# C(a-(c-b)z)2F1-ac(1-z)2F1(a+1)+(c-a)(c-b)z2F1(c+1)=0

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