∑k=0n1-a⁢pk⁢qk1-a⁢(a;p)k⁢(c;q)k(q;q)k⁢(a⁢p/c;p)k⁢c-k=(a⁢p;p)n⁢(c⁢q;q)n(q;q)n⁢(a⁢p/c;p)n⁢c-n.