二项式系数相关

  • (dbinom{n}{k}=dfrac{n!}{k!(n-k)!}(nge k ge 0, nin mathbb{Z}))

  • (dbinom{n}{k}=dbinom{n}{n-k}(nge0quad n,kinmathbb{Z}))

  • (dbinom{r}{k}=dfrac{r}{k}dbinom{r-1}{k-1}(kinmathbb{Z},k e0))

  • (dbinom{r}{k}=dbinom{r-1}{k}+dbinom{r-1}{k-1}(kinmathbb{Z}))

  • (sumlimits_{0leq k< n}dbinom{k}{m}=dbinom{n+1}{m+1}(m,nge0))

  • (sumlimits_{ kleq n}dbinom{r+k}{k}=dbinom{r+n+1}{n}(m,nge0))

((x+y)^r=sumlimits_{1leq kleq r}dbinom{r}{k}x^ky^{r-k})

(dbinom{r}{k}=(-1)^kdbinom{k-r-1}{k})

(dbinom{r}{m}dbinom{m}{k}=dbinom{r}{k}dbinom{r-k}{m-k})

(sumlimits_kdbinom{r}{k}dbinom{s}{n-k}=dbinom{r+s}{n})

(sumlimits_kdbinom{r}{m+k}dbinom{s}{n-k}=dbinom{r+s}{m+n})

(sumlimits_kdbinom{l}{m+k}dbinom{s}{n+k}=dbinom{l+s}{l-m+n})

(sumlimits_kdbinom{l}{m+k}dbinom{s+k}{n}(-1)^k=(-1)^{l+m}dbinom{s-m}{n-l})

(sumlimits_kdbinom{l-k}{m}dbinom{s}{k-n}(-1)^k=(-1)^{l+m}dbinom{s-m-1}{l-n-m})

原文地址:https://www.cnblogs.com/defense/p/12619093.html