I said
Notice the argument of q^2 vs Jackson's q. However, this is not sufficient to explain the failure of the linear combination trick.
The explanation turns out to be simple and elegant: Yours truly is an idiot. Simply repeating the computation with additional consciousness enabled, ScoobyJackson(n) = n n a (1 - q ) (1 - a q ) J(n) - ----------------------------- J(n - 1), n 2 n a q a q b c d (1 - -----) (1 - -----) b c d b c d where J(n) := Jackson(n). Furthermore, if S := ScoobyJackson, then ScooobyJackson(n) = 2 2 n n n a q a (1 - q ) (1 - a q ) (1 - -------) b c d S(n) - ---------------------------------------------------------- S(n - 1) 2 2 (n - 1) n - 1 2 n - 1 a q a q a q b c d q (1 - -------------) (1 - --------) (1 - ---------) b c d b c d b c d i.e., 2 2 n 2 2 n - 1 b c d q a q a q - a (1 - -------) (1 - -------) (1 - -----------) a b c d b c d 2 n - 1 1 a q 3 k - 1 inf (a, b, c, d, - sqrt(a) q, sqrt(a) q, --, ---------; q) q ==== n b c d k \ q ( > ----------------------------------------------------------------) / a q a q a q b c d n + 1 ==== (- sqrt(a), sqrt(a), q, ---, ---, ---, --------, a q ; q) k = 0 b c d n - 2 k a q n - 1 2 n - 1 a q a q /(b c d (1 - --------) (1 - ---------)) = b c d b c d n n n a q a q a q n a q a q a q (a q, ---, ---, ---; q) (- a (1 - q ) (1 - ----) (1 - ----) (1 - ----) b c b d c d n b c d n - 1 n - 1 n - 1 n - 1 a q a q a q a (1 - q ) (1 - --------) (1 - --------) (1 - --------) 2 n - 1 b c d a q ((- ----------------------------------------------------------- - --------- n - 1 n - 1 n - 1 b c d a q a q a q b c d (1 - --------) (1 - --------) (1 - --------) b c b d c d 2 2 n 2 2 (n - 1) 2 n - 1 a q a q a q + 1) (1 - -------)/(q (1 - -------------) (1 - ---------)) + 1) b c d b c d b c d n n n 2 n a q a q a q a q /(b c d (1 - ----) (1 - ----) (1 - ----)) - ----- + 1) b c b d c d b c d a a q a q a q /(-----, ---, ---, ---; q) , b c d b c d n where there are now seven extra powers of q downstairs, and the argument is q^3. It looks fairly easy to prove that these go on forever (reminiscent of the bottom of a Goo...ogle hits page), and indeed, Mizan says ("we" referring to George Gasper) Let me try to answer your "why" part.We were aware that a 8W7 series with an argument q^(r+1)(and appropriately altered parameter,e.g.,n+1 replaced by n+1-r),can be expressed,via Watson's and then Sears' transformation formulas,into r+1 terms. ... Apparently, both procedures (weighted difference or Watson/Sears) apply to a,b,c, and d as well as n, yielding an infinite tree of identities instead of just a sequence. --rwg MISMANAGE SINE GAMMA