2025.9.6 闲话

\[\sum_{i=0}^n\dbinom ni(-1)^i\dfrac1{m+i+1} \]

5k 问能不能一等到底：

\[\begin{aligned}\sum_{i=0}^n\dbinom ni(-1)^i\dfrac1{m+i+1}&=\sum_{i=0}^n\dbinom ni(-1)^i\int_0^1(1-x)^{m+i}\mathrm dx\\&=\int_0^1\sum_{i=0}^n\dbinom ni(-1)^i(1-x)^{m+i}\mathrm dx\\&=\int_0^1(1-x)^mx^n\mathrm dx\\&=\dfrac{n!m!}{(n+m+1)!}\end{aligned} \]

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