View source for 2Phi1(q,-1;-q;z)=1+2Sum z^k/(1+q^k)

==Theorem==
The following formula holds:
$${}_2\phi_1(q,-1;-q;z)=1+2\displaystyle\sum_{k=1}^{\infty} \dfrac{z^k}{1+q^k},$$
where ${}_2\phi_1$ denotes [[basic hypergeometric phi]].

==Proof==

==References==
* {{BookReference|Higher Transcendental Functions Volume I|1953|Harry Bateman|prev=(z/(1-q))2Phi1(q,q;q^2;z)=Sum z^k/(1-q^k)|next=z/(1-sqrt(q))2Phi1(q,sqrt(q);sqrt(q^3);z)=Sum z^k/(1-q^(k-1/2))}}: $4.8 (7)$

[[Category:Theorem]]
[[Category:Unproven]]