具體數學之二項式係數2

又是一大堆公式來襲~~

二項級數的部分和另一種有意思的關係式子：
$\sum_{k \leqslant m} \left( \begin{array}{c}{m+r} \\ {k}\end{array}\right) x^{k} y^{m-k}=\sum_{k \leq m} \left( \begin{array}{c}{-r} \\ {k}\end{array}\right)(-x)^{k}(x+y)^{m-k}, \quad m是整數$
當m小於0時，兩邊均爲0；當m=0時，兩邊都是1。
令左邊部分爲Sm，右邊部分爲Tm

$S_{m}=\sum_{k \leq m} \left( \begin{array}{c}{m-1+r} \\ {k}\end{array}\right) x^{k} y^{m-k}+\sum_{k \leqslant m} \left( \begin{array}{c}{m-1+r} \\ {k-1}\end{array}\right) x^{k} y^{m-k}$
其中，
$\sum_{k \leqslant m} \left( \begin{array}{c}{m-1+r} \\ {k}\end{array}\right) x^{k} y^{m-k}=\left( \begin{array}{c}{m-1+r} \\ {m}\end{array}\right) x^{m}+ \sum_{k \leqslant m-1} \left( \begin{array}{c}{m+1+r} \\ {k}\end{array}\right) x^{k} y^{m-k}=y S_{m-1}+\left( \begin{array}{c}{m-1+r} \\ {m}\end{array}\right) x^{m}$

$\sum_{k \leq m} \left( \begin{array}{c}{m-1+r} \\ {k-1}\end{array}\right) x^{k} y^{m-k}=x\sum_{k \leq m} \left( \begin{array}{c}{m-1+r} \\ {k-1}\end{array}\right) x^{k-1} y^{(m-1)-(k-1)}=x S_{m-1}$

因此，$S_{m}=(x+y) S_{m-1}+\left( \begin{array}{c}{-r} \\ {m}\end{array}\right)(-x)^{m}$

因此，可證明Sm和Tm是相等的！

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$\sum_{k \leqslant m} \left( \begin{array}{c}{m+r} \\ {k}\end{array}\right)(-1)^{k}=\left( \begin{array}{c}{-r} \\ {m}\end{array}\right)$，整數 $m \geqslant 0$
相當於（5.16）

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$\left( \begin{array}{c}{r} \\ {m}\end{array}\right) \left( \begin{array}{l}{m} \\ {k}\end{array}\right)=\left( \begin{array}{l}{r} \\ {k}\end{array}\right) \left( \begin{array}{l}{r-k} \\ {m-k}\end{array}\right), \quad m, k是整數$

$\begin{aligned} \left( \begin{array}{c}{r} \\ {m}\end{array}\right) \left( \begin{array}{l}{m} \\ {k}\end{array}\right) &=\frac{r !}{m !(r-m) !} \frac{m !}{k !(m-k) !} \\ &=\frac{r !}{k !(m-k) !(r-m) !} \\ &=\frac{r !}{k !(r-k) !} \frac{(r-k) !}{k !(r-k) !} \frac{(r-k) !}{(m-k) !(r-m) !}=\left( \begin{array}{l}{r} \\ {k}\end{array}\right) \left( \begin{array}{l}{r-k} \\ {m-k}\end{array}\right) \end{aligned}$

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$\left( \begin{array}{c}{a+b+c} \\ {a, b, c}\end{array}\right)=\frac{(a+b+c) !}{a ! b ! c !}$

多項式係數：

$\left( \begin{array}{c}{a_{1}+a_{2}+\cdots+a_{m}} \\ {a_{1}, a_{2}, \cdots, a_{m}}\end{array}\right)=\frac{\left(a_{1}+a_{2}+\cdots+a_{m}\right) !}{a_{1} ! a_{2} ! \cdots a_{m} !}$$=\left( \begin{array}{c}{a_{1}+a_{2}+\cdots+a_{m} !} \\ {a_{2}+\cdots+a_{m}}\end{array}\right) \cdot \left( \begin{array}{c}{a_{m-1}+a_{m}} \\ {a_{m}}\end{array}\right)$

例子：$\sum_{k} \left( \begin{array}{c}{r} \\ {m+k}\end{array}\right) \left( \begin{array}{c}{s} \\ {n-k}\end{array}\right)$
令$k_{1}=m+k$
=$\sum_{k_{1}-m} \left( \begin{array}{c}{r} \\ {k_{1}}\end{array}\right) \left( \begin{array}{c}{s} \\ {n+m+k1}\end{array}\right)$=$\sum_{k1-m} \left( \begin{array}{l}{r} \\ {k_{1}}\end{array}\right) \left( \begin{array}{c}{S} \\ {n-(k1-m)}\end{array}\right)$

令$K_{2}=n-k$
$\sum_{n-k_{2}} \left( \begin{array}{c}{r} \\ {m+n-k_{2}}\end{array}\right) \left( \begin{array}{l}{S} \\ {k_{2}}\end{array}\right)$

上述表中的式子根本記不住的，其實就是範德蒙德卷積：

$\sum_{k} \left( \begin{array}{c}{r} \\ {k}\end{array}\right) \left( \begin{array}{c}{s} \\ {n-k}\end{array}\right)=\left( \begin{array}{c}{r+s} \\ {n}\end{array}\right), \quad n是整數$

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