今天学习组合数
定理1:上指标求和
\[\sum^n_{i=0}\dbinom{m+i}{i}=\dbinom{m+n+1}{m+1}
\]
证明
\[\begin{aligned}
\sum^n_{i=0}\dbinom{m+i}{i}&=\sum^n_{i=0}\dbinom{m+i}{m}\\
&=\dbinom{m}{m}+\dbinom{m+1}{m}+\dbinom{m+2}{m}+\dots+\dbinom{m+n}{m}\\
&=\dbinom{m+1}{m+1}+\dbinom{m+1}{m}+\dbinom{m+2}{m}+\dots+\dbinom{m+n}{m}\\
&=\dbinom{m+2}{m+1}+\dbinom{m+2}{m}+\dots+\dbinom{m+n}{m}\\
&\dots\\
&=\dbinom{m+n+1}{m+1}
\end{aligned}
\]
定理2
\[\sum_{i=0}^{n} i\dbinom{n}{i}=n2^{n-1}
\]
证明
\[\begin{aligned}
\sum_{i=0}^{n} i\dbinom{n}{i} &= 0\dbinom{n}{0}+1\dbinom{n}{1}+\dots+(n-1)\dbinom{n}{n-1}+n\dbinom{n}{n}\\
&= 0\dbinom{n}{0}+1\dbinom{n}{1}+\dots+(n-1)\dbinom{n}{1}+n\dbinom{n}{0}\\
&= \frac{n}{2}\dbinom{n}{0} +\frac{n}{2}\dbinom{n}{1}+\dots+\frac{n}{2}\dbinom{n}{1}+\frac{n}{2}\dbinom{n}{0}\\
&= \frac{n}{2}\dbinom{n}{0} +\frac{n}{2}\dbinom{n}{1}+\dots+\frac{n}{2}\dbinom{n}{n-1}+\frac{n}{2}\dbinom{n}{n}\\
&= \sum_{i=0}^n \frac{n}{2}\dbinom{n}{i} = \frac{n}{2} \sum_{i=0}^n \dbinom{n}{i}\\
&= \frac{n}{2}2^n = n2^{n-1}
\end{aligned}
\]
定理3
\[\dbinom{n}{m}:\dbinom{n-1}{m-1}:\dbinom{n-1}{m}=n:m:n-m
\]
证明
\[\begin{aligned}
\dbinom{n}{m}&=\dfrac{n^\underline{m}}{m^\underline{m}}\\
\dbinom{n-1}{m-1}&=\dfrac{(n-1)^\underline{m-1}}{(m-1)^\underline{m-1}}\\
\dbinom{n-1}{m}&=\dfrac{(n-1)^\underline{m}}{m^\underline{m}}\\
\dbinom{n-1}{m-1}:\dbinom{n-1}{m}&=\dfrac{(n-1)^\underline{m-1}}{(m-1)^\underline{m-1}}:\dfrac{(n-1)^\underline{m}}{m^\underline{m}}\\
&=\dfrac{m^\underline{m}}{(m-1)^\underline{m-1}}:\dfrac{(n-1)^\underline{m}}{(n-1)^\underline{m-1}}\\
&=m:n-m\\
\dbinom{n}{m}:\dbinom{n-1}{m-1}:\dbinom{n-1}{m}&=\dbinom{n-1}{m-1}+\dbinom{n-1}{m}:\dbinom{n-1}{m-1}:\dbinom{n-1}{m}\\
&=n:m:n-m
\end{aligned}
\]
推论
\[\dfrac{\dbinom{n-1}{m-1}}{m}=\dfrac{\dbinom{n}{m}}{n}
\]