rt，《组合数学》习题三T3
求证：
∑i=0n(ni)2xi=∑i=0n(ni)(2n−in)(x−1)i\sum_{i=0}^{n}\left(\begin{array}{c}n\\i\end{array}\right)^2x^i=\sum_{i=0}^{n}\left( \begin{array}{c} n \\ i \end{array} \right) \left( \begin{array}{c} 2n-i \\ n \end{array} \right) (x-1)^i∑i=0n(ni)2xi=∑i=0n(ni)(2n−in)(x−1)i
目前我的思路如下：
右式=
∑i=0n∑k=0i(ni)(2n−in)(ik)(−1)i−kxk=\sum_{i=0}^{n}\sum_{k=0}^{i}\left( \begin{array}{c} n \\ i \end{array} \right) \left( \begin{array}{c} 2n-i \\ n \end{array} \right) \left( \begin{array}{c} i \\ k \end{array} \right) (-1)^{i-k}x^k=∑i=0n∑k=0i(ni)(2n−in)(ik)(−1)i−kxk=(二项式定理)
∑k=0nxk∑i=kn(ni)(2n−ii)(ik)(−1)i−k\sum_{k=0}^{n}x^k\sum_{i=k}^{n}\left( \begin{array}{c} n \\ i \end{array} \right)\left( \begin{array}{c} 2n-i \\ i \end{array} \right) \left( \begin{array}{c} i \\ k \end{array} \right) (-1)^{i-k}∑k=0nxk∑i=kn(ni)(2n−ii)(ik)(−1)i−k
即证：
(ni)2=∑i=kn(ni)(2n−ii)(ik)(−1)i−k\left( \begin{array}{c} n \\ i \end{array} \right) ^2=\sum_{i=k}^{n}\left( \begin{array}{c} n \\ i \end{array} \right)\left( \begin{array}{c} 2n-i \\ i \end{array} \right) \left( \begin{array}{c} i \\ k \end{array} \right) (-1)^{i-k}(ni)2=∑i=kn(ni)(2n−ii)(ik)(−1)i−k
然后就不会了，似乎要套一个容斥原理的拓展（
