∑k=0n(1-a⁢pk⁢qk)⁢(1-b⁢pk⁢q-k)(1-a)⁢(1-b)⁢(a,b;p)k⁢(c,a/(b⁢c);q)k(q,a⁢q/b;q)k⁢(a⁢p/c,b⁢c⁢p;p)k⁢qk=(a⁢p,b⁢p;p)n⁢(c⁢q,a⁢q/(b⁢c);q)n(q,a⁢q/b;q)n⁢(a⁢p/c,b⁢c⁢p;p)n,