证明 $\sum_{k=0}^{n} C_{k}^m = C_{m+1}^{n+1} $

前置芝士：证明 \(C(m,n)=C(m,n-1)+C(m-1,n-1)\)

\(∵C(m,n)=C(m,n-1)+C(m-1,n-1)\)
\(∴C(m+1,n+1)=C(m+1,n)+C(m,n)\)

证明 $\sum_{k=0}^{n} C_{k}^m = C_{m+1}^{n+1} $

\(\sum_{k=0}^{n} C_{k}^m = C_{m}^{m}+C_{m+1}^{m}+\dots+C_{n}^{m}\)

\(=(C_{m+1}^{m+1}+C_{m+1}^{m})+\dots+C_{n}^{m}\)

\(=(C_{m+2}^{m+1}+C_{m+2}^{m})+\dots+C_{n}^{m}\)

合并。。。合并

\(=C_{n+1}^{m+1}\)