# (b-a)2F1+a2F1(a+1)-b2F1(b+1)=0

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