如何证明
∫01xk(1−x)n−k∫01xk−1(1−x)n−k=∑i=0n−k(n−ki)(−1)i1i+k+1∑i=0n−k(n−ki)(−1)i1i+k=kn+1\frac{\int _0^1 x^k (1 - x)^{n - k}}{\int _0^1 x^{k - 1} (1 - x)^{n - k}} = \frac{ \sum_{i = 0}^{n - k} \binom{n - k}{i} (-1)^i \frac{1}{i + k + 1} }{ \sum_{i = 0}^{n - k} \binom{n - k}{i} (-1)^i \frac{1}{i + k} } = \frac{k}{n + 1}∫01xk−1(1−x)n−k∫01xk(1−x)n−k=∑i=0n−k(in−k)(−1)ii+k1∑i=0n−k(in−k)(−1)ii+k+11=n+1k
(第一个等号是展开 第二个是要证的 /kel)