$$\sum_{i=0}^n{m+i \choose m}={m+n+1 \choose m+1}$$

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lucas定理

$$\mathrm{C}_{n}^{m}\equiv \mathrm{C}_{n/p}^{m/p}·\mathrm{C}_{n\%p}^{m\%p}(mod\ p)$$

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$$a^b\equiv a^{b\%{\varphi(c)}+\varphi(c)}\quad mod (c), b\ge\varphi(c)$$

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$$\int_0^t \sqrt{\frac{1}{1-x^2}} \, dx=$$ $$\frac{t \sqrt{\frac{1}{1-t^2}} \sqrt{1-t^2}}{\sin }+\pi ,\left(t\notin \mathbb{R}\land \left(\frac{\text{Re}}{t}=0\lor \frac{\text{Re}}{t}\geq 1\lor \frac{\text{Re}}{t}+1\leq 0\lor \frac{1}{t}\notin \mathbb{R}\right)\right)$$ $$\lor \left(\text{Re} t<0\land \text{Re} t+1>0\land \left(\frac{\text{Re}}{t}+1<0\lor \frac{1}{t}\notin \mathbb{R}\right)\right)$$ $$\lor \left(\text{Re} t<1\land \text{Re} t>0\land \left(\frac{\text{Re}}{t}>1\lor \frac{1}{t}\notin \mathbb{R}\right)\right)$$

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$$\int \sqrt{\frac{1}{1-x^2}} \, dx=\sqrt{\frac{1}{1-x^2}} \sqrt{1-x^2} \arcsin(x)$$

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