二项式系数相关

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

  • \(\dbinom{n}{k}=\dbinom{n}{n-k}(n\ge0\quad n,k\in\mathbb{Z})\)

  • \(\dbinom{r}{k}=\dfrac{r}{k}\dbinom{r-1}{k-1}(k\in\mathbb{Z},k\ne0)\)

  • \(\dbinom{r}{k}=\dbinom{r-1}{k}+\dbinom{r-1}{k-1}(k\in\mathbb{Z})\)

  • \(\sum\limits_{0\leq k< n}\dbinom{k}{m}=\dbinom{n+1}{m+1}(m,n\ge0)\)

  • \(\sum\limits_{ k\leq n}\dbinom{r+k}{k}=\dbinom{r+n+1}{n}(m,n\ge0)\)

\((x+y)^r=\sum\limits_{1\leq k\leq r}\dbinom{r}{k}x^ky^{r-k}\)

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

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

\(\sum\limits_k\dbinom{r}{k}\dbinom{s}{n-k}=\dbinom{r+s}{n}\)

\(\sum\limits_k\dbinom{r}{m+k}\dbinom{s}{n-k}=\dbinom{r+s}{m+n}\)

\(\sum\limits_k\dbinom{l}{m+k}\dbinom{s}{n+k}=\dbinom{l+s}{l-m+n}\)

\(\sum\limits_k\dbinom{l}{m+k}\dbinom{s+k}{n}(-1)^k=(-1)^{l+m}\dbinom{s-m}{n-l}\)

\(\sum\limits_k\dbinom{l-k}{m}\dbinom{s}{k-n}(-1)^k=(-1)^{l+m}\dbinom{s-m-1}{l-n-m}\)

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转载自www.cnblogs.com/defense/p/12619093.html