##思路
∑0≤k1,k2,k3,⋯km≤n∏1≤j<mC(kj+1,kj)\sum_{0\leq k_1,k_2,k_3,\cdots k_m\leq n}\prod _{1\leq j<m}C(k_{j+1},k_j)0≤k1,k2,k3,⋯km≤n∑1≤j<m∏C(kj+1,kj)
=∑km=0nC(n,km)∑km−1=0kmC(km,km−1)⋯∑k1=0k2C(k2,k1)=\sum_{k_m=0}^nC(n,k_m)\sum_{k_{m-1}=0}^{k_m}C(k_m,k_{m-1})\cdots \sum_{k_1=0}^{k_2}C(k_2,k_1)=km=0∑nC(n,km)km−1=0∑kmC(km,km−1)⋯k1=0∑k2C(k2,k1)
=∑km=0nC(n,km)∑km−1=0kmC(km,km−1)⋯∑k2=0k3C(k3,k2)×2k2=\sum_{k_m=0}^nC(n,k_m)\sum_{k_{m-1}=0}^{k_m}C(k_m,k_{m-1})\cdots\sum_{k_2=0}^{k_3}C(k_3,k_2)\times2^{k_2}=km=0∑nC(n,km)km−1=0∑kmC(km,km−1)⋯k2=0∑k3C(k3,k2)×2k2
=∑km=0nmkm=\sum_{k_m=0}^{n}m^{k_m}=km=0∑nmkm
=mn+1−1m−1=\frac{m^{n+1}-1}{m-1}=m−1mn+1−1
###没有想到吧k序列反向求和,找不到合适的公式。