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Here again the two methods diverge. The WZ method operates without a
hitch as has already been shown by Zeilberger and his computer [7].
Namely, if a+f=-m and b+c+d+e=m+1, then
The WZ-method produces the certificate function
and it is then immediate that
and consequently by summing (4.3) from r=0 to m+1 we find
In contrast, Pfaff again is unable to prove the required result by
itself. In this instance, one is forced to prove a much less known
companion identity due to A. Lakin [10].
Namely if a+f=-m and b+c+d+e=m, then
where
is the third elementary symmetric
function of a,b,c,d,e and f.
The Pfaff method now proceeds as before. In this case,
and
To conclude the simultaneous proof of Dougall's (4.1) and Lakin's
(4.5) theorems, one verifies that the right-hand sides satisfy (4.6)
and (4.7) also.
In this section then, the WZ method has been fast and elegant. The
Pfaff method has required the unearthing of an almost forgotten result
[10]. It has produced more but with twice the effort. If Lakin's
result had been totally forgotten, considerable computer time would
have been required to find it.
It is perhaps instructive to refer to the WZ method applied to Lakin's
series. Note the complexity that
introduces. In
this instance
and the relevant final recurrence is
It should be pointed out here that Pfaff's method has been able to act
with simpler recurrences in Sections 3 and 4 because it relies on
shifts of parameters in addition to the shift of the index m.

Contents
Next: The 5F4 Summation.
Up: Pfaff's Method (III): Comparison
Previous: Bailey's Theorem.