View source for (c-a-b)2F1+a(1-z)2F1(a+1)-(c-b)2F1(b-1)=0

==Theorem==
The following formula holds:
$$(c-a-b){}_2F_1(a,b;c;z)+a(1-z){}_2F_1(a+1,b;c;z)-(c-b){}_2F_1(a,b-1;c;z)=0,$$
where ${}_2F_1$ denotes [[hypergeometric 2F1]]. 

==Proof==

==References==
* {{BookReference|Higher Transcendental Functions Volume I|1953|Arthur Erdélyi|author2=Wilhelm Magnus|author3=Fritz Oberhettinger|author4=Francesco G. Tricomi|prev=(b-a)2F1+a2F1(a+1)-b2F1(b+1)=0|next=c(a-(c-b)z)2F1-ac(1-z)2F1(a+1)+(c-a)(c-b)z2F1(c+1)=0}}: $\S 2.8 (33)$

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