二项式系数本文重定向自 二项式系数

定义及概念

${\displaystyle (1+x)^{n}=\sum _{k=0}^{n}{\binom {n}{k}}x^{k}={\binom {n}{0}}+{\binom {n}{1}}x+\cdots +{\binom {n}{n}}x^{n}}$

${\displaystyle (x+y)^{n}=\sum _{k=0}^{n}{\binom {n}{k}}x^{n-k}y^{k}}$

计算二项式系数

递归公式

${\displaystyle {\binom {n}{k}}={\binom {n-1}{k-1}}+{\binom {n-1}{k}}\quad \forall n,k\in \mathbb {N} }$

${\displaystyle {\binom {n}{0}}=1\quad \forall n\in \mathbb {N} \cup \{0\},}$
${\displaystyle {\binom {0}{k}}=0\quad \forall k\in \mathbb {N} .}$

乘数公式

${\displaystyle {\binom {n}{k}}={\frac {n^{\underline {k}}}{k!}}={\frac {n(n-1)(n-2)\cdots [n-(k-1)]}{k(k-1)(k-2)\cdots 1}}=\prod _{i=1}^{k}{\frac {n-(k-i)}{i}},}$

阶乘公式

${\displaystyle {\binom {n}{k}}={\frac {n!}{k!\,(n-k)!}}\quad {\mbox{for }}\ 0\leq k\leq n.}$

${\displaystyle {\binom {n}{k}}={\binom {n}{n-k}}\quad {\mbox{for }}\ 0\leq k\leq n.}$

(1)

一般化形式及其与二项式级数的关系

${\displaystyle {\binom {\alpha }{k}}={\frac {\alpha ^{\underline {k}}}{k!}}={\frac {\alpha (\alpha -1)(\alpha -2)\cdots (\alpha -k+1)}{k(k-1)(k-2)\cdots 1}}\quad {\mbox{for }}k\in \mathbb {N} {\mbox{ and arbitrary }}\alpha .}$

${\displaystyle (1+X)^{\alpha }=\sum _{k=0}^{\infty }{\alpha \choose k}X^{k}.}$

(2)

${\displaystyle (1+X)^{\alpha }(1+X)^{\beta }=(1+X)^{\alpha +\beta }\quad {\mbox{and}}\quad ((1+X)^{\alpha })^{\beta }=(1+X)^{\alpha \beta }.}$

${\displaystyle \alpha }$是一非负整数${\displaystyle n}$，则所有${\displaystyle k>n}$的项为零，此无穷级数变成有限项的和，还原为二项式公式，但对于${\displaystyle \alpha }$的其他值，包括负数和有理数，此级数为无穷级数。

帕斯卡三角形 (杨辉三角)

${\displaystyle {n \choose k}+{n \choose k+1}={n+1 \choose k+1},}$

(3)

 0: 1 1: 1 1 2: 1 2 1 3: 1 3 3 1 4: 1 4 6 4 1 5: 1 5 10 10 5 1 6: 1 6 15 20 15 6 1 7: 1 7 21 35 35 21 7 1 8: 1 8 28 56 70 56 28 8 1

${\displaystyle n}$横行列出${\displaystyle {\tbinom {n}{k}}}$${\displaystyle k=0,\ldots ,n}$项，其建构方法为在外边填上1，然后将上一行中每两个相邻数相加的和填在其下，此方法可快速地计算二项式系数而不涉及乘法或分数，例如从第5横行可马上得出

${\displaystyle (x+y)^{5}={\boldsymbol {1}}x^{5}+{\boldsymbol {5}}x^{4}y+{\boldsymbol {10}}x^{3}y^{2}+{\boldsymbol {10}}x^{2}y^{3}+{\boldsymbol {5}}xy^{4}+{\boldsymbol {1}}y^{5}}$

组合数学和统计学

• 共有${\displaystyle {\tbinom {n}{k}}}$种方式从${\displaystyle n}$元素中选取${\displaystyle k}$项。见组合
• 共有${\displaystyle {\tbinom {n+k-1}{k}}}$种方式从一个${\displaystyle n}$元素集合中选取（容许重复选取）${\displaystyle k}$元素建立多重集
• 共有${\displaystyle {\tbinom {n+k}{k}}}$字符串包含${\displaystyle k}$个1和${\displaystyle n}$个零。
• 共有${\displaystyle {\tbinom {n+1}{k}}}$个字符串包含${\displaystyle k}$个1和${\displaystyle n}$个零，且没有两个1相邻。
• 卡塔兰数${\displaystyle {\frac {\tbinom {2n}{n}}{n+1}}}$
• 统计学中的二项式分布${\displaystyle {\tbinom {n}{k}}p^{k}(1-p)^{n-k}\!}$
• 贝兹曲线的公式。

以多项式表达二项式系数

${\displaystyle {\binom {t}{k}}={\frac {(t)_{k}}{k!}}={\frac {(t)_{k}}{(k)_{k}}}={\frac {t(t-1)(t-2)\cdots (t-k+1)}{k(k-1)(k-2)\cdots (2)(1)}};\,\!}$

${\displaystyle {\binom {t}{k}}=\sum _{i=0}^{k}{\frac {s_{k,i}}{k!}}t^{i}}$

${\displaystyle {\tbinom {t}{k}}}$导数可以对数微分计算：

${\displaystyle {\frac {\mathrm {d} }{\mathrm {d} t}}{\binom {t}{k}}={\binom {t}{k}}\sum _{i=0}^{k-1}{\frac {1}{t-i}}\,.}$

以二项式系数为多项式空间之基底

${\displaystyle a_{k}=\sum _{i=0}^{k}(-1)^{k-i}{\binom {k}{i}}p(i).}$

(3.5)

整数值多项式

${\displaystyle 9{\tbinom {t}{2}}+6{\tbinom {t}{1}}+0{\tbinom {t}{0}}}$

${\displaystyle t=1,2,3}$${\displaystyle {\frac {3t(3t+1)}{2}}=6,21,45}$帕斯卡矩阵可算出：

${\displaystyle {\begin{pmatrix}{\tbinom {t-1}{0}}&{\tbinom {t-1}{1}}&{\tbinom {t-1}{2}}\end{pmatrix}}{\begin{pmatrix}1&0&0\\-1&1&0\\1&-2&1\end{pmatrix}}{\begin{pmatrix}6\\21\\45\end{pmatrix}}={\begin{pmatrix}{\tbinom {t-1}{0}}&{\tbinom {t-1}{1}}&{\tbinom {t-1}{2}}\end{pmatrix}}{\begin{pmatrix}6\\15\\9\end{pmatrix}}}$
${\displaystyle =6{\tbinom {t-1}{0}}+15{\tbinom {t-1}{1}}+9{\tbinom {t-1}{2}}=6{\tbinom {t}{1}}+9{\tbinom {t}{2}}}$

有关二项式系数的恒等式

关系式

• ${\displaystyle {\binom {n+1}{k}}={\binom {n}{k}}+{\binom {n}{k-1}}}$
• ${\displaystyle {\binom {n}{k}}={\frac {n}{k}}{\binom {n-1}{k-1}}}$
• ${\displaystyle {\binom {n-1}{k}}-{\binom {n-1}{k-1}}={\frac {n-2k}{n}}{\binom {n}{k}}.}$

${\displaystyle {\binom {n}{i}}{\binom {i}{m}}={\binom {n}{m}}{\binom {n-m}{i-m}}}$

一阶求和公式

• ${\displaystyle \sum _{r=0}^{n}{\binom {n}{r}}=2^{n}}$
• ${\displaystyle \sum _{r=0}^{k}{\binom {n+r-1}{r}}={\binom {n+k}{k}}}$
• ${\displaystyle \sum _{r=0}^{n-k}{\frac {(-1)^{r}(n+1)}{k+r+1}}{\binom {n-k}{r}}={\binom {n}{k}}^{-1}}$
• ${\displaystyle \sum _{r=0}^{n}{\binom {dn}{dr}}={\frac {1}{d}}\sum _{r=1}^{d}(1+e^{\frac {2\pi ri}{d}})^{dn}}$
• ${\displaystyle \sum _{i=m}^{n}{\binom {a+i}{i}}={\binom {a+n+1}{n}}-{\binom {a+m}{m-1}}}$
${\displaystyle {\binom {a+m}{m-1}}+{\binom {a+m}{m}}+{\binom {a+m+1}{m+1}}+...+{\binom {a+n}{n}}={\binom {a+n+1}{n}}}$
• ${\displaystyle F_{n}=\sum _{i=0}^{\infty }{\binom {n-i}{i}}}$
${\displaystyle F_{n-1}+F_{n}=\sum _{i=0}^{\infty }{\binom {n-1-i}{i}}+\sum _{i=0}^{\infty }{\binom {n-i}{i}}=1+\sum _{i=1}^{\infty }{\binom {n-i}{i-1}}+\sum _{i=1}^{\infty }{\binom {n-i}{i}}=1+\sum _{i=1}^{\infty }{\binom {n+1-i}{i}}=\sum _{i=0}^{\infty }{\binom {n+1-i}{i}}=F_{n+1}}$
• ${\displaystyle \sum _{i=m}^{n}{\binom {i}{a}}={\binom {n+1}{a+1}}-{\binom {m}{a+1}}}$
${\displaystyle {\binom {m}{a+1}}+{\binom {m}{a}}+{\binom {m+1}{a}}...+{\binom {n}{a}}={\binom {n+1}{a+1}}}$

二阶求和公式

• ${\displaystyle \sum _{r=0}^{n}{\binom {n}{r}}^{2}={\binom {2n}{n}}}$
• ${\displaystyle \sum _{i=0}^{n}{\binom {r_{1}+n-1-i}{r_{1}-1}}{\binom {r_{2}+i-1}{r_{2}-1}}={\binom {r_{1}+r_{2}+n-1}{r_{1}+r_{2}-1}}}$
${\displaystyle (1-x)^{-r_{1}}(1-x)^{-r_{2}}=(1-x)^{-r_{1}-r_{2}}}$
${\displaystyle (1-x)^{-r_{1}}(1-x)^{-r_{2}}=(\sum _{n=0}^{\infty }{\binom {r_{1}+n-1}{r_{1}-1}}x^{n})(\sum _{n=0}^{\infty }{\binom {r_{2}+n-1}{r_{2}-1}}x^{n})=\sum _{n=0}^{\infty }(\sum _{i=0}^{n}{\binom {r_{1}+n-1-i}{r_{1}-1}}{\binom {r_{2}+i-1}{r_{2}-1}})x^{n}}$
${\displaystyle (1-x)^{-r_{1}-r_{2}}=\sum _{n=0}^{\infty }{\binom {r_{1}+r_{2}+n-1}{r_{1}+r_{2}-1}}x^{n}}$
• ${\displaystyle \sum _{i=0}^{k}{\binom {n}{i}}{\binom {m}{k-i}}={\binom {n+m}{k}}}$

三阶求和公式

• ${\displaystyle {\binom {n+k}{k}}^{2}=\sum _{j=0}^{k}{\binom {k}{j}}^{2}{\binom {n+2k-j}{2k}}}$

备注

1. ^
2. ^ Lilavati 第6节，第4章（见Knuth (1997)）。
3. ^
4. ^ 见（Graham，Knuth & Patashnik 1994），其中就${\displaystyle k<0}$定义了${\displaystyle {\tbinom {n}{k}}=0}$，其他一般化形式包括考虑两参数为实数或复数时以伽玛函数${\displaystyle k<0}$时定义${\displaystyle {\tbinom {n}{k}}}$，但此举会令大部分二项式系数的恒等式失效，故未有被广泛采用。然而，此方法替不等于零的参数下定义则可得出如Hilton, Holton and Pedersen, Mathematical reflections: in a room with many mirrors, Springer, 1997中那种好看的“帕斯卡风车”，但是却会令帕斯卡法则在原点失效。
5. ^ 此可视作泰勒定理的离散形式，亦与牛顿多项式有关，此式的交错项之和可以Nörlund–Rice积分表示。