今天学习组合数

定理1:上指标求和

\sum_{i = 0}^{n} (\binom{m + i}{i}) = (\binom{m + n + 1}{m + 1})

$\sum^n_{i=0}\dbinom{m+i}{i}=\dbinom{m+n+1}{m+1}$

证明

\begin{aligned} \sum_{i = 0}^{n} (\binom{m + i}{i}) & = \sum_{i = 0}^{n} (\binom{m + i}{m}) \\ = (\binom{m}{m}) + (\binom{m + 1}{m}) + (\binom{m + 2}{m}) + \dots + (\binom{m + n}{m}) \\ = (\binom{m + 1}{m + 1}) + (\binom{m + 1}{m}) + (\binom{m + 2}{m}) + \dots + (\binom{m + n}{m}) \\ = (\binom{m + 2}{m + 1}) + (\binom{m + 2}{m}) + \dots + (\binom{m + n}{m}) \\ \dots \\ = (\binom{m + n + 1}{m + 1}) \end{aligned}

$\begin{aligned} \sum^n_{i=0}\dbinom{m+i}{i}&=\sum^n_{i=0}\dbinom{m+i}{m}\\ &=\dbinom{m}{m}+\dbinom{m+1}{m}+\dbinom{m+2}{m}+\dots+\dbinom{m+n}{m}\\ &=\dbinom{m+1}{m+1}+\dbinom{m+1}{m}+\dbinom{m+2}{m}+\dots+\dbinom{m+n}{m}\\ &=\dbinom{m+2}{m+1}+\dbinom{m+2}{m}+\dots+\dbinom{m+n}{m}\\ &\dots\\ &=\dbinom{m+n+1}{m+1} \end{aligned}$

定理2

\sum_{i = 0}^{n} i (\binom{n}{i}) = n 2^{n - 1}

$\sum_{i=0}^{n} i\dbinom{n}{i}=n2^{n-1}$

证明

\begin{aligned} \sum_{i = 0}^{n} i (\binom{n}{i}) & = 0 (\binom{n}{0}) + 1 (\binom{n}{1}) + \dots + (n - 1) (\binom{n}{n - 1}) + n (\binom{n}{n}) \\ = 0 (\binom{n}{0}) + 1 (\binom{n}{1}) + \dots + (n - 1) (\binom{n}{1}) + n (\binom{n}{0}) \\ = \frac{n}{2} (\binom{n}{0}) + \frac{n}{2} (\binom{n}{1}) + \dots + \frac{n}{2} (\binom{n}{1}) + \frac{n}{2} (\binom{n}{0}) \\ = \frac{n}{2} (\binom{n}{0}) + \frac{n}{2} (\binom{n}{1}) + \dots + \frac{n}{2} (\binom{n}{n - 1}) + \frac{n}{2} (\binom{n}{n}) \\ = \sum_{i = 0}^{n} \frac{n}{2} (\binom{n}{i}) = \frac{n}{2} \sum_{i = 0}^{n} (\binom{n}{i}) \\ = \frac{n}{2} 2^{n} = n 2^{n - 1} \end{aligned}

$\begin{aligned} \sum_{i=0}^{n} i\dbinom{n}{i} &= 0\dbinom{n}{0}+1\dbinom{n}{1}+\dots+(n-1)\dbinom{n}{n-1}+n\dbinom{n}{n}\\ &= 0\dbinom{n}{0}+1\dbinom{n}{1}+\dots+(n-1)\dbinom{n}{1}+n\dbinom{n}{0}\\ &= \frac{n}{2}\dbinom{n}{0} +\frac{n}{2}\dbinom{n}{1}+\dots+\frac{n}{2}\dbinom{n}{1}+\frac{n}{2}\dbinom{n}{0}\\ &= \frac{n}{2}\dbinom{n}{0} +\frac{n}{2}\dbinom{n}{1}+\dots+\frac{n}{2}\dbinom{n}{n-1}+\frac{n}{2}\dbinom{n}{n}\\ &= \sum_{i=0}^n \frac{n}{2}\dbinom{n}{i} = \frac{n}{2} \sum_{i=0}^n \dbinom{n}{i}\\ &= \frac{n}{2}2^n = n2^{n-1} \end{aligned}$

定理3

(\binom{n}{m}) : (\binom{n - 1}{m - 1}) : (\binom{n - 1}{m}) = n : m : n - m

$\dbinom{n}{m}:\dbinom{n-1}{m-1}:\dbinom{n-1}{m}=n:m:n-m$

证明

\begin{aligned} (\binom{n}{m}) & = \frac{n^{\underline{m}}}{m^{\underline{m}}} \\ (\binom{n - 1}{m - 1}) & = \frac{(n - 1)^{\underline{m - 1}}}{(m - 1)^{\underline{m - 1}}} \\ (\binom{n - 1}{m}) & = \frac{(n - 1)^{\underline{m}}}{m^{\underline{m}}} \\ (\binom{n - 1}{m - 1}) : (\binom{n - 1}{m}) & = \frac{(n - 1)^{\underline{m - 1}}}{(m - 1)^{\underline{m - 1}}} : \frac{(n - 1)^{\underline{m}}}{m^{\underline{m}}} \\ = \frac{m^{\underline{m}}}{(m - 1)^{\underline{m - 1}}} : \frac{(n - 1)^{\underline{m}}}{(n - 1)^{\underline{m - 1}}} \\ = m : n - m \\ (\binom{n}{m}) : (\binom{n - 1}{m - 1}) : (\binom{n - 1}{m}) & = (\binom{n - 1}{m - 1}) + (\binom{n - 1}{m}) : (\binom{n - 1}{m - 1}) : (\binom{n - 1}{m}) \\ = n : m : n - m \end{aligned}

$\begin{aligned} \dbinom{n}{m}&=\dfrac{n^\underline{m}}{m^\underline{m}}\\ \dbinom{n-1}{m-1}&=\dfrac{(n-1)^\underline{m-1}}{(m-1)^\underline{m-1}}\\ \dbinom{n-1}{m}&=\dfrac{(n-1)^\underline{m}}{m^\underline{m}}\\ \dbinom{n-1}{m-1}:\dbinom{n-1}{m}&=\dfrac{(n-1)^\underline{m-1}}{(m-1)^\underline{m-1}}:\dfrac{(n-1)^\underline{m}}{m^\underline{m}}\\ &=\dfrac{m^\underline{m}}{(m-1)^\underline{m-1}}:\dfrac{(n-1)^\underline{m}}{(n-1)^\underline{m-1}}\\ &=m:n-m\\ \dbinom{n}{m}:\dbinom{n-1}{m-1}:\dbinom{n-1}{m}&=\dbinom{n-1}{m-1}+\dbinom{n-1}{m}:\dbinom{n-1}{m-1}:\dbinom{n-1}{m}\\ &=n:m:n-m \end{aligned}$

推论

\frac{(\binom{n - 1}{m - 1})}{m} = \frac{(\binom{n}{m})}{n}

$\dfrac{\dbinom{n-1}{m-1}}{m}=\dfrac{\dbinom{n}{m}}{n}$

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本文作者：ASnown

本文链接：https://www.cnblogs.com/As-Snow/p/17485820.html

posted @ 2023-06-16 16:09 ASnown 阅读(36) 评论(1) 编辑收藏举报

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As-Snow

今天学习组合数

定理1:上指标求和

证明

定理2

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定理3

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