View source for 2F1(a,b;a+b+1/2;z)^2=3F2(2a,a+b,2b;a+b+1/2,2a+2b;z)

==Theorem==
The following formula (known as Clausen's formula) holds:
$${}_2F_1 \left( a,b;a+b +\dfrac{1}{2}; z \right)^2 = {}_3F_2 \left( 2a, a+b, 2b;a+b+\dfrac{1}{2}, 2a+2b;z \right),$$
where ${}_2F_1$ denotes [[hypergeometric 2F1]] and ${}_3F_2$ denotes [[hypergeometric 3F2]].

==Proof==

==References==
* {{BookReference|Higher Transcendental Functions Volume I|1953|Harry Bateman|prev=findme|next=0F1(;r;z)0F1(;s;z)=2F1(r/2+s/2, r/2+s/2-1/2;r,s,r+s-1;4z)}}: $4.2 (1)$ (formula incorrect in printed edition, but fixed in errata)

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