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

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## 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

- 1953: Arthur Erdélyi, Wilhelm Magnus, Fritz Oberhettinger and Francesco G. Tricomi:
*Higher Transcendental Functions Volume I*... (previous) ... (next): $4.2 (1)$ (formula incorrect in printed edition, but fixed in errata)