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AFewSuns

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$\displaystyle\sum_{n \geq 0}{(x+y)^n\frac{u^n}{n!}}=\sum_{k \geq 0}\frac{x}{x-kz}(x-kz)^k\frac{u^k}{k!}\sum_{l \geq 0}\frac{(y+kz)^l}{l!}u^l$

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