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$$ \sum_{n=0}^\infty \frac{(x)_n}{2^n\left(\frac{1}{2}+x\right)_n}=\sum_{n=0}^\infty \frac{(6n+4x+3)(4n^2+6n+1+2x)}{1-2n}\frac{(-1)^n}{2^{3n+2}}\frac{\binom{2n}{n}(x)_n}{\left(\frac{1}{2}+x\right)_{2n+2}} $$
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Thu, 10 Nov 2022 13:27 GMT