Summary of correlation of two-polynomial coefficients

Note: There are no special instructions, the two-item indicator k is an integer (without special instructions, K can take all integers), for the above indicator, R represents all numbers (integers and real numbers), and n is an integer.

1, combinatorial number definition: $\begin{pmatrix}r\\ k\end{pmatrix}=\left\{\begin{matrix}\frac{r (r-1) ... (r-k+1)} {k (k-1) ... (2) (1)}; K\geq 0 \ 0; K<0 \end{matrix}\right.$ here k is an integer, R can be an integer or a real number.

2, $\begin{pmatrix}n\\ k\end{pmatrix}=\frac{n!} K! (N-K)!} =\frac{n!} {(n (n-k))! (N-K)!} =\begin{pmatrix}n\\ n-k\end{pmatrix}$ here n,k is an integer, and $n\geq 0$. Note that there is a limit to n here. K does not matter, if K is less than 0, both sides are 0.

3, $k \begin{pmatrix}r\\ k\end{pmatrix}=r\begin{pmatrix}r-1\\ k-1\end{pmatrix}$,$ (r-k) \begin{pmatrix}r\\ k\end{ Pmatrix}=r\begin{pmatrix}r-1\\ k\end{pmatrix}$. This is true of all the numbers, and it is understood that for the first equation, both sides are k-th polynomial of R, and when R takes any integer, the polynomial on both sides is equal. Two k-th polynomial has an infinite number of points equal, so even a polynomial identity. A second explanation is similar.

4. Addition formula: $\begin{pmatrix}r\\ k\end{pmatrix}=\begin{pmatrix}r-1\\ k\end{pmatrix}+\begin{pmatrix}r-1\\ k-1\end{ pmatrix}$.

5, $\begin{pmatrix}r+n+1\\ n\end{pmatrix}=\begin{pmatrix}r+n\\ n\end{pmatrix}+\begin{pmatrix}r+n\\ N-1\end{pmatrix } $,$\begin{pmatrix}r+n\\ n-1\end{pmatrix}=\begin{pmatrix}r+n-1\\ n-1\end{pmatrix}+\begin{pmatrix}r+n-1\\ n-2\end{ pmatrix}$, in this way each time to expand the back of this item, and finally get $\begin{pmatrix}r+n+1\\ n\end{pmatrix}=\sum_{k=0}^{n}\begin{pmatrix}r+k\\ k\end{ Pmatrix}=\sum _{k\leq n}\begin{pmatrix}r+k\\ K\end{pmatrix} $. Similarly, if each of the preceding items is expanded, the last one can be obtained: $\begin{pmatrix}n+1\\ m+1\end{pmatrix}=\sum_{k=0}^{n}\begin{pmatrix}k\\ m\end{pmatrix}$

6, $\begin{pmatrix}r\\ k\end{pmatrix}= ( -1) ^{k}\begin{pmatrix}k-r-1\\ k\end{pmatrix}$

7, $\sum _{k\leq m}\begin{pmatrix}r\\ K\end{pmatrix} ( -1) ^{k}=\sum _{k\leq m}\begin{pmatrix}k-r-1\\ k\end{pmatrix}=\ begin{pmatrix}-r+m\\ m\end{pmatrix}= ( -1) ^{k}\begin{pmatrix}r-1\\ m\end{pmatrix}$ The first third equal sign runs Equation 6, and the second equals sign uses the first formula in 5.

8, $\sum _{k\leq m}\begin{pmatrix}r\\ K\end{pmatrix} (\frac{r}{2}-k) =\sum_{k=0}^{m}\begin{pmatrix}r\\ K\end{pmatrix} (\FRAC{R}{2}-K) $
$=\sum_{k=0}^{m} (\begin{pmatrix}r\\ k\end{pmatrix}\frac{r}{2}-k\begin{pmatrix}r\\ K\end{pmatrix}) =\sum_{k=0}^{m} (\begin{pmatrix}r\\ k\end{pmatrix}\frac{r}{2}-r\begin{pmatrix}r-1\\ K-1\end{pmatrix}) $
$=\frac{r}{2}\sum_{k=0}^{m} (\begin{pmatrix}r\\ k\end{pmatrix}-2\begin{pmatrix}r-1\\ K-1\end{pmatrix}) $
$=\frac{r}{2}\sum_{k=0}^{m} (\begin{pmatrix}r-1\\ k\end{pmatrix}-\begin{pmatrix}r-1\\ K-1\end{pmatrix}) $
$=\frac{r}{2}\begin{pmatrix}r-1\\ m\end{pmatrix}=\frac{m+1}{2}\begin{pmatrix}r\\ m+1\end{pmatrix}$ The third equals sign uses the first of Equation 3, the fifth equals to equation four, and the last equals to the first of equation three.

9, $ (x+y) ^{n}=\sum_{k=0}^{n}\begin{pmatrix}n\\ k\end{pmatrix}x^{k}y^{n-k}=\sum_{k}\begin{pmatrix}n\\ k\end{ pmatrix}x^{k}y^{n-k}$, where n is a non-negative integer.

10, $\sum_{k\leq m}\begin{pmatrix}m+r\\ k\end{pmatrix}x^{k}y^{m-k}=\sum_{k\leq m}\begin{pmatrix}-r\\ K\end{pmatrix} (-X) ^{k} (X+y) ^{m-k}$,m is an integer.
Proof: First, when M is negative, both sides are obviously 0. So I just need to prove that M is a non-negative integer case. Here r can take all the numbers. We assume that R is an integer that satisfies the $-m\leq r\leq 0$, $r +m\geq 0$
$ (x+y) ^{m+r}y^{-r}=y^{-r}\sum_{k=0}^{m+r}\begin{pmatrix}m+r\\ k\end{pmatrix}x^{k}y^{m+r-k}=\sum_{k=0}^{m+r}\ Begin{pmatrix}m+r\\ k\end{pmatrix}x^{k}y^{m-k}$= left (by Equation 9 only $r+m\geq 0$ can be expanded)
$ (x+y) ^{m+r}y^{-r}= (x+y) ^{m+r} (-x+x+y) ^{-r}= (x+y) ^{m+r}\sum_{k=0}^{-r}\begin{pmatrix}-r\\ K\end{pmatrix} (-X) ^{k } (X+y) ^{-r-k}$
$=\sum_{k=0}^{-r}\begin{pmatrix}-r\\ K\end{pmatrix} (-X) ^{k} (x+y) ^{m-k}$= right
All when R takes 0,-1,-2,..,-m the left and right sides are equal, and both sides are the M-polynomial of R, but there are at least m+1 points equal, so the two sides are identical.

11, in Formula 10, make X=-1,y=1, can get $\sum _{k\leq m}\begin{pmatrix}m+r\\ K\end{pmatrix} ( -1) ^{k}=\begin{pmatrix}-r\\ m\end{ pmatrix}$. If you take x=y=1,r=m+1: $\sum _{k\leq m}\begin{pmatrix}2m+1\\ k\end{pmatrix}=\sum _{k\leq m}\begin{pmatrix}m+k\\ k\end{pmatrix}2^{m-k}$, because $\sum _{k}\begin{pmatrix}2m+1\\ k\end{pmatrix}=2^{2m+1}$, so $\sum _{k\leq M}\begin{pmatrix }2m+1\\ k\end{pmatrix}=2^{2m}$, so $\sum _{k\leq m}\begin{pmatrix}m+k\\ k\end{pmatrix}2^{-k}=2^{m}$

12, $\begin{pmatrix}r\\ m\end{pmatrix}\begin{pmatrix}m\\ k\end{pmatrix}=\begin{pmatrix}r\\ k\end{pmatrix}\begin{ Pmatrix}r-k\\ m-k\end{pmatrix}$, this can be proved right by defining the direct expansion.

13, $\sum _{k}\begin{pmatrix}r\\ m+k\end{pmatrix}\begin{pmatrix}s\\ n-k\end{pmatrix}=\begin{pmatrix}r+s\\ m+n\end{ Pmatrix}$,m,n is an integer.
Proof: replace n with k-m instead of k,n-m, can get: $\sum _{k}\begin{pmatrix}r\\ k\end{pmatrix}\begin{pmatrix}s\\ n-k\end{pmatrix}=\begin{ Pmatrix}r+s\\ n\end{pmatrix}$, where n is the original n+m. This formula can be understood from the meaning, it means that from the R girls Choose K Girls, from S boys choose N-k Boys all the situation, is from R+s medium select N.

14, $\sum _{k}\begin{pmatrix}l\\ m+k\end{pmatrix}\begin{pmatrix}s\\ n+k\end{pmatrix}=\begin{pmatrix}l+s\\ l-m+n\end {pmatrix}$, $l \geq 0$,m,n is an integer.
Proof: The formula 2 is available: $\begin{pmatrix}l\\ m+k\end{pmatrix}=\begin{pmatrix}l\\ l-m-k\end{pmatrix}$, so replace it becomes equation 13.

15, $\sum _{k}\begin{pmatrix}l\\ m+k\end{pmatrix}\begin{pmatrix}s+k\\ N\end{pmatrix} ( -1) ^{k}= ( -1) ^{l+m}\begin{ Pmatrix}s-m\\ N-l\end{pmatrix},l\geq 0$,m,n is an integer
Proof: Mathematical induction, making $f (L) =\sum _{k}\begin{pmatrix}l\\ m+k\end{pmatrix}\begin{pmatrix}s+k\\ N\end{pmatrix} ( -1) ^{k}$,l=0 in addition to k= The remainder of-m is 0, so $f (0) =\begin{pmatrix}s-m\\ N\end{pmatrix} ( -1) ^{m}$. Assume that all numbers smaller than L $p$ have $f (p) =\begin{pmatrix}s-m\\ n-p\end {Pmatrix} ( -1) ^{p+m}$.
$f (l) =\sum _{k}\begin{pmatrix}l\\ m+k\end{pmatrix}\begin{pmatrix}s+k\\ N\end{pmatrix} ( -1) ^{k}$
$=\sum _{k} (\begin{pmatrix}l-1\\ M+k\end{pmatrix} +\begin{pmatrix}l-1\\ M+k-1\end{pmatrix}) \begin{pmatrix}s+k\\ n\ End{pmatrix} ( -1) ^{k}$
$=\sum _{k}\begin{pmatrix} l-1\\ M+k\end{pmatrix} \begin{pmatrix}s+k\\ N\end{pmatrix} ( -1) ^{k}+\sum _{k} \begin{ pmatrix}l-1\\ M+k-1\end{pmatrix} \begin{pmatrix}s+k\\ N\end{pmatrix} ( -1) ^{k}$
$=f (L-1) + ( -1) ^{l+m}\begin{pmatrix}s-m+1\\ n-l+1\end{pmatrix}$
$= ( -1) ^{l+m-1}\begin{pmatrix}s-m\\ n-l+1\end{pmatrix}+ ( -1) ^{l+m}\begin{pmatrix}s-m+1\\ n-l+1\end{pmatrix}$
$= ( -1) ^{l+m} (\begin{pmatrix}s-m+1\\ n-l+1\end{pmatrix}-\begin{pmatrix}s-m\\ N-l+1\end{pmatrix}) $
$= ( -1) ^{l+m}\begin{pmatrix}s-m\\ n-l\end{pmatrix}$


16, $\sum _{k\leq l}\begin{pmatrix}l-k\\ m\end{pmatrix}\begin{pmatrix}s\\ K-n\end{pmatrix} ( -1) ^{k}= ( -1) ^{l+m}\begin {pmatrix}s-m-1\\ L-m-n\end{pmatrix},l,m,n\geq 0$,l,m,n are integers. The same is used in mathematical induction, the addition of the formula to take apart the first left.

17, $\sum _{-q\leq k\leq l}\begin{pmatrix}l-k\\ m\end{pmatrix}\begin{pmatrix}q+k\\ n\end{pmatrix}=\begin{pmatrix}l+q +1\\ M+n+1\end{pmatrix},n,m\geq 0,l+q\geq 0$, all integers.
Proof: $\sum _{-q\leq k\leq l}\begin{pmatrix}l-k\\ m\end{pmatrix}\begin{pmatrix}q+k\\ n\end{pmatrix}$
$=\sum _{0\leq K+q\leq l+q}\begin{pmatrix}l+q-(k+q) \ m\end{pmatrix}\begin{pmatrix}q+k\\ n\end{pmatrix}=\sum _{0\leq K\leq l+q}\begin{pmatrix}l+q-k\\ m\end{pmatrix}\begin{pmatrix}k\\ n\end{pmatrix}$, so now just proof: $\sum _{0\leq k\leq l+q}\ begin{pmatrix}l+q-k\\ m\end{pmatrix}\begin{pmatrix}k\\ n\end{pmatrix}=\begin{pmatrix}l+q+1\\ M+n+1\end{pmatrix},n , M\geq 0,l+q\geq 0$, make $p=l+q$, then just prove it now: $\sum _{0\leq k\leq p}\begin{pmatrix}p-k\\ m\end{pmatrix}\begin{pmatrix}k\\ n\ end{pmatrix}=\begin{pmatrix}p+1\\ M+n+1\end{pmatrix},n,m,p\geq 0$. Now, using the mathematical induction method, the first item can be opened with the addition formula.

Summary of correlation of two-polynomial coefficients