具体数学之二项式系数1

本章讲述的是二项式系数,包含了一大堆记不住的公式@<@

1. ( r k ) = { r ( r 1 ) ( r k + 1 ) k ( k 1 ) ( 1 ) = r k k ! , k 0 0 , k &lt; 0 \left( \begin{array}{l}{r} \\ {k}\end{array}\right)=\left\{\begin{array}{l}{\frac{r(r-1) \cdots(r-k+1)}{k(k-1) \cdots(1)}=\frac{r^{k}}{k ! } , k \geqslant 0} \\ {0}, k&lt;0\end{array}\right.
当k=0时,上述结果为1
r为上指标,k为下指标,表示从r个数里面取k个的排序

2.帕斯卡三角形(杨辉三角形)
在这里插入图片描述
( r 0 ) = 1 , ( r 1 ) = r , ( r 2 ) = r ( r 1 ) 2 \left( \begin{array}{l}{r} \\ {0}\end{array}\right)=1, \left( \begin{array}{l}{r} \\ {1}\end{array}\right)=r, \left( \begin{array}{l}{r} \\ {2}\end{array}\right)=\frac{r(r-1)}{2}


☆ 考虑式子 ( 1 k ) = ? ( 1 1 k ) \left( \begin{array}{c}{-1} \\ {k}\end{array}\right) \stackrel{?}{=} \left( \begin{array}{c}{-1} \\ {-1-k}\end{array}\right)

( 1 k ) = ( 1 ) ( 2 ) ( k ) k ! = ( 1 ) k \left( \begin{array}{c}{-1} \\ {k}\end{array}\right)=\frac{(-1)(-2) \cdots(-k)}{k !}=(-1)^{k}

( 1 1 k ) = ( 1 ) 1 k \left( \begin{array}{c}{-1} \\ {-1-k}\end{array}\right)=(-1)^{-1-k} (由上式可以得出),它是1或者-1
因此上述等式总是相等是错误的!

3.吸收等式(5.5)

( r k ) = r k ( r 1 k 1 ) \left( \begin{array}{l}{r} \\ {k}\end{array}\right)=\frac{r}{k} \left( \begin{array}{l}{r-1} \\ {k-1}\end{array}\right) k 0 整数 k \neq 0

4.相伴恒等式(5.7)

( r k ) ( r k ) = r ( r 1 k ) (r-k) \left( \begin{array}{l}{r} \\ {k}\end{array}\right)=r \left( \begin{array}{c}{r-1} \\ {k}\end{array}\right)

( r k ) ( r k ) = ( r k ) ( r r k ) = r ( r 1 r k 1 ) = r ( r 1 k ) \begin{aligned}(r-k) \left( \begin{array}{c}{r} \\ {k}\end{array}\right) &amp;=(r-k) \left( \begin{array}{c}{r} \\ {r-k}\end{array}\right) ,对称性\\ &amp;=r \left( \begin{array}{c}{r-1} \\ {r-k-1}\end{array}\right),吸收等式 \\ &amp;=r \left( \begin{array}{c}{r-1} \\ {k}\end{array}\right) ,对称性\end{aligned}

5.加法公式(杨辉三角的性质)(5.8)

( r k ) = ( r 1 k ) + ( r 1 k 1 ) \left( \begin{array}{l}{r} \\ {k}\end{array}\right)=\left( \begin{array}{c}{r-1} \\ {k}\end{array}\right)+\left( \begin{array}{l}{r-1} \\ {k-1}\end{array}\right) ,k是整数

利用定义证明:
k n ( r + k k ) = ( r 0 ) + ( r + 1 1 ) + + ( r + n n ) \sum_{k \leqslant n} \left( \begin{array}{c}{r+k} \\ {k}\end{array}\right)=\left( \begin{array}{l}{r} \\ {0}\end{array}\right)+\left( \begin{array}{c}{r+1} \\ {1}\end{array}\right)+\cdots+\left( \begin{array}{c}{r+n} \\ {n}\end{array}\right) = ( r + n + 1 n ) =\left( \begin{array}{c}{r+n+1} \\ {n}\end{array}\right) ,n是整数

6.关于上指标求和

0 k n ( k m ) = ( 0 m ) + ( 1 m ) + + ( n m ) \sum_{0 \leqslant k \leqslant n} \left( \begin{array}{l}{k} \\ {m}\end{array}\right)=\left( \begin{array}{l}{0} \\ {m}\end{array}\right)+\left( \begin{array}{l}{1} \\ {m}\end{array}\right)+\cdots+\left( \begin{array}{l}{n} \\ {m}\end{array}\right) = ( n + 1 m + 1 ) =\left( \begin{array}{l}{n+1} \\ {m+1}\end{array}\right) ,整数 m , n 0 m, n \geqslant 0

k n ( m + k k ) = m k n ( m + k k ) = m k n ( m + k m ) = 0 k m + n ( k m ) = ( m + n + 1 m + 1 ) = ( m + n + 1 n ) \begin{aligned} \sum_{k \leq n} \left( \begin{array}{c}{m+k} \\ {k}\end{array}\right) &amp;=\sum_{-m \leq k \leqslant n} \left( \begin{array}{c}{m+k} \\ {k}\end{array}\right) \\ &amp;=\sum_{-m \leq k \leqslant n} \left( \begin{array}{c}{m+k} \\ {m}\end{array}\right) \\ &amp;=\sum_{0 \leqslant k \leqslant m+n} \left( \begin{array}{c}{k} \\ {m}\end{array}\right) \\ &amp;=\left( \begin{array}{c}{m+n+1} \\ {m+1}\end{array}\right)=\left( \begin{array}{c}{m+n+1} \\ {n}\end{array}\right) \end{aligned}

7.上指标反转(5.14)

( r k ) = ( 1 ) k ( k r 1 k ) \left( \begin{array}{l}{r} \\ {k}\end{array}\right)=(-1)^{k} \left( \begin{array}{c}{k-r-1} \\ {k}\end{array}\right) ,k是整数

8. ( 1 ) m ( n 1 m ) = ( 1 ) n ( m 1 n ) (-1)^{m} \left( \begin{array}{c}{-n-1} \\ {m}\end{array}\right)=(-1)^{n} \left( \begin{array}{c}{-m-1} \\ {n}\end{array}\right) = ( m + n n ) =\left( \begin{array}{c}{m+n} \\ {n}\end{array}\right) ,整数 m , n 0 m, n \geqslant 0

利用上指标公式,也可以推导出下列式子(帕斯卡三角形一行的部分交替求和):
k m ( r k ) ( 1 ) k = ( r 0 ) ( r 1 ) + + ( 1 ) m ( r m ) \sum_{k≤m} \left( \begin{array}{l}{r} \\ {k}\end{array}\right)(-1)^{k}=\left( \begin{array}{l}{r} \\ {0}\end{array}\right)-\left( \begin{array}{l}{r} \\ {1}\end{array}\right)+\cdots+(-1)^{m} \left( \begin{array}{l}{r} \\ {m}\end{array}\right) = ( 1 ) m ( r 1 m ) , m =(-1)^{m} \left( \begin{array}{c}{r-1} \\ {m}\end{array}\right), \quad m是整数

k m ( r k ) ( 1 ) k = k m ( k r 1 k ) = ( r + m m ) = ( 1 ) m ( r 1 m ) \begin{aligned} \sum_{k \leqslant m} \left( \begin{array}{c}{r} \\ {k}\end{array}\right)(-1)^{k}=\sum_{k \leqslant m} \left( \begin{array}{c}{k-r-1} \\ {k}\end{array}\right) \\ =\left( \begin{array}{c}{-r+m} \\ {m}\end{array}\right) \\=(-1)^{m} \left( \begin{array}{c}{r-1} \\ {m}\end{array}\right) \end{aligned}

发布了129 篇原创文章 · 获赞 105 · 访问量 23万+

猜你喜欢

转载自blog.csdn.net/sunshine_lyn/article/details/90049010