关于\(\dfrac{1}{(1-z)^c}\)的幂级数。
首先，\(G(z)\dfrac{1}{1-z}\)表示\(\{\sum_{i=0}^ng_i\}\)的生成函数
\(c=0,G_0(z)=1z^0+0z^1+0z^2+0z^3+\cdots\)
\(c=1,G_1(z)=1z^0+1z^1+1z^2+1z^3+\cdots\)
\(c=2,G_2(z)=1z^0+2z^1+3z^2+4z^3+\cdots\)
\(c=3,G_3(z)=1z^0+3z^1+6z^2+10z^3+\cdots\)
\(\cdots\cdots\cdots\cdots\cdots\cdots\cdots\)
容易发现，也很好证明\([x^n]G_c(z)=[x^{n-1}]G_c(z)+[x^n]G_{c-1}(z)\)
暂且表示为\(g(c,n)=g(c-1,n)+g(c,n-1)\)
\(g(c,n)=\)从\((1,0)\)到\((c,n)\)的方案数\(=\dbinom{c+n-1}{n}\)
\(\dfrac{1}{(1-z)^c}=\sum_{n\geq0}\dbinom{c+n-1}{n}\)