3楼: Originally posted by hank612 at 2016-06-07 11:33:08
感觉楼主对二项式系数挺熟悉的，那么二项式展开式应该不会陌生吧

1. (1-x)^k=\sum_{i=0}^k (-1)^{k-i}\binom{k}{i} x^{k-i}

2. \frac{1}{(1-x)^{k}}=\sum_{j=0}^{\infty}\binom{k+j-1}{k-1}x^j

那么， \fr ...

2楼: Originally posted by gold2007 at 2016-06-07 07:25:24
j=1没意义吧

4楼: Originally posted by i维数 at 2016-06-07 21:37:46
厉害！我也来给一个比较初等的证明

组合恒等式.png
...

3楼: Originally posted by hank612 at 2016-06-07 11:33:08
感觉楼主对二项式系数挺熟悉的，那么二项式展开式应该不会陌生吧

1. (1-x)^k=\sum_{i=0}^k (-1)^{k-i}\binom{k}{i} x^{k-i}

2. \frac{1}{(1-x)^{k}}=\sum_{j=0}^{\infty}\binom{k+j-1}{k-1}x^j

那么， \fr ...

4楼: Originally posted by i维数 at 2016-06-07 21:37:46
厉害！我也来给一个比较初等的证明

组合恒等式.png
...

6楼: Originally posted by cmxregister9 at 2016-06-08 06:29:54
非常感谢，谢谢大牛！您的硬推导中f(x)构造得很巧妙！...