^2,q^2) \\
&+ \frac{f(a,b;q)_\infty (f,ac^2/bf;q^2)_\infty}{ac(1-ac)(1-c/b) (1-f/c)(1-
bf/ac)(cq^2,acq^2/b;q^2)_\infty (acq/f,bfq/c;q)_\infty} \\
&\cdot \sumj \frac{(bf^2/ac^2,bdf/ac^2,bef/ac^2,f/c,bf/ac;q^2)_j (1-
bf^2q^{4j}/ac^2)}{(q^2, fq^2/d,fq^2/e, bfq^2/ac,fq^2/c;q^2)_j (1-bf^2/ac^2)}\, 
\left(-\frac{fq^2}{ab}\right)^j q^{j^2} \endalign$$
by the $n=\infty$ case of (2.7).   Hence, 
$$\align
&{}_{10}W_9 (ac^2/b;c,d,e,f,ac/b,cq/b,cq^2/b;q^2,q^2) \tag 5.10 \\
&-  \frac{bf(1-c)(1-ac/b)}{ac^2(1-f/c)(1-bf/ac)}\, 
\frac{(ac^2q^2/b,d,e,fq^2/d,fq^2/e,bfq^2/ac^2;q^2)_\i}{(bf^2\q/ac^2,ac^2\q/be,
ac^2\q/bd,bdf/ac^2, bef/ac^2, ac^2\q/bf;\q)_\i} \\
&\cdot \frac{(bf/c,cq/b;q)_\i}{(ac,fq/ac;q)_\i}\, {}_{10}W_9 \left( 
\frac{bf^2}{ac^2}; \frac{bdf}{ac^2}, \frac{bef}{ac^2}, \frac fc, 
\frac{bf}{ac}, \frac{fq}{ac}\, \frac{fq^2}{ac^2}; q^2,q^2 \right) \\
&= \frac{(acq/d, acq/e;q)_\i 
(ac^2\q/b,abq,bf/ac^2,ac^2q^2/bde;\q)_\i}{(acq,f/ac; q)_\i(ac^2\q/bd, 
ac^2\q/be, bdf/ac^2, bef/ac^2; \q)_\i} \\
&\cdot \sumk \frac{1-acq^{3k}}{1-ac}\, \frac{(a,b,cq/b;q)_k 
(d,e,f;q^2)_k}{(cq^2,ac\q/b,abq;q)_k (acq/d,acq/e,acq/f;q)_k}\, q^k \\
&+ \frac{bf^2(a,b,cq/b;q)_\i}{a^2c^3(1-f/c)(1-bf/ac) (ac,f/ac,acq/f;q)_\i} \\
&\cdot \frac{(f,ac^2/bf,ac^2\q/b,d,e,fq^2/d, f\q/e,bf\q/ac^2;\q)_\i}{(c\q, 
ac\q/b,ac^2\q/bf, ac^2\q/be, ac^2\q/bd,bdf/ac^2, bef/ac^2, 
bf^2\q/ac^2;\q)_\i} \\
&\cdot \sumj \frac{(bf^2/ac^2,bdf/ac^2, bef/ac^2, f/c, bf/ac;\q)_j (1-
bf^2q^{4j}/ac^2)}{(\q,f\q/d,f\q/e,bf\q/ac,f\q/c;q^2)_j (1-bf^2/ac^2)} \,
\left(-\frac{fq^2}{ab}\right)^j q^{j^2} \endalign$$
when (5.6) holds.   By applying the transformation formula [7, (5.14)]
$$\align
&{} \sumj \frac{(a,q\sa,-q\sa,c,e,d,f;q)_j}{(q,\sa,-
\sa,aq/c,aq/d,aq/e,aq/f;q)_j}\, \left(-\frac{a^2\q}{cdef}\right)^j 
q^{\binom j2} \tag 5.11 \\
&= \frac{(aq,aq/ef;q)_\infty}{(aq/e,aq/f;q)_\i}\, {}_3\phi_2 \left[\matrix 
aq/cd,e,f \\ aq/c,aq/d \endmatrix ;q,\frac{aq}{ef}\right] \endalign$$
to the sum over  $j$  in (5.10), we find that it equals
$$\frac{(\q,bf^2\q/ac^2;\q)_\i}{(bf\q/ac,f\q/c;\q)_\i}\, {}_3\phi_2 
\left[\matrix f/c,bf/ac,ac^2\q/bde \\ fq^2/d,fq^2/e \endmatrix ;q^2,q^2 
\right] \tag 5.12 $$
and hence, by setting $e=a^2c^2q/df$ in (5.10), we obtain the quadratic 
transformation formula
$$\align
&{}_{10}W_9 (ac^2/b;f,ac/b,c,cq/b,cq^2/b,d,a^2c^2q/df;q^2,q^2) \tag 5.13 \\
&+ 
\frac{(ac^2q^2/b,bf/ac^2,ac/b,c,cq/b,cq^2/b,bf\q/ac;\q)_\i}{(bf^2\q/ac^2,
ac^2/bf,ac^2q^2/bd, dfq/ab,bdf/ac^2,abq/d,c\q;\q)_\i} \\
&\cdot \frac{(fq^2/c,bf/c,bfq/c,fq^2/d,df^2q/a^2c^2,d,a^2c^2q/df; 
q^2)_\i}{(ac\q/b,f/c,bf/ac,ac, acq,fq/ac,f\q/ac; \q)_\i} \\
&\cdot {}_{10}W_9 (bf^2/ac^2;f,bdf/ac^2, abq/d, f/c,bf/ac,fq/ac, 
fq^2/ac;\q,\q) \\
&- \frac{(a,b,cq/f;q)_\i (f,d,a^2c^2q/df, bf/ac^2,ac^2q^2/b,fq^2/d, 
df^2q/a^2c^2,\q;\q)_\i}{(ac, ac/f,fq/ac;q)_\i (bf/ac, f/c, cq^2,ac\q/b, 
ac^2\q/bf, dfq/ab,bdf/ac^2, abq/d;\q)_\i} \\
&\cdot {}_3\phi_2 \left[\matrix f/c,bf/ac,fq/ab \\ fq^2/d, df^2q/a^2c^2 
\endmatrix ; q^2,q^2 \right] \\
&= \frac{(acq/d, df/ac;q)_\i(ac^2\q/b, abq,bf/ac^2,fq/ab;q^2)_\i}{(acq, 
f/ac;q)_\i (ac^2q^2/bd, dfq/ab,bdf/ac^2,abq/d;q^2)_\i} \\
&\cdot \sumk \frac{1-acq^{3k}}{1-ac}\, \frac{(a,b,cq/b;q)_k 
(d,f,a^2c^2q/df;q^2)_k}{(cq^2,acq^2/b, abq;q^2)_k (acq/d, acq/f, df/ac;q)_k} 
\, q^k. \endalign$$

Formula (5.1) follows from the case  $c=1$ of (5.13) by applying Sears' 
transformation formula [11, (5.2)] to the ${}_3\phi_2$ series.  If, as in the 
derivation of (3.21) from (3.19), we set $b=q/a$ in (5.13) and use (3.9), we 
obtain a quadratic summation formula which can also be derived by 
replacing  $a$, $b$, $c$, $d$ in (5.1) by $ac$, $a$, $d$, $a^2c^2q/df$, 
respectively.   The quadratic transformation formula (3.12) in [10] can be 
derived by multiplying both sides of (5.13) by $(f/ac;q)_\i$ and then setting  
$f = ac$.   

It should be observed that the first two terms on the left side of (5.13) 
containing the  ${}_{10}W_9$ series can be transformed to another pair of 
${}_{10}W_9$ series by applying Bailey's four term transformation formula 
[4, (7.2)].   Also, since the ${}_3\phi_2$ series in (5.13) is balanced 
(Saalsch\"utzian) it can be summed by [12, (3.3.2.2)] whenever it terminates.  
In addition, since $(q^{-n};q)_\i=0$  for  $n=0,1,2,\dots$,  the case  
$d=q^{-2n}$ of (5.13) gives
$$\align
&{}\sumnk \frac{1-acq^{3k}}{1-ac}\, \frac{(a,b,cq/b;q)_k (f,a^2c^2q^{2n+1}/f, 
q^{-2n};q^2)_k}{(cq^2,ac\q/b,abq;\q)_k (acq/f, f/acq^{2n}, acq^{2n+1};q)_k} \, 
q^k \tag 5.14 \\
&= \frac{(acq;q)_{2n} (ac^2\q/bf,abq/f;\q)_n}{(acq/f;q)_{2n}(abq, ac^2q^2/b; 
\q)_n} \\
&\cdot {}_{10}W_9 (ac^2/b; f,ac/b,c,cq/b,cq^2/b,a^2c^2q^{2n+1}/f, q^{-2n}; 
\q,\q) \endalign$$
and the case $b=cq^{n+1}$ gives
$$\align
&{}\sumnk \frac{1-acq^{3k}}{1-ac}\, \frac{(d,f,a^2c^2q/df;q^2)_k(a,cq^{n+1}, 
q^{-n};q)_k}{(acq/d, acq/f,df/ac;q)_k (cq^2,aq^{1-n},acq^{n+2};q^2)_k}\, q^k
\tag 5.15 \\
&= \frac{(acq^2/d,q/ac,acq/f,dfq/ac;q^2)_n}{(acq^2,fq/ac,dq/ac,acq/df;q^2)_n} 
\\
&\cdot {}_{10}W_9 (acq^{-n-1};c,d,f,a^2c^2q/df, aq^{-n-1},q^{1-n},q^{-n};
q^2,\q), \endalign$$
where  $n=0,1,2,\dots$.

\subheading{6.  A quartic transformation formula with three parameters}
By starting with the special case
$$\align
&{}  \sumnk \frac{(1-acq^{5k})(1-b/cq^{3k})}{(1-ac)(1-b/c)}\, 
\frac{(a,b;q)_k(q^{-4n},ac^2q^{4k}/b;q^4)_k}{(cq^4,acq^4/b;q^4)_k 
(acq^{4n+1},b/cq^{4n-1};q)_k}\, q^{4k} 
\tag 6.1 \\
&= \frac{(1-c)(1-ac/b)(1-acq^{4n})(1-cq^{4n}/b)}{(1-ac)(1-c/b)(1-cq^{4n})
(1-acq^{4n}/b)}, \qquad n=0,1,2,\dots, \endalign$$
of (2.8),  multiplying both sides of (6.1) by
$$\frac{(1-ac^2q^{8n}/b)(ac^2/b, a^2b^2/q^2;q^4)_n (c/b;q)_{4n}}{(1-ac^2/b) 
(q^4,c^2q^6/ab^3;q^4)_n (acq;q)_{4n}}\, q^{4n}, $$
summing over $n$ and changing the order of summation, we get the 
expansion formula
$$\align
&{}_{10}W_9 (ac^2/b; a^2b^2/q^2,ac/b, c,cq/b,cq^2/b, cq^3/b,cq^4/b;q^4, 
q^4) \tag 6.2 \\
&= \sumk \frac{(a,b;q)_k (a^2b^2/\q;q^4)_k (ac^2q^4)_{2k} 
(cq/b;q)_{3k}}{(cq^4,acq^4/b, c^2q^6/ab^3;q^4)_k (ac;q)_{5k}}\, \cb^k 
q^{4k+3\binom k2} \\
&\cdot {}_8W_7 (ac^2q^{8k}/b; a^2b^2q^{4k}/q^2, cq^{3k}/b, cq^{3k+1}/b, 
cq^{3k+2}/b, cq^{3k+3}/b; q^4,q^4). \endalign $$
Since the above ${}_8W_7$ equals
$$\align
&{} \frac{(abq^{2k};q)_\i (abq^{2k-1};q^2)_\i (ac^2q^{8k+4}/b, 
ab^3/c^2q^{4k+2}; q^4)_\i}{(acq^{5k+1},ab^2/cq^{k+2};q)_\i} \tag 6.3 \\
&- \frac{(cq^{3k}/b, a^2b^3q^{k-1}/c;q)_\i(ac^2q^{8k+4}/b, 
ab^3/c^2q^{4k+2};q^4)_\i}{(acq^{5k+1},ab^2/cq^{k+2};q)_\i (a^3b^5/c^2, 
c^2q^{4k+2}/ab^3;q^4)_\i} \\
&\cdot {}_8W_7 (a^3b^5/c^2q^4; a^2b^2q^{4k-2}, ab^2/cq^{k+2}, ab^2/cq^{k+1}, 
ab^2/cq^k, ab^2/cq^{k-1}; q^4,q^4) \endalign$$
by (3.9), it follows that
$$\align
&{}_{10}W_9 (ac^2/b;a^2b^2/q^2,ac/b,c,cq/b,cq^2/b,cq^3/b,cq^4/b;q^4,q^4)
\tag 6.4 \\
&= \frac{(ab;q)_\i (ab/q,\q)_\i (ac^2q^4/b,ab^3/c^2\q;q^4)_\i}{(acq, 
ab^2/cq^2;q)_\i} \\
&\cdot \sumk \frac{1-acq^{5k}}{1-ac}\, \frac{(a,b;q)_k (cq/b,cq^2/b, 
cq^3/b;q^3)_k (a^2b^2/q^2;q^4)_k}{(cq^4,acq^4/b;q^4)_k (abq,ab,ab/q;q^2)_k 
(cq^3/ab^2;q)_k}\, q^k \\
&- \frac{(a^2b^3/cq,c/b;q)_\i (ac^2q^4/b,ab^3/c^2q^2;q^4)_\i}{(acq, 
ab^2/cq^2;q)_\i (a^3b^5/c^2, c^2q^2/ab^3;q^4)_\i} \\
&\cdot \sumj \frac{(1-a^3b^5q^{8j-4}/c^2) (a^3b^5/c^2q^4, a^2b^2/q^2; q^4)_j 
(ab^2/cq^2;q)_{4j}}{(1-a^3b^5/c^2q^4) (q^4,ab^3q^2/c^2;q^4)_j 
(a^2b^3/cq;q)_{4j}}\, q^{4j} \\
&\cdot \sumk \frac{(1-acq^{5k})(1-b/cq^{3k})}{(1-ac)(1-b/c)}\, \frac{(a,b;q)_k 
(a^2b^2q^{4j-2}, c^2/ab^3q^{4j-2};q^4)_k}{(cq^4,acq^4/b;q^4)_k (c/ab^2q^{4j-
3},a^2b^3q^{4j-1}/c;q)_k}\, q^{4k}. \endalign$$

By (2.7), the above sum over  $k$ equals
$$\align
&{} \frac{(1-c)(1-ac/b)(1-a^2b^3q^{4j-2}/c) (1-ab^2q^{4j-2}/c)}{(1-ac)(1-
c/b)(1-a^2b^2q^{4j-2}/c) (1-ab^3q^{4j-2}/c)} \\
&- \frac{ab^3(a,b;q)_\i (a^2b^2/q^2, c^2q^2/ab^3;q^4)_\i (-a^3b^5/c^2)^j 
q^{12\binom j2}}{c^2\q(1-ac)(1-b/c)(cq^3/ab^2,a^2b^3/cq; q)_\i 
(cq^4,acq^4/b;q^4)_\i} \\
&\cdot \frac{(ab^3\q/c^2;q^4)_j (a^2b^3/cq;q)_{4j}}{(1-a^2b^2q^{4j-2}/c) (1-
ab^3q^{4j-2}/c) (a^2b^2/q^2;q^4)_j (ab^2/cq^2;q)_{4j}} \endalign$$
and so, from (6.4) and the transformation formula [7,(5.27)], we obtain the 
quartic transformation formula
$$\align
&{}_{10}W_9 (ac^2/b;a^2b^2/q^2,ac/b,c,cq/b,cq^2/b,cq^3/b,cq^4/b; q^4,q^4)
\tag 6.5 \\
&+ \frac{(1-c)(1-ac/b)(a^2b^3/cq^2,cq/b;q)_\i (ac^2q^4/b, 
ab^3/c^2\q;q^4)_\i}{(1-a^2b^2/cq^2)(1-ab^3/cq^2) (ab^2/cq,ac;q)_\i 
(a^3b^5/c^2,c^2q^2/ab^3;q^4)_\i} \\
&\cdot {}_{10}W_9 (a^3b^5/c^2q^4; a^2b^2/q^2,a^2b^2/cq^2, ab^3/cq^2,ab^2/cq, 
ab^2/c,ab^2q/c, ab^2q^2/c;q^4,q^4) \\
&+ \frac{ab^2(a,b,cq/b;q)_\i (ac^2q^4/b,ab^3/c^2\q,a^2b^2/\q;q^4)_\i}{c\q (1-
a^2b^2/cq^2) (ac,ab^2/c\q,cq^3/ab^2;q)_\i (ab^3/cq^2,cq^4,acq^4/b;q^4)_\i} \\
&\cdot {}_1\phi_1 \left[\matrix a^2b^2/cq^2 \\ a^2b^2q^2/c \endmatrix ;q^4, 
\frac{ab^3q^2}c\right] \\
&= \frac{(ab;q)_\i (ab/q;q^2)_\i (ac^2q^4/b,ab^3/c^2q^2;q^4)_\i}{(acq, 
ab^2/cq^2;q)_\i} \\
&\cdot \sumk \frac{1-acq^{5k}}{1-ac}\, \frac{(a,b;q)_k 
(cq/b,cq^2/b,cq^3/b;q^3)_k (a^2b^2/q^2;q^4)_k}{(cq^4,acq^4/b;q^4)_k 
(abq,ab,ab/q;\q)_k (cq^3/ab^2;q)_k}\,q^k. \endalign$$

When  $c=1$, (6.5) reduces to the quartic summation formula  [7, (5.28)].   
When  $b=q^2/a$, the sum of the two ${}_{10}W_9$ series on the left side of 
(6.5) reduces to a sum of two ${}_8W_7$ series which can be summed by (3.9) to 
give
$$\align
&{}\sumk \frac{1-acq^{5k}}{1-ac}\, \frac{(a,q^2/a;q)_k (ac/q,ac,acq;q^3)_k 
(q^2;q^4)_k}{(cq^4,a^2cq^2;q^4)_k (q^3,q^2,q;q^2)_k (ac/q;q)_k}\, q^k 
\tag 6.6 \\
&- \frac{\q(a,\q/a,acq;q)_\i (\q;q^4)_\i}{ac(1-q^2/c)(q^2,ac;q)_\i (q;q^2)_\i 
(cq^4,a^2cq^2, q^4/a^2c;q^4)_\i} \\
&\cdot {}_1\phi_1 \left[\matrix q^2/c \\ q^6/c \endmatrix 
;q^4,\frac{q^8}{a^2c}\right] \\
&= \frac{(acq^2,q^2/ac,aq,q^3/a;q^2)_\i}{(q,q^3;q^2)_\i 
(cq^4,q^2/c,a^2cq^2,q^4/a^2c;q^4)_\i} . \endalign$$
It should also be observed that by multiplying both sides of (6.5) by  
$(ab^2/cq^2;q)_\i$ and then setting  $c=ab^2/q^2$ we get the quartic 
transformation formula
$$\align
&{} \sumk \frac{1-a^2b^2q^{5k-2}}{1-a^2b^2/q^2}\, \frac{(a,b;q)_k 
(ab/q,ab,abq;q^3)_k (a^2b^2/q^2;q^4)_k}{(ab^2q^2,a^2bq^2;q^4)_k 
(abq,ab,ab/q;q^2)_k (q;q)_k}\, q^k \tag 6.7 \\
&= \frac{(aq,b;q)_\i (a^2b^2q^2;q^4)_\i}{(q;q)_\i (abq;q^2)_\i 
(b,ab^2q^2,a^2bq^2;q^4)_\i}\, {}_1\phi_1 \left[\matrix a \\ aq^4 \endmatrix
;q^4,bq^4 \right]. \endalign$$

Additional summation, transformation, and expansion formulas will be 
considered elsewhere.


\cl{\bf References} \bs

\item{1.}  W. A. Al-Salam and A. Verma, On quadratic transformations of basic 
series, SIAM J. Math\. Anal\. 15 (1984), 414--420. \ss

\item{2.} R. Askey, The $q$-gamma and $q$-beta functions, Applicable Analysis 
8 (1978), 125--141. \ss

\item{3.}  W. N. Bailey, Generalized Hypergeometric Series, Cambridge 
University Press, London, 1935 (reprinted by Stechert-Hafner, New York, 1964).
\ss

\item{4.}  \blank, Well-poised basic hypergeometric series, Quart. J. Math\. 
(Oxford) 18 (1947), 157--166. \ss

\item{5.} L. Carlitz, Some inverse relations, Duke Math\. J. 40 (1973), 
803--901. \ss

\item{6.}  J. L. Fields and J. Wimp, Expansions of hypergeometric functions in 
hypergeometric functions, Math\. Comp\. 15 (1961), 390--395. \ss

\item{7.} G. Gasper, Summation, transformation, and expansion formulas for 
bibasic series, to appear. \ss

\item{9.}  G. Gasper and Mizan Rahman, Basic Hypergeometric Series, Cambridge 
University Press, to appear. \ss

\item{9.} I. Gessel and D. Stanton, Strange evaluations of hypergeometric 
series, SIAM J. Math\. Anal\. 13 (1982), 295--308. \ss

\item{10.}  Mizan Rahman, Some quadratic and cubic summation formulas for basic 
hypergeometric series,  to appear. \ss

\item{11.} D. B. Sears, Transformations of basic hypergeometric functions of 
special type, Proc\. London Math\. Soc\. (2) 52 (1951), 467--483. \ss

\item{12.} L. J. Slater, Generalized Hypergometric Functions, Cambridge 
University Press, London, 1966. \ss

\item{13.} A. Verma, Some transformations of series with arbitrary terms, 
Instituto Lombardo (Rend\. Sc.) A 106 (1972), 342--353.

\bye
