一些组合恒等式不会证：
1.∑k=0nkC2∗n+1n−k=(2n+1)C2n−1n−22n−1\sum^n_{k=0}kC^{n-k}_{2*n+1}=(2n+1)C^n_{2n-1}-2^{2n-1}∑k=0nkC2∗n+1n−k=(2n+1)C2n−1n−22n−1
2.∑k=0nkC2nn−k=nC2n−1n\sum^n_{k=0}kC^{n-k}_{2n}=nC^n_{2n-1}∑k=0nkC2nn−k=nC2n−1n
3.∑k=12n−1(−1)k−1n−kC2nk=0\sum^{2n-1}_{k=1}(-1)^{k-1} \frac{n-k}{C^k_{2n}}=0∑k=12n−1(−1)k−1C2nkn−k=0
必关！！！
