Application and Popularization of Formal Calculation

Ji Peng

Department of Electronic Information, Nanjing University, Nanjing, China.

DOI: 10.4236/oalib.1109483                

Abstract

Formal Calculation, formerly known as the Shape of Numbers, is suitable for calculating some nested sums. The formula has been obtained, and the calculation problem of various combinations of arithmetic sequences has been solved. This paper analysis the coefficients of the formulas and obtain some simplified identities. Furthermore, the Formal Calculation is extended from binomial coefficient to Gaussian coefficient, and the application is extended from two parameters forms to multiparameter forms.

Keywords

Formal Calculation, Shape of Numbers, Calculation Formula, Combinatorics, Gaussian Coefficient, Stirling Number

1. Introduction

Peng, J. has introduced the Shape of Numbers and three forms of calculation in [1]: $K_i,D_i\in \text{CommutativeRing}$.

M series: $Seri{e}_i=\left\{K_i,K_i+D_i,K_i+2D_i,\cdots ,K_i+\left(N-1\right)D_i\right\}$, $i\in \left[1,M\right]$

Use $PS=\left[K_1:D_1,\cdots ,K_M:D_M\right]$ to represent the series.

$\left[K_1:1,\cdots ,K_M:1\right]$ is abbreviated as $\left[K_1,\cdots ,K_M\right]$.

$\left[K_1:D,\cdots ,K_M:D\right]$ is abbreviated as $\left[K_1,\cdots ,K_M\right]:D$.

Use $PT=\left[T_1,T_2,\cdots ,T_M\right]$ to indicate some items in M series (the Shape).

By default, the following uses:

$PS=\left[K_1:D_1,K_2:D_2,\cdots ,K_M:D_M\right],PT=\left[T_1,T_2,\cdots ,T_M\right]$

$PSA=\left[K_1:D_1,\cdots ,K_M:D_M,K_{M+1}:D_{M+1}\right]=\left[PS,K_{M+1}:D_{M+1}\right],$

$PTA=\left[PT,T_{M+1}=T_M+2-p\right]$

Recursive definie operator ${\nabla }^P$, $p\in \mathbb{Z}$ :

${\nabla }^{0}f\left(n\right)=f\left(n\right),{\displaystyle {\sum }_{n=0}^{N-1}{\nabla }^{1}f\left(n+1\right)}=f\left(N\right),{\displaystyle {\sum }_{n=0}^{N-1}f\left(n+1\right)}={\nabla }^{-1}f(\; N\; )$

Recursive define SUN(N, PS, PT), abbreviated as SUM(N):

$SUM\left(N,\left[K_1:D_1\right],\left[T_1=1\right]\right)={\displaystyle {\sum }_{n=0}^{N-1}\left(K_1+n\times D_1\right)}$

$SUM\left(N,PSA,PTA\right)={\displaystyle {\sum }_{n=0}^{N-1}\left(K_{M+1}+n\times D_{M+1}\right)\times {\nabla }^{p}SUM\left(n+1\right)}$

For example:

$SUM\left(N,PS,\left[1,2,\cdots ,M\right]\right)={\displaystyle {\sum }_{n=0}^{N-1}{\displaystyle {\prod }_{i=1}^M\left(K_i+n{D}_i\right)}}$

$\begin{array}{l}SUM\left(N,PS,\left[1,3,\cdots ,2M-1\right]\right)\\ ={\displaystyle {\sum }_{n_M=0}^{N-1}}\left(K_M+n_M{D}_M\right)\cdots {\displaystyle {\sum }_{n_1=0}^{n_2}\left(K_2+n_2{D}_2\right)}{\displaystyle {\sum }_{n_1=0}^{n_2}\left(K_1+n_1{D}_1\right)}\end{array}$

$SUM\left(N,PS,\left[1,2,4\right]\right)={\displaystyle {\sum }_{n_3=0}^{N-1}\left(K_3+n_3{D}_3\right)}{\displaystyle {\sum }_{n=0}^{n_3}\left(K_2+n{D}_1\right)\left(K_1+n{D}_1\right)}$

$SUM\left(N,PS,\left[1,3,4\right]\right)={\displaystyle {\sum }_{n=0}^{N-1}\left(K_3+n{D}_3\right)\left(K_2+n{D}_2\right)}{\displaystyle {\sum }_{n_1=0}^n\left(K_1+n_1{D}_1\right)}$

$SUM\left(N,PS,\left[1,4\right]\right)={\displaystyle {\sum }_{n=0}^{N-1}\left(K_2+n{D}_2\right)}{\displaystyle {\sum }_{n_2=0}^n{\displaystyle {\sum }_{n_1=0}^{n_2}\left(K_1+n_1{D}_1\right)}}$

The following use K to represent set $\left\{K_1,K_2,\cdots ,K_M\right\}$, T to represent set $\left\{T_1,T_2,\cdots ,T_M\right\}$.

Use the Form: $\left(T_1+K_1\right)\left(T_2+K_2\right)\cdots \left(T_M+K_M\right)={\displaystyle \sum {\displaystyle {\prod }_{i=1}^M{X}_i}}$, $X_i=T_i$ or $K_i$

$X\left(T\right)=\text{Countof}\left\{X_1,\cdots ,X_M\right\}\in T$,

$X_{T-1}=\text{Countof}\left\{X_1,\cdots ,X_{i-1}\right\}\in T$, $X_{K-1}=\text{Countof}\left\{X_1,\cdots ,X_{i-1}\right\}\in K$.

Don’t swap the factors, then each ${\displaystyle \prod X_i}$ corresponds to one expression in the SUM().

1.1) $H\left(q\right)={\displaystyle {\sum }_{{\displaystyle \prod X_i}\text{with}X\left(T\right)=q}{\displaystyle {\prod }_{i=1}^M{B}_i}}$, $q=X\left(T\right)$, $SUM\left(N\right)=$

$\stackrel{{\text{Form}}_{\text{1}}}{\to }{\displaystyle {\sum }_{q=0}^M{H}_1\left(q\right)}\left(\begin{array}{c}N+T_M-M\\ N-1-q\end{array}\right)={\displaystyle {\sum }_{q=0}^M{H}_1\left(q\right)}\left(\begin{array}{c}N+T_M-M\\ T_M-M+1+q\end{array}\right)$

$\stackrel{{\text{Form}}_{\text{2}}}{\to }{\displaystyle {\sum }_{q=0}^M{H}_2\left(q\right)}\left(\begin{array}{c}N+T_M-M+q\\ N-1\end{array}\right)={\displaystyle {\sum }_{q=0}^M{H}_2\left(q\right)}\left(\begin{array}{c}N+T_M-M+q\\ T_M-M+1+q\end{array}\right)$

$\stackrel{{\text{Form}}_{\text{3}}}{\to }{\displaystyle {\sum }_{q=0}^M{H}_3\left(q\right)}\left(\begin{array}{c}N+T_M-q\\ N-1-q\end{array}\right)={\displaystyle {\sum }_{q=0}^M{H}_3\left(q\right)}\left(\begin{array}{c}N+T_M-q\\ T_M+1\end{array}\right)$

$\stackrel{{\text{Form}}_{\text{1}}}{\to }{B}_i=\{\begin{array}{l}\left(T_i-X_{K-1}\right)D_i;X_i=T_i\\ K_i+X_{T-1}{D}_i;X_i=K_i\end{array}$

$\stackrel{{\text{Form}}_{\text{2}}}{\to }{B}_i=\{\begin{array}{l}\left(T_i-X_{K-1}\right)D_i;X_i=T_i\\ K_i+\left(X_{K-1}-T_i\right)D_i;X_i=K_i\end{array}$

$\stackrel{{\text{Form}}_{\text{3}}}{\to }{B}_i=\{\begin{array}{l}-K_i+\left(T_i-X_{T-1}\right)D_i;X_i=T_i\\ K_i+X_{T-1}{D}_i;X_i=K_i\end{array}$

H₁(q), H₂(q), H₃(q), short for H(q, PS, PT), is also defined above.

Sometimes use H(q) to represent these three coefficients.

If $f\left(n\right)={\displaystyle \sum A_i}\left(\begin{array}{c}{N}_i\\ m_i\end{array}\right)$, $m_i$ is not changed with n, then ${\nabla }^{p}f\left(n\right)={\displaystyle \sum A_i}\left(\begin{array}{c}{N}_i-p\\ m_i-p\end{array}\right)$

Sometimes ÑSUM(N) and sometimes SUM(N) are listed below,

The corresponding SUM(N) and ÑSUM(N) are easily obtained.

In particular, S₁(), S₂() is unsigned Stirling number:

1.2) $SUM\left(N,\left[1,2,\cdots ,M\right],\left[1,3,\cdots ,2M-1\right]\right)=S_1\left(N+M,N\right)$.

1.3) $SUM\left(N,\left[1,1,\cdots ,1\right],\left[1,3,\cdots ,2M-1\right]\right)=S_2\left(N+M,N\right)$.

1.4) $SUM\left(N,\left[1,1,\cdots ,1\right],\left[1,2,\cdots ,M\right]\right)=1^M+2^M+\cdots +N^M$

1.5) $\nabla SUM\left(N,PS,\left[1,2,\cdots ,M\right]\right)={\displaystyle {\prod }_{i=1}^M\left(K_i+\left(N-1\right)D_i\right)}={\displaystyle {\prod }_{i=1}^M\left(K_i+n{D}_i\right)}$

1.6) In $SUM\left(N,\left[\cdots PS\cdots \right],\left[\cdots ,T+1,T+2,\cdots ,T+M,\cdots \right]\right)$, $K_i$ can exchange order

1.7) $SUM\left(N,\left[L_1,\cdots ,L_P,PS\right],\left[L_1,\cdots ,L_P,PT\right]\right)={\displaystyle {\prod }_{i=1}^P{L}_{i}SUM\left(N,PS,PT\right)}$

This indicates that T₁ can be greater than 1, T is defined in ℕ.

1.8) $SUM\left(N,PT,PT\right)={\displaystyle {\prod }_{i=1}^M{T}_i}\left(\begin{array}{c}N+T_M\\ T_M+1\end{array}\right)$

1.9) ${\displaystyle {\sum }_{q=0}^M{H}_1\left(q\right)}\left(\begin{array}{c}A\\ B-q\end{array}\right)={\displaystyle {\sum }_{q=0}^M{H}_2\left(q\right)}\left(\begin{array}{c}A+q\\ B\end{array}\right)={\displaystyle {\sum }_{q=0}^M{H}_3\left(q\right)}\left(\begin{array}{c}A+M-q\\ B-q\end{array}\right)$

This indicates Form₁ = Form₂ = Form₃. If regardless of the actual meaning, PT’s domain can be extended to $ℂ$.

The Shape of Numbers of [1] has nothing to do with triangle numbers, square numbers, etc. This paper calls them Formal Calculation.

2. Simplified Formula

$H\left(q\right)={\displaystyle \sum {\displaystyle \prod X}},X\in T$ or $K={\displaystyle \sum \left({\displaystyle \prod X\in T}\right)\left({\displaystyle \prod X\in K}\right)}$

Sometimes simple expressions can be obtained.

Define

$F_q^{K=\left\{K_1,K_2,\cdots \right\}}={\displaystyle \sum {\displaystyle {\prod }_{i=1}^q{I}_i}}$, $I_i\in K$ and $I_i\ne I_j$. $F_q^{\left\{1,2,\cdots ,N\right\}}$ abbreviated as $F_q^N$

$E_q^{K=\left\{K_1,K_2,\cdots \right\}}={\displaystyle \sum {\displaystyle {\prod }_{i=1}^q{I}_i}}$, $I_i\in K$, $E_q^{\left\{1,2,\cdots ,N\right\}}$ abbreviated as $E_q^N$

$F_q^N=\underset{1\le {\lambda }_1<{\lambda }_2<\cdots <{\lambda }_q\le N}{{\displaystyle \sum }}{\displaystyle \prod \lambda }=S_1\left(N+q,N\right)$

$E_q^N=\underset{1\le {\lambda }_1<{\lambda }_2<\cdots <{\lambda }_q\le N}{{\displaystyle \sum }}{\displaystyle \prod \lambda }=\underset{{\lambda }_1+{\lambda }_2+\cdots +{\lambda }_N=q}{{\displaystyle \sum }}{1}^{{\lambda }_1}{2}^{{\lambda }_2}\cdots N^{{\lambda }_N}=S_2\left(N+q,N\right)$

${\left[A:D\right]}_q=A\left(A-D\right)\left(A-2D\right)\cdots \left(A-\left(q-1\right)D\right),{\left[A:D\right]}_0=1$

${\left[A:D\right]}^q=A\left(A+D\right)\left(A+2D\right)\cdots \left(A+\left(q-1\right)D\right),{\left[A:D\right]}^0=1$

$H\left(q,T\right)=\left({\displaystyle \prod X\in T}\right)\text{of}H\left(q\right),H\left(q,K\right)=\left({\displaystyle \prod X\in K}\right)\text{of}H(\; q\; )$

$H\left(q,{\displaystyle \sum T}\right)={\displaystyle \sum H}\left(q,T\right),H\left(q,{\displaystyle \sum K}\right)={\displaystyle \sum H}\left(q,K\right)$

$PT=\left[T_1=T+1,\cdots ,T_M=T+M\right]\to H_1\left(q,T\right)=H_2\left(q,T\right)=D^q{T}_1\cdots T_q$

$PT=\left[1,2,\cdots ,M\right],D=1\to H_1\left(q,{\displaystyle \sum K}\right)=F_{M-q}^K{E}_0^q+F_{M-q-1}^K{E}_1^q+\cdots +F_0^K{E}_{M-q}^q$

$f\left(n\right)=a_0+a_{1}n+a_2{n}^2+\cdots +a_M{n}^M=a_M\left(K_1+n\right)\left(K_2+n\right)\cdots \left(K_M+n\right)$

$F\left(N\right)={\displaystyle {\sum }_{n=0}^{N-1}f\left(n\right)}=a_{M}SUM\left(N,\left[K_1,K_2,\cdots ,K_M\right],\left[1,2,\cdots ,M\right]\right)$

According to Vieta’s formulas, $F_i^K=\frac{a_{M-i}}{a_M}$

$H_1\left(q\right)=q!{\displaystyle {\sum }_{i=0}^{M-q}{F}_{M-q-i}^K{E}_i^q}=q!{\displaystyle {\sum }_{i=0}^{M-q}\frac{a_{q+i}}{a_M}{E}_i^q}$

$a_M{H}_1\left(0\right)=0!a_0$

$a_M{H}_1\left(1\right)=1!\left(a_1+a_2+\cdots +a_M\right)$

$\begin{array}{c}{a}_M{H}_1\left(2\right)=2!\left(a_2+3a_3+7a_4+\cdots \right)\\ =2!\left(S_2\left(2,2\right)a_2+S_2\left(3,2\right)a_3+S_2\left(4,2\right)a_4+\cdots \right)\end{array}$

2.1) $F\left(N\right)={\displaystyle {\sum }_{q=0}^M{H}_1\left(q\right)}\left(\begin{array}{c}N\\ q+1\end{array}\right)$

$a_M\left[\begin{array}{c}{H}_1\left(0\right)\\ ⋮\\ H_1\left(M\right)\end{array}\right]=\left[\begin{array}{c}0!\\ ⋮\\ M!\end{array}\right]\left[\begin{array}{ccc}{S}_2\left(0,0\right)& \cdots & S_2\left(M,0\right)\\ ⋮& \ddots & ⋮\\ S_2\left(0,M\right)& \cdots & S_2\left(M,M\right)\end{array}\right]\left[\begin{array}{c}{a}_0\\ ⋮\\ a_M\end{array}\right]$

It’s the same as: $F\left(N\right)={\displaystyle {\sum }_{q=0}^M{a}_q}{\displaystyle {\sum }_{n=0}^{N-1}{n}^q}={\displaystyle {\sum }_{q=0}^M{a}_q}{\displaystyle {\sum }_{i=1}^{q}i!S_2\left(q,i\right)\left(\begin{array}{c}N\\ i+1\end{array}\right)}$

2.1. PT = [T + 1, T + 2, …, T + M], PS = [P − (M − 1)D, P − (M − 2)D, …, P]:D

PS can exchange order $=\left[P,P-D,P-2D,\cdots \right]:D$

$H_1\left(q,K\right)={\left[P:D\right]}_{M-q}$,

$H_1\left(q,{\displaystyle \sum K}\right)=\left(\begin{array}{c}M\\ M-q\end{array}\right)H_1\left(q,K\right)\stackrel{D=1}{\to }\left(\begin{array}{c}M\\ q\end{array}\right)\left(\begin{array}{c}P\\ M-q\end{array}\right)\left(M-q\right)!$

2.1.1) $\left(\begin{array}{c}n+A\\ A\end{array}\right)\left(\begin{array}{c}n+M+B\\ M\end{array}\right)$

① $={\displaystyle {\sum }_{q=0}^M\left(\begin{array}{c}A+q\\ q\end{array}\right)\left(\begin{array}{c}M+B\\ M-q\end{array}\right)\left(\begin{array}{c}n+A\\ A+q\end{array}\right)}$,

② $={\displaystyle {\sum }_{q=0}^M{\left(-1\right)}^{M-q}}\left(\begin{array}{c}A-B\\ M-q\end{array}\right)\left(\begin{array}{c}A+q\\ q\end{array}\right)\left(\begin{array}{c}n+A+q\\ A+q\end{array}\right)$,

③ $={\displaystyle {\sum }_{q=0}^M\left(\begin{array}{c}A-B\\ q\end{array}\right)\left(\begin{array}{c}M+B\\ M-q\end{array}\right)\left(\begin{array}{c}n+A+M-q\\ A+M\end{array}\right)}$.

[Proof]

$\begin{array}{l}=\frac{1}{A!M!}\nabla SUM(N,\left[1,2,\cdots ,A,B+1,B+2,\cdots ,B+M\right],\\ \left[1,2,\cdots ,A,A+1,\cdots ,A+M\right])\\ =\frac{1}{M!}\nabla SUM\left(N,\left[B+1,B+2,\cdots ,B+M\right],\left[A+1,A+2,\cdots ,A+M\right]\right)\\ =\frac{1}{M!}\nabla SUM\left(N,\left[B+M,\cdots ,B+2,B+1\right],\left[A+1,A+2,\cdots ,A+M\right]\right)\end{array}$

${\text{Form}}_1=\frac{1}{M!}\nabla {\displaystyle {\sum }_{q=0}^M{H}_1\left(q\right)}\left(\begin{array}{c}N+A\\ A+1+q\end{array}\right)=\frac{1}{M!}{\displaystyle {\sum }_{q=0}^M{H}_1\left(q\right)}\left(\begin{array}{c}n+A\\ A+q\end{array}\right)$

$\begin{array}{l}{H}_1\left(q\right)\stackrel{H_1\left(q,{\displaystyle \sum K}\right)=\left(\begin{array}{c}M\\ M-q\end{array}\right)\left(\begin{array}{c}M+B\\ M-q\end{array}\right)\left(M-q\right)!,H_1\left(q,T\right)=q!\left(\begin{array}{c}A+q\\ q\end{array}\right)}{\to }\\ q!\left(\begin{array}{c}A+q\\ q\end{array}\right)\left(\begin{array}{c}M\\ q\end{array}\right)\left(\begin{array}{c}M+B\\ M-q\end{array}\right)\left(M-q\right)!\to ①\end{array}$

$\begin{array}{l}{\text{Form}}_2\stackrel{H_2\left(q,K\right)={\left[B-A\right]}^{M-q}}{\to }\frac{1}{M!}\nabla {\displaystyle {\sum }_{q=0}^M{\left[B-A\right]}^{M-q}{\left[A+1\right]}^q\left(\begin{array}{c}M\\ q\end{array}\right)\left(\begin{array}{c}N+A+q\\ A+q+1\end{array}\right)}\\ =\frac{1}{M!}{\displaystyle {\sum }_{q=0}^M{\left(-1\right)}^{M-q}}\left(\begin{array}{c}A-B\\ M-q\end{array}\right)\left(M-q\right)!\left(\begin{array}{c}A+q\\ q\end{array}\right)q!\left(\begin{array}{c}M\\ q\end{array}\right)\left(\begin{array}{c}n+A+q\\ A+q\end{array}\right)\to ②\end{array}$

$\begin{array}{l}{\text{Form}}_3\stackrel{H_3\left(q,T\right)={\left[A-B\right]}_q}{\to }\\ \frac{1}{M!}{\displaystyle {\sum }_{q=0}^M{\left[A-B\right]}_q}\left(\begin{array}{c}M\\ q\end{array}\right)\left(\begin{array}{c}M+B\\ M-q\end{array}\right)\left(M-q\right)!\left(\begin{array}{c}n+A+M-q\\ A+M\end{array}\right)\to ③\end{array}$

q.e.d.

$A=M,B=0\to {\left(\begin{array}{c}n+M\\ M\end{array}\right)}^2\stackrel{{\text{Form}}_{\text{1}}}{\to }{\displaystyle {\sum }_{q=0}^M\left(\begin{array}{c}M+q\\ q\end{array}\right)}\left(\begin{array}{c}M\\ q\end{array}\right)\left(\begin{array}{c}n+M\\ M+q\end{array}\right)$,

$A=M,B=0\to {\left(\begin{array}{c}n+M\\ M\end{array}\right)}^2\stackrel{{\text{Form}}_{\text{3}}}{\to }{\displaystyle {\sum }_{q=0}^M{\left(\begin{array}{c}M\\ q\end{array}\right)}^2}\left(\begin{array}{c}n+2M-q\\ 2M\end{array}\right)$, record at [2]: (6.32).

$B=0\to \left(\begin{array}{c}n+A\\ A\end{array}\right)\left(\begin{array}{c}n+M\\ M\end{array}\right)\stackrel{{\text{Form}}_{\text{3}}}{\to }{\displaystyle {\sum }_{q=0}^M\left(\begin{array}{c}A\\ q\end{array}\right)}\left(\begin{array}{c}M\\ q\end{array}\right)\left(\begin{array}{c}n+A+M-q\\ A+M\end{array}\right)$, record at [2]: (6.21).

$A=0\to \left(\begin{array}{c}x+y\\ M\end{array}\right)={\displaystyle {\sum }_{q=0}^M\left(\begin{array}{c}x\\ M-q\end{array}\right)(\; y\; q\; )}$

if $0<B<M$, $PS=\left[0,-1,\cdots ,-\left(B-1\right),A,A+1,\cdots ,A+\left(M-B-1\right)\right]$

$H_1\left(q\right)\ne 0\to X_1,X_2,\cdots ,X_B\in T\to H_1\left(q<B,{\displaystyle \sum K}\right)=0$

Method of 2.1.1) $\to H_1\left(q\ge B,{\displaystyle \sum K}\right)=\left(\begin{array}{c}M-B\\ M-q\end{array}\right){\left[A+\left(M-1\right)\right]}_{M-q}$.

2.1.2) $\left(\begin{array}{c}n+X\\ A\end{array}\right)\left(\begin{array}{c}n+Y\\ M\end{array}\right)={\displaystyle {\sum }_{g=0}^A\left(\begin{array}{c}M+g\\ g\end{array}\right)}\left(\begin{array}{c}M+X-Y\\ A-g\end{array}\right)\left(\begin{array}{c}n+Y\\ M+g\end{array}\right),0\le Y\le M$

[Proof]

$\begin{array}{l}=\frac{1}{A!M!}\nabla SUM\left(N,\left[X,X-1,\cdots ,X-A+1,Y,Y-1,\cdots ,Y-M+1\right],\left[1,2,\cdots \right]\right)\\ =\frac{1}{A!M!}\nabla SUM\left(N,\left[1,2,\cdots ,Y,0,-1,\cdots ,-\left(M-Y\right)+1,X,\cdots ,X-A+1\right],\left[1,2,\cdots \right]\right)\\ =\frac{Y!}{A!M!}SUM(N,\left[0,-1,\cdots ,-\left(M-Y\right)+1,X,\cdots ,X-A+1\right],\\ \left[Y+1,Y+2,\cdots ,M+A\right])\end{array}$

$H_1\left(q\ge M-Y,{\displaystyle \sum K}\right)\stackrel{Z=M+A-Y}{\to }\left(\begin{array}{c}Z-\left(M-Y\right)\\ Z-q\end{array}\right){\left[X+\left(M-Y\right)\right]}_{Z-q}$

$\begin{array}{l}\frac{Y!}{A!M!}{H}_1\left(q\right)\\ =\frac{Y!}{A!M!}\left(\begin{array}{c}A\\ M+A-Y-q\end{array}\right){\left[M+X-Y\right]}_{M+A-Y-q}{\left[Y+1\right]}^q\stackrel{-g=M-Y-q}{\to }\\ =\frac{Y!}{A!M!}\left(\begin{array}{c}A\\ A-g\end{array}\right)\left(\begin{array}{c}M+X-Y\\ A-g\end{array}\right)\left(A-g\right)!\left(\begin{array}{c}M+g\\ M-y+g\end{array}\right)\left(M-y+g\right)!\\ =\left(\begin{array}{c}M+g\\ g\end{array}\right)\left(\begin{array}{c}M+X-Y\\ A-g\end{array}\right)\end{array}$

$\begin{array}{c}\left(\begin{array}{c}n+X\\ A\end{array}\right)\left(\begin{array}{c}n+Y\\ M\end{array}\right)=\frac{Y!}{A!M!}\nabla {\displaystyle {\sum }_{q=M-Y}^Z{H}_1\left(q\right)}\left(\begin{array}{c}N+\left(M+A\right)-Z\\ \left(M+A\right)-Z+q+1\end{array}\right)\\ =\frac{Y!}{A!M!}{\displaystyle {\sum }_{g=0}^A{H}_1\left(g\right)}\left(\begin{array}{c}n+Y\\ M+g\end{array}\right)\end{array}$

q.e.d.

$M=Y,X=0\to \left(\begin{array}{c}n\\ A\end{array}\right)\left(\begin{array}{c}n+M\\ M\end{array}\right)=\left(\begin{array}{c}M+A\\ A\end{array}\right)\left(\begin{array}{c}n+M\\ M+A\end{array}\right)$

${\displaystyle {\sum }_{n=0}^{N-1}\left(\begin{array}{c}n\\ A\end{array}\right)\left(\begin{array}{c}n+M\\ M\end{array}\right)}=\left(\begin{array}{c}M+A\\ A\end{array}\right)\left(\begin{array}{c}N+M\\ M+A+1\end{array}\right)=\left(\begin{array}{c}N+M\\ M\end{array}\right)\left(\begin{array}{c}N\\ A\end{array}\right)\frac{N-A}{M+A+1}$, record at [3].

2.1.1) is for $\left(\begin{array}{c}n+A\\ A\end{array}\right)\left(\begin{array}{c}n+Y\\ M\end{array}\right)$, 2.1.2) is for $\left(\begin{array}{c}n+X\\ A\end{array}\right)\left(\begin{array}{c}n+Y\\ M\end{array}\right)$, $0\le Y\le M$.

There has no formula for $\left(\begin{array}{c}n+X\\ A\end{array}\right)\left(\begin{array}{c}n+Y\\ M\end{array}\right)$, $Y>M$, $X\ne A$.

2.1.3) $\left(\begin{array}{c}n\\ B\end{array}\right)\left(\begin{array}{c}A-n\\ M-B\end{array}\right)={\displaystyle {\sum }_{g=0}^{M-B}{\left(-1\right)}^{M-B-g}}\left(\begin{array}{c}A-M+g\\ g\end{array}\right)\left(\begin{array}{c}M-g\\ B\end{array}\right)\left(\begin{array}{c}n\\ M-g\end{array}\right),n\ge 0$

[Proof]

$PS=\left[0,-1,\cdots ,-\left(B-1\right),A-\left(M-B-1\right):-1,\cdots ,A:-1\right]$

${\displaystyle \sum H_1\left(q,K\right)}=\{\begin{array}{l}0,q<B\\ {\left(-1\right)}^q\left(\begin{array}{c}M-B\\ M-q\end{array}\right){\left[A-\left(M-B-1\right)\right]}^{M-q},q\ge B\end{array}$

q.e.d.

2.1.4) ${\left[P+nD:D\right]}_M={\displaystyle {\sum }_{q=0}^M{D}^q{\left[M\right]}_q{\left[P:D\right]}_{M-q}(\; n\; q\; )}$

2.1.5) $\left(\begin{array}{c}A-n\\ M\end{array}\right)=\frac{1}{M!}{\displaystyle {\sum }_{q=0}^M{\left(-1\right)}^{q}q!\left(\begin{array}{c}M\\ q\end{array}\right){\left[A-M+1\right]}^{M-q}(\; n\; q\; )}$

[Proof]

$PS=\left[A-M+1,\cdots ,A-1,A\right]:-1$,

$PT=\left[1,2,\cdots ,M\right]\to \left(\begin{array}{c}A-n\\ M\end{array}\right)=\frac{1}{M!}\nabla SUM\left(N\right)$,

$H_1\left(q,{\displaystyle \sum K}\right)=\left(\begin{array}{c}M\\ q\end{array}\right){\left[A-M+1\right]}^{M-q},H_1\left(q,T\right)={\left(-1\right)}^{q}q!$

q.e.d.

$A=2M\to \left(\begin{array}{c}2M-n\\ M\end{array}\right)={\displaystyle {\sum }_{q=0}^M{\left(-1\right)}^q}\left(\begin{array}{c}n\\ q\end{array}\right)\left(\begin{array}{c}2M-q\\ M\end{array}\right)$, record at [2]: (3.50).

2.1.6) $\left(\begin{array}{c}n\\ A\end{array}\right)\left(\begin{array}{c}n\\ B\end{array}\right)={\displaystyle {\sum }_{q=0}^B\left(\begin{array}{c}B\\ q\end{array}\right)}\left(\begin{array}{c}A+B-q\\ B-q\end{array}\right)\left(\begin{array}{c}n\\ A+B-q\end{array}\right)$, record at [2]: (6.44).

[Proof]

$PS=\left[0,-1,\cdots ,-A+1,0,-1,\cdots ,-B+1\right],PT=\left[1,2,\cdots ,A+B\right]$,

$\left\{X_1,X_2,\cdots ,X_A\right\}\in T\to H_1\left(q<A\right)=0,H_1\left(q\ge A\right)=q!\left(\begin{array}{c}B\\ q-A\end{array}\right){\left[A\right]}_{A+B-q}$

$\left(\begin{array}{c}n\\ A\end{array}\right)\left(\begin{array}{c}n\\ B\end{array}\right)=\frac{1}{A!B!}{\displaystyle {\sum }_{q=A}^{A+B}q!\left(\begin{array}{c}B\\ q-A\end{array}\right){\left[A\right]}_{A+B-q}(\; n\; q\; )}$

$\stackrel{q:=A+q}{\to }\frac{1}{A!B!}{\displaystyle {\sum }_{q=0}^B\left(A+q\right)!}\left(\begin{array}{c}B\\ q\end{array}\right){\left[A\right]}_{B-q}\left(\begin{array}{c}n\\ A+q\end{array}\right)\stackrel{q:=B-q}{\to }\text{conclusion}$

q.e.d.

2.2. P ≥ 0, PT = [P + 1, P + 2, …, P + M], PS = [P + 2, P + 4, …, P + 2M]

$\begin{array}{c}{H}_2\left(q,{{\displaystyle \sum }}^{\text{}}K\right)=SUM\left(q+1,\left[1,3,\cdots ,2\left(M-q\right)-1\right],\left[1,3,\cdots ,2\left(M-q\right)-1\right]\right)\\ =\left(2\left(M-q\right)-1\right)!!\left(\begin{array}{c}2M-q\\ q\end{array}\right)\end{array}$

$\begin{array}{c}SUM\left(N\right)=\frac{1}{P!}SUM\left(N,\left[1,2,\cdots ,P,PS\right],\left[1,2,\cdots ,P,P+1,\cdots ,P+M\right]\right)\\ ={\displaystyle {\sum }_{n=0}^{N-1}\left(\begin{array}{c}n+P\\ P\end{array}\right)}\left(P+2+n\right)\left(P+4+n\right)\cdots \left(P+2M+n\right)\\ ={\displaystyle {\sum }_{q=0}^M\left(2\left[M-q\right]-1\right)!!\left(\begin{array}{c}2M-q\\ q\end{array}\right){\left[P+1\right]}^q\left(\begin{array}{c}N+P+q\\ P+q+1\end{array}\right)}\end{array}$

2.2.1)

$\left(\begin{array}{c}n+P\\ P\end{array}\right){\displaystyle {\prod }_{i=1}^M\left(P+2i+n\right)}={\displaystyle {\sum }_{q=0}^M\left(2\left[M-q\right]-1\right)!!}\left(\begin{array}{c}2M-q\\ q\end{array}\right){\left[P+1\right]}^q\left(\begin{array}{c}n+P+q\\ P+q\end{array}\right)$

2.2.2) ${\displaystyle {\prod }_{i=1}^M\left(2i+n\right)}={\displaystyle {\sum }_{q=0}^M\left(2\left[M-q\right]-1\right)!!\left(\begin{array}{c}2M-q\\ q\end{array}\right)q!\left(\begin{array}{c}n+q\\ q\end{array}\right)}$

$SUM\left(N,\left[1,3,\cdots ,2M-1\right],\left[1,2,\cdots ,M\right]\right)=SUM\left(N,\left[3,\cdots ,2M-1\right],\left[2,\cdots ,M\right]\right)$

Change M to M − 1 and q! to (q + 1)!à

2.2.3) ${\displaystyle {\prod }_{i=1}^M\left(2i+n-1\right)}={\displaystyle {\sum }_{q=0}^{M-1}\left(2\left[M-q\right]-3\right)!!\left(\begin{array}{c}2M-q-2\\ q\end{array}\right)\left(q+1\right)!\left(\begin{array}{c}n+1+q\\ q+1\end{array}\right)}$

2.3. PT = [1, 3, …, 2M − 1], PS = [P + D, P + 3D, …, P + T_MD]:D

$H_2\left(q,K\right)={\left[P:D\right]}^{M-q}$

$H_2\left(q,{\displaystyle \sum T}\right)=D^{q}SUM\left(M-q+1,\left[1,3,\cdots ,2q-1\right],\left[1,3,\cdots ,2q-1\right]\right)$

2.3.1) $SUM\left(N\right)={\displaystyle {\sum }_{q=0}^M{\left[P:D\right]}^{M-q}{D}^q\left(2q-1\right)!!\left(\begin{array}{c}M+q\\ 2q\end{array}\right)\left(\begin{array}{c}N+M-1+q\\ M+q\end{array}\right)}$

$P=1,D=1\to PS=\left[2,4,\cdots ,2M\right]$

2.3.2) $\begin{array}{c}SUM\left(N\right)={\displaystyle {\sum }_{n_M=0}^{N-1}\left(2M+n_M\right)}\cdots {\displaystyle {\sum }_{n_2=0}^{n_3}\left(4+n_2\right)}{\displaystyle {\sum }_{n_1=0}^{n_2}\left(2+n_1\right)}\\ ={\displaystyle {\sum }_{q=0}^M\frac{\left(M+q\right)!}{\left(q\right)!2^q}\left(\begin{array}{c}N+M-1+q\\ M+q\end{array}\right)}={\displaystyle {\sum }_{q=0}^M\left(\begin{array}{c}N-1+q\\ q\end{array}\right)\frac{{\left[N+q\right]}^M}{2^q}}\end{array}$

(*) $N=1\to 2^{M}M!={\displaystyle {\sum }_{q=0}^M\frac{\left(M+q\right)!}{\left(q\right)!2^q}}={\displaystyle {\sum }_{q=0}^M\frac{{\left[1+q\right]}^M}{2^q}}$

This can also be obtained from 2.2.2).

2.4. PT = [1, 3, …, 2M − 1], PS = [P, P + D, …, P + (M − 1)D]:2D

2.4.1) $SUM\left(N\right)=\left(\begin{array}{c}M+N-1\\ M\end{array}\right){\left[P+\left(M+N-2\right)D:D\right]}_M$

[Proof]

$\begin{array}{l}{H}_1\left(q,{\displaystyle \sum K},\left[P,P+D,\cdots ,P+\left(M-1\right)D\right]:D,\left[1,2,\cdots ,M\right]\right)\\ =SUM\left(q+1,\left[P,P+D,\cdots ,P+\left(M-1-q\right)D\right]:2D,\left[1,3,\cdots ,2\left(M-q\right)-1\right]\right)\\ =H_1\left(q,{{\displaystyle \sum }}^{\text{}}K,\left[P+\left(M-1\right)D,\cdots ,P+D,P\right]:D,\left[1,2,\cdots ,M\right]\right)\\ =\left(\begin{array}{c}M\\ q\end{array}\right)H_1\left(q,K,\cdots \right)\stackrel{3.2)}{\to }\left(\begin{array}{c}M\\ q\end{array}\right){\left[P+\left(M-1\right)D:D\right]}_{M-q}\end{array}$

$\begin{array}{l}SUM\left(q+1,\left[P,P+D,\cdots ,P+\left(M-1-q\right)D\right]:2D,\left[1,3,\cdots ,2\left(M-q\right)-1\right]\right)\\ =\left(\begin{array}{c}M\\ q\end{array}\right){\left[P+\left(M-1\right)D:D\right]}_{M-q}\stackrel{M:=M-q}{\to }\end{array}$

$\begin{array}{l}SUM\left(q+1,\left[P,P+D,\cdots ,P+\left(M-1\right)D\right]:2D,\left[1,3,\cdots ,2M-1\right]\right)\\ =\left(\begin{array}{c}M+q\\ q\end{array}\right){\left[P+\left(M+q-1\right)D:D\right]}_M\\ \stackrel{N:=q+1}{\to }\left(\begin{array}{c}M+N-1\\ M\end{array}\right){\left[P+\left(M+N-2\right)D:D\right]}_M\end{array}$

q.e.d.

2.4.2) $\begin{array}{l}SUM\left(N,\left[P+1,P+2,\cdots \right]:2,\left[1,3,\cdots ,2M-1\right]\right)\\ =M!\left(\begin{array}{c}N+M-1\\ M\end{array}\right)\left(\begin{array}{c}N+M+P-1\\ M\end{array}\right)\end{array}$

2.4.3) $SUM\left(N,\left[1,2,\cdots ,M\right]:2,\left[1,3,\cdots ,2M-1\right]\right)=M!{\left(\begin{array}{c}N+M-1\\ M\end{array}\right)}^2$

$M=1\to 1+3+\cdots +\left(2N-1\right)=N^2$

$\begin{array}{l}M=2\to SUM\left(N,\left[1,2\right]:2,\left[1,3\right]\right)={\displaystyle {\sum }_{n=0}^{N-1}\left(2+2n\right)SUM\left(n+1,\left[1\right]:2,\left[1\right]\right)}\\ ={\displaystyle {\sum }_{n=0}^{N-1}\left(2+2n\right){\left(1+n\right)}^2}=2{\displaystyle {\sum }_{n=0}^{N-1}{\left(1+n\right)}^3}\to {\displaystyle {\sum }_{n=1}^N{n}^3}={\left(\begin{array}{c}N+1\\ 2\end{array}\right)}^2\end{array}$

2.4.2) and 2.1.1)à

$\begin{array}{l}H\left(q,\left[P+1,P+2,\cdots \right]:2,\left[1,3,\cdots ,2M-1\right]\right)\\ =M!H\left(q,\left[P+1,P+2,\cdots \right],\left[M+1,M+2,\cdots ,2M\right]\right)\end{array}$

2.4.5) $\left(\begin{array}{c}n\\ A\end{array}\right)\left(\begin{array}{c}n\\ B\end{array}\right)={\displaystyle {\sum }_{q=0}^B\left(\begin{array}{c}A\\ B-q\end{array}\right)\left(\begin{array}{c}B\\ q\end{array}\right)\left(\begin{array}{c}n+B-q\\ A+B\end{array}\right)}$, record at [2]: (6.45).

[Proof]

$PS=\left[0,-1,\cdots ,-A+1,0,-1,\cdots ,-B+1\right],PT=\left[1,2,\cdots ,A+B\right]$, use Form₃

$\left\{X_1,X_2,\cdots ,X_A\right\}\in T\to H_3\left(q<A\right)=0,q\ge A\to {\displaystyle {\prod }_{i\le A}{X}_i}\in T=A!$

$\begin{array}{c}{{\displaystyle \sum }}^{\text{}}{\displaystyle {\prod }_{i>A}{X}_i\in T}=SUM\left(A+B+1-q,\left[1,2,\cdots ,q-A\right]:2,\left[1,3,\cdots ,2\left(q-A\right)-1\right]\right)\\ =\left(q-A\right)!{\left(\begin{array}{c}A+B-q+q-A\\ q-A\end{array}\right)}^2\end{array}$

$H_3\left(q\ge A\right)=A!\left(q-A\right)!\left(\begin{array}{c}B\\ q-A\end{array}\right)\left(\begin{array}{c}B\\ q-A\end{array}\right){\left[A\right]}_{A+B-q}$

$\left(\begin{array}{c}n\\ A\end{array}\right)\left(\begin{array}{c}n\\ B\end{array}\right)=\frac{1}{A!B!}{\displaystyle {\sum }_{q=A}^{A+B}{H}_3\left(q\right)}\left(\begin{array}{c}n+A+B-q\\ A+B\end{array}\right)$

$\stackrel{q:=A+q}{\to }\frac{1}{A!B!}{\displaystyle {\sum }_{q=0}^{B}A!q!}{\left(\begin{array}{c}B\\ q\end{array}\right)}^2{\left[A\right]}_{B-q}\left(\begin{array}{c}n+B-q\\ A+B\end{array}\right)$

q.e.d.

2.5. P ≥ 0, PT = [1, 3, …, 2M − 1], PS = [P + 2, P + 4, …, P + 2M]:3

[1] has obtained: $H_1\left(q\right)={\displaystyle {\sum }_{x=q}^M{H}_2\left(x\right)}\left(\begin{array}{c}x\\ q\end{array}\right)={\displaystyle {\sum }_{x=0}^q{H}_3\left(x\right)}\left(\begin{array}{c}M-x\\ M-q\end{array}\right)$

2.5.1) $SUM\left(N\right)={\displaystyle {\sum }_{q=0}^M\left(\begin{array}{c}P+N-1+q\\ q\end{array}\right)\left(\begin{array}{c}N+M-1-q\\ M-q\end{array}\right)\frac{{\left[M+N-q\right]}^M}{2^{M-q}}}$

[Proof]

For $SUM\left(N,\left[P+2,P+4,\cdots ,P+2M\right],\left[P+1,P+2,\cdots ,P+M\right]\right)$

$H_1\left(q,T\right)={\left[P+1\right]}^q$

$H_1\left(q,{\displaystyle \sum K}\right)=SUM\left(q+1,\left[P+2,P+4,\cdots \right]:3,\left[1,3,\cdots ,2\left(M-q\right)-1\right]\right)$

$\begin{array}{l}{H}_1\left(q\right)={\displaystyle {\sum }_{x=q}^M{H}_2\left(x\right)}\left(\begin{array}{c}x\\ q\end{array}\right)\\ \stackrel{\mathrm{2.3.1}),D=1}{\to }{\displaystyle {\sum }_{x=q}^M\left(2\left[M-x\right]-1\right)!!}\left(\begin{array}{c}2M-x\\ x\end{array}\right){\left[P+1\right]}^x(\; x\; q\; )\end{array}$

$\begin{array}{l}SUM\left(q+1,\left[P+2,P+4,\cdots ,P+2\left(M-q\right)\right]:3,\left[1,3,\cdots ,2\left(M-q\right)-1\right]\right)\\ ={\displaystyle {\sum }_{x=q}^M\frac{\left(2M-x\right)!}{\left(m-x\right)!2^{M-x}}\frac{{\left[P+1\right]}^x}{q!\left(x-q\right)!{\left[P+1\right]}^q}}\\ ={\displaystyle {\sum }_{x=q}^M\frac{\left(2M-x\right)!}{\left(m-x\right)!2^{M-x}}\frac{\left(P+x\right)!}{q!\left(x-q\right)!\left(P+q\right)!}}\\ \stackrel{x:=q+x}{\to }{\displaystyle {\sum }_{x=0}^{M-q}\frac{\left(2M-q-x\right)!}{\left(M-q-x\right)!2^{M-q-x}}\frac{\left(P+x+q\right)!}{q!x!\left(P+q\right)!}}\end{array}$

$\begin{array}{l}SUM\left(q+1,\left[P+2,P+4,\cdots ,P+2M\right]:3,\left[1,3,\cdots ,2M-1\right]\right)\\ ={\displaystyle {\sum }_{x=0}^M\frac{\left(2M+q-x\right)!}{\left(M-x\right)!2^{M-x}}\frac{\left(P+x+q\right)!}{q!x!\left(P+q\right)!}}\to \text{conclusion}\end{array}$

q.e.d.

$\begin{array}{l}SUM\left(N,\left[2,4,\cdots ,2M\right]:3,\left[1,3,\cdots ,2M-1\right]\right)\\ ={\displaystyle {\sum }_{x=0}^M\frac{\left(2M+N-1-x\right)!}{\left(M-x\right)!2^{M-x}}\frac{\left(P+x+N-1\right)!}{x!\left(N-1\right)!\left(P+N-1\right)!}}\to \end{array}$

2.5.2)

$\begin{array}{l}SUM\left(N,\left[2,4,\cdots ,2M\right]:3,\left[1,3,\cdots \right]\right)\\ ={\displaystyle {\sum }_{q=0}^M\left(\begin{array}{c}N-1+q\\ q\end{array}\right)}\left(\begin{array}{c}N+M-1-q\\ M-q\end{array}\right)\frac{{\left[M+N-q\right]}^M}{2^{M-q}}\end{array}$

2.5.3)

$\begin{array}{l}SUM\left(N,\left[3,5,\cdots ,2M-1\right]:3,\left[3,5,\cdots ,2M-1\right]\right)\\ ={\displaystyle {\sum }_{q=0}^{M-1}\left(\begin{array}{c}N+q\\ q\end{array}\right)}\left(\begin{array}{c}N+M-2-q\\ M-q-1\end{array}\right)\frac{{\left[M+N-q-1\right]}^{M-1}}{2^{M-q-1}}\end{array}$

2.6. Summary

3. r-Flod Sum

Define ${\displaystyle {\sum }_{\left(r\right)}^{N}f\left(k\right)}={\displaystyle {\sum }_{k_r=1}^N\cdots {\displaystyle {\sum }_{k_2=1}^{k_3}{\displaystyle {\sum }_{k=1}^{k_2}f\left(k\right)}}}={\displaystyle {\sum }_{k_r=0}^{N-1}\cdots {\displaystyle {\sum }_{k_2=0}^{k_3}{\displaystyle {\sum }_{k=0}^{k_2}f\left(k+1\right)}}}$

${\displaystyle {\sum }_{\left(r\right)}^{N}1}=SUM\left(N,\left[1,1,\cdots ,1\right]:0,\left[1,3,\cdots ,2r-1\right]\right)\stackrel{H_1\left(q>0\right)=0}{\to }\left(\begin{array}{c}N+r-1\\ r\end{array}\right)$

3.1) ${\displaystyle {\sum }_{\left(r\right)}^N\nabla SUM\left(k+1,PS,PT\right)}={\displaystyle {\sum }_{q=0}^M{H}_1\left(q,PS,PT\right)}\left(\begin{array}{c}N+T_M-M+r-1\\ T_M-M+r+q\end{array}\right)$

[Proof]

$PS1=\left[PS,1:0,1:0,\cdots \right],PT1=\left[PT,T_M+2,T_M+4,\cdots ,T_M+2\left(r-1\right)\right]$

$\begin{array}{l}{\displaystyle {\sum }_{\left(r\right)}^N\nabla SUM\left(k+1,PS,PT\right)}=SUM\left(N,PS1,PT1\right)\\ ={\displaystyle {\sum }_{q=0}^{M+r-1}{H}_1\left(q,PS1,PT1\right)}\left(\begin{array}{c}N+T_M-M+r-1\\ T_M-M+r+q\end{array}\right)\end{array}$

In $H_1\left(q,PS1,PT1\right)$, $i>M$, $\{\begin{array}{l}\left(T_i-X_{K-1}\right)D_i=0;X_i=T_i\\ K_i+X_{T-1}{D}_i=1;X_i=K_i\end{array}$

$H_1\left(q>M,PS1,PT1\right)=0,H_1\left(q\le M,PS1,PT1\right)=H_1\left(q,PS,PT\right)$

q.e.d.

Another way: understand the definition of ${\nabla }^p$ and use three Forms.

3.2) ${\displaystyle {\sum }_{\left(r\right)}^N\nabla SUM\left(k+1,PS,PT\right)}={\nabla }^{-\left(r-1\right)}SUM\left(N,PS,PT\right)$

$\begin{array}{l}{\displaystyle {\sum }_{\left(r\right)}^N{K}^2}={\displaystyle {\sum }_{\left(r\right)}^N\nabla SUM\left(k+1,\left[1,1\right],\left[1,2\right]\right)}\\ =SUM\left(N,\left[1,1,1:0,1:0,\cdots \right],\left[1,2,4,6,\cdots ,2r\right]\right)\\ =2\left(\begin{array}{c}N+r-1\\ r+2\end{array}\right)+3\left(\begin{array}{c}N+r-1\\ r+1\end{array}\right)+\left(\begin{array}{c}N+r-1\\ r\end{array}\right)=2\left(\begin{array}{c}N+r\\ r+2\end{array}\right)+\left(\begin{array}{c}N+r\\ r+1\end{array}\right)\end{array}$

$=\frac{2\left(N+r\right)!\left(N-1\right)}{\left(r+2\right)!\left(N-1\right)!}+\frac{\left(N+r\right)!\left(r+2\right)}{\left(r+2\right)!\left(N-1\right)!}=\frac{\left(N+r\right)!\left(2N+r\right)}{\left(r+2\right)!\left(N-1\right)!}\to$ record at [2].

$\begin{array}{l}{\displaystyle {\sum }_{\left(r\right)}^N{K}^3}={\nabla }^{-\left(r-1\right)}SUM\left(N,\left[1,1\right],\left[2,3\right]\right)\\ ={\nabla }^{-\left(r-1\right)}\left[6\left(\begin{array}{c}N+1\\ 4\end{array}\right)+6\left(\begin{array}{c}N+1\\ 3\end{array}\right)+\left(\begin{array}{c}N+1\\ 2\end{array}\right)\right]\end{array}$

$=6\left(\begin{array}{c}N+r\\ r+3\end{array}\right)+6\left(\begin{array}{c}N+r\\ r+2\end{array}\right)+\left(\begin{array}{c}N+r\\ r+1\end{array}\right)$

$=\frac{\left(6N^2+r\left(6N+r-1\right)\right)\left(N+r\right)!}{\left(r+3\right)!\left(N-1\right)!}\to$ record at [2].

3.3) ${\displaystyle {\sum }_{k_r=0}^{N-1}\cdots {\displaystyle {\sum }_{k_2=0}^{k_3}{\displaystyle {\sum }_{k=0}^{k_2}\left(\begin{array}{c}{k}_i\\ p\end{array}\right)}}}=\left(\begin{array}{c}p+i-1\\ p\end{array}\right)\left(\begin{array}{c}N+r-1\\ r+p\end{array}\right)\stackrel{p=1}{\to }i\left(\begin{array}{c}N+r-1\\ r+1\end{array}\right)$

[Proof]

${\displaystyle {\sum }_{k_r=0}^{N-1}\cdots {\displaystyle {\sum }_{k_2=0}^{k_3}{\displaystyle {\sum }_{k=0}^{k_2}\left(\begin{array}{c}{k}_i\\ p\end{array}\right)}}}={\displaystyle {\sum }_{k_r=0}^{N-1}\cdots {\displaystyle {\sum }_{k_i=0}^{k_{i+1}}\left(\begin{array}{c}{k}_i\\ p\end{array}\right)\left(\begin{array}{c}{k}_i+i-1\\ i-1\end{array}\right)}}$

$PS1=\left[1,2,\cdots ,i-1,0,-1,-2,\cdots ,-p+1,1:0,1:0,\cdots ,1:0\right]$

$\begin{array}{c}PT1=\left[1,2,\cdots ,i-1,i,i+1,\cdots ,i+p-1,i+p+1,i+p+3,\cdots ,i+p+2\left(r-i\right)-1\right]\\ =\frac{1}{p!\left(i-1\right)!}SUM\left(N,PS1,PT1\right)\end{array}$

$PS=\left[0,-1,-2,\cdots ,-p+1,1:0,1:0,\cdots ,1:0\right]$

$\begin{array}{l}PT=\left[i,i+1,i+2,\cdots ,i+p-1,i+p+1,i+p+3,\cdots ,i+p+2\left(r-i\right)-1\right]\\ =\frac{1}{p!}SUM\left(N,PS,PT\right)\stackrel{H_1\left(q\ne p\right)=0,\{X_1,X_2,\cdots ,X_P\}\in T}{\to }\left(\begin{array}{c}p+i-1\\ p\end{array}\right)\left(\begin{array}{c}N+r-1\\ r+p\end{array}\right)\end{array}$

q.e.d.

This is the conclusion of [4] and the proof is simpler.

4. ${\displaystyle {\sum }_{n=0}^{N-1}{\nabla }^{P_1}SUM\left(n+1,\left[K_1\right],\left[T_1\right]\right){\nabla }^{P_2}SUM\left(n+1,\left[K_2\right],\left[T_2\right]\right)}\cdots$

4.1) $P\le T,{\nabla }^{P}SUM\left(N,\left[K:D\right],\left[T\right]\right),D\ne 0$

$=\frac{TD}{\left(T+1-P\right)!}\left(n+1\right)\left(n+2\right)\cdots \left(n+T-P\right)\left(n+\frac{T+1-P}{TD}K\right)$

[Proof]

$\begin{array}{l}{\nabla }^{P}SUM\left(N,\left[K:D\right],\left[T\right]\right)={\nabla }^P\left[K\left(\begin{array}{c}N+T-1\\ T\end{array}\right)+TD\left(\begin{array}{c}N+T-1\\ T+1\end{array}\right)\right]\\ =\left[K\left(\begin{array}{c}n+T-P\\ T-P\end{array}\right)+TD\left(\begin{array}{c}n+T-P\\ T+1-P\end{array}\right)\right]\\ =\frac{1}{\left(T-P\right)!}\left(n+1\right)\left(n+2\right)\cdots \left(n+T-P\right)\left(\frac{TD}{T+1-P}n+K\right)\\ =\frac{TD}{\left(T+1-P\right)!}\left(n+1\right)\left(n+2\right)\cdots \left(n+T-P\right)\left(n+\frac{T+1-P}{TD}K\right)\end{array}$

q.e.d.

This leads to:

4.2) $P_1\le T_1,P_2\le T_2,D_1\ne 0,D_2\ne 0$

$PS=\left[1,2,\cdots ,T_1-P_1,\frac{K_1\left(T_1+1-P_1\right)}{T_1{D}_1},1,2,\cdots ,T_2-P_2,\frac{K_2\left(T_2+1-P_2\right)}{T_2{D}_2}\right]$

$PT=\left[1,2,3,\cdots ,T_1+T_2+2-P_1-P_2\right]$

$\begin{array}{l}{\displaystyle {\sum }_{n=0}^{N-1}{\nabla }^{P_1}SUM\left(n+1,\left[K_1:D_1\right],\left[T_1\right]\right){\nabla }^{P_2}SUM\left(n+1,\left[K_2:D_2\right],\left[T_2\right]\right)}\\ =\frac{T_1{T}_2{D}_1{D}_2}{\left(T_1+1-P_1\right)!\left(T_2+1-P_2\right)!}SUM\left(N,PS,PT\right)\end{array}$

4.3)

$\begin{array}{l}{\displaystyle {\sum }_{n=0}^{N-1}\nabla SUM\left(n+1,\left[K_1\right],\left[T_1\right]\right)\nabla SUM\left(n+1,\left[K_2\right],\left[T_2\right]\right)}\\ ={\displaystyle {\sum }_{n=0}^{N-1}\left[K_1\left(\begin{array}{c}n+T_1-1\\ T_1-1\end{array}\right)+T_1\left(\begin{array}{c}n+T_1-1\\ T_1\end{array}\right)\right]\left[K_2\left(\begin{array}{c}n+T_2-1\\ T_2-1\end{array}\right)+T_2\left(\begin{array}{c}n+T_2-1\\ T_2\end{array}\right)\right]}\\ =\frac{1}{\left(T_1-1\right)!}\frac{1}{\left(T_2-1\right)!}SUM\left(N,\left[1,2,\cdots ,T_1-1,K_1,1,2,\cdots ,T_2-1,K_2\right],\left[1,2,\cdots ,T_1+T_2\right]\right)\end{array}$

$T_1=T_2=1\to {\displaystyle {\sum }_{n=0}^{N-1}\left(n+K_1\right)\left(n+K_2\right)}=SUM\left(N,\left[K_1,K_2\right],\left[1,2\right]\right)$

4.4)

$\begin{array}{l}{\displaystyle {\sum }_{n=0}^{N-1}SUM\left(n+1,\left[K_1\right],\left[T_1\right]\right)SUM\left(n+1,\left[K_2\right],\left[T_2\right]\right)}\\ =\frac{T_1{T}_2}{\left(T_1+1\right)!\left(T_2+1\right)!}SUM(N,\left[1,\cdots ,T_1,\frac{K_1\left(T_1+1\right)}{T_1},1,\cdots ,T_2,\frac{K_2\left(T_2+1\right)}{T_2}\right],\\ \left[1,2,\cdots ,T_1+T_2+2\right])\end{array}$

The following calculation problems have been solved:

${\displaystyle {\sum }_{n=0}^{N-1}{\nabla }^{P_1}SUM\left(n+1,\left[K_1:D_1\right],\left[T_1\right]\right){\nabla }^{P_2}(\dots ){\nabla }^{P_3}(\dots ){\nabla }^{P_4}(\dots )\cdots }$

$\nabla {\displaystyle {\sum }_{n=0}^{N-1}\left(K+n\right)}=\nabla SUM\left(N,\left[K\right],\left[1\right]\right)=K+n$

$\begin{array}{l}\nabla {\displaystyle {\sum }_{n,n_1,n_2=0,n_1\le n_2=n}^{N-1}\left(K+n_1+n_2\right)}\stackrel{n_1\in \left[0,n\right],\text{countofitems}=n+1}{\to }\\ =\left(K+n_2\right)\left(\begin{array}{c}n+1\\ 1\end{array}\right)+{\displaystyle {\sum }_{n_1=0}^n{n}_1}=\left(K+n\right)\left(\begin{array}{c}n+1\\ 1\end{array}\right)+\left(\begin{array}{c}n+1\\ 2\end{array}\right)\\ =K\left(\begin{array}{c}n+1\\ 1\end{array}\right)+3\left(\begin{array}{c}n+1\\ 2\end{array}\right)={\nabla }^{2}SUM\left(N,\left[K\right],\left[3\right]\right)\end{array}$

$\begin{array}{l}\nabla {{\displaystyle \sum }}^{\text{}}\left(K+n_1+\cdots +n_M\right)\stackrel{\text{Repeatableselection}M-1\text{in}\left[0,n\right],\text{countofitems}=\left(\begin{array}{c}n+1+M-1-1\\ M-1\end{array}\right)}{\to }\\ =\left(K+n\right)\left(\begin{array}{c}n+M-1\\ M-1\end{array}\right)+{\displaystyle {\sum }_{n_{M-1}=0}^n\cdots {\displaystyle {\sum }_{n_1=0}^{n_2}\left(n_1+n_2+\cdots +n_{M-1}\right)}}\end{array}$

$\begin{array}{l}\stackrel{3.3),P=1}{\to }\left(K+n\right)\left(\begin{array}{c}n+M-1\\ M-1\end{array}\right)+{\displaystyle {\sum }_{i=1}^{M-1}i}\left(\begin{array}{c}n+M-1\\ M+1\end{array}\right)\\ =\left(K+n\right)\left(\begin{array}{c}n+M-1\\ M-1\end{array}\right)+\left(\begin{array}{c}M\\ 2\end{array}\right)\left(\begin{array}{c}n+M-1\\ M\end{array}\right)\end{array}$

$\begin{array}{l}=\left(K+n\right)\left(\begin{array}{c}n+M-1\\ M-1\end{array}\right)+\left(\begin{array}{c}M\\ 2\end{array}\right)\left(\begin{array}{c}n+M-1\\ M\end{array}\right)\\ =K\left(\begin{array}{c}n+M-1\\ M-1\end{array}\right)+n\left(\begin{array}{c}n+M-1\\ M-1\end{array}\right)+\left(\begin{array}{c}M\\ 2\end{array}\right)\left(\begin{array}{c}n+M-1\\ M\end{array}\right)\\ =K\left(\begin{array}{c}n+M-1\\ M-1\end{array}\right)+\left(M+\left(\begin{array}{c}M\\ 2\end{array}\right)\right)\left(\begin{array}{c}n+M-1\\ M\end{array}\right)\end{array}$

$=K\left(\begin{array}{c}n+M-1\\ M-1\end{array}\right)+\left(\begin{array}{c}M+1\\ 2\end{array}\right)\left(\begin{array}{c}n+M-1\\ M\end{array}\right)$

$\begin{array}{l}=K\left(\begin{array}{c}N+M-2\\ M-1\end{array}\right)+\left(\begin{array}{c}M+1\\ 2\end{array}\right)\left(\begin{array}{c}N+M-2\\ M\end{array}\right)\\ ={\nabla }^{\left(\begin{array}{c}M\\ 2\end{array}\right)+1}SUM\left(N,\left[K\right],\left[\left(\begin{array}{c}M+1\\ 2\end{array}\right)\right]\right)\end{array}$

4.5) ${\displaystyle {\sum }_{n_1,n_2,\cdots ,n_M=0,n_1\le n_2\le \cdots \le n_M}^{N-1}\left(K+n_1+\cdots +n_M\right)}={\nabla }^{\left(\begin{array}{c}M\\ 2\end{array}\right)}SUM\left(N,\left[K\right],\left[\left(\begin{array}{c}M+1\\ 2\end{array}\right)\right]\right)$

4.6) $\begin{array}{l}{\displaystyle {\sum }_{n_1,n_2,\cdots ,n_M=0,n_1\le n_2\le \cdots \le n_M}^{N-1}\left(K+n_1{d}_1+\cdots +n_M{d}_M\right)}\\ ={\nabla }^{\left(\begin{array}{c}M\\ 2\end{array}\right)}SUM\left(N,\left[K:\left(d_1+2d_2+\cdots +M{d}_M\right){\left(\begin{array}{c}M+1\\ 2\end{array}\right)}^{-1}\right],\left[\left(\begin{array}{c}M+1\\ 2\end{array}\right)\right]\right)\end{array}$

Use 4.1), the following calculation problems have been solved:

${\displaystyle {\sum }_0^{N-1}{\nabla }^{P_1}}\left({{\displaystyle \sum }}^{\text{}}\left(K+n_1{d}_{1,1}+\cdots +n_{M_1}{d}_{1,M}\right)\right){\nabla }^{P_2}\left({{\displaystyle \sum }}^{\text{}}\left(K+n_1{d}_{2,1}+\cdots \right)\right){▽}^{P_3}(\dots )\cdots$

Investigation

${\displaystyle {\sum }_{n_1,n_2,\cdots ,n_M=0,n_1\le n_2\le \cdots \le n_M}^{N-1}\left(K_1+n_1{d}_{1,1}+\cdots +n_M{d}_{M,1}\right)\left(K_2+n_1{d}_{1,2}+\cdots +n_M{d}_{M,2}\right)}$

$\begin{array}{l}{\displaystyle {\sum }_{n=0}^{N-1}\left(K_1+n{d}_{1,1}\right)\left(K_2+n{d}_{1,2}\right)}=SUM\left(N,\left[K_1:d_{1,1},K_2:d_{1,2}\right],\left[1,2\right]\right)\\ =2d_{1,1}{d}_{12}\left(\begin{array}{c}N\\ 3\end{array}\right)+\left(K_1{d}_{1,2}+K_2{d}_{1,1}+d_{1,1}{d}_{1,2}\right)\left(\begin{array}{c}N\\ 2\end{array}\right)+K_1{K}_2(\; N\; 1\; )\end{array}$

Suppose

$\begin{array}{c}F\left(N,M\right)={\displaystyle {\sum }_{n_1,n_2,\cdots ,n_M=0,n_1\le n_2\le \cdots \le n_M}^{N-1}\left(K_1+n_1{d}_{1,1}+\cdots \right)\left(K_2+n_1{d}_{1,2}+\cdots \right)}\\ =A\left(M\right)\left(\begin{array}{c}N+M-1\\ M+2\end{array}\right)+B\left(M\right)\left(\begin{array}{c}N+M-1\\ M+1\end{array}\right)+K_1{K}_2\left(\begin{array}{c}N+M-1\\ M\end{array}\right)\end{array}$

Let $D_i=d_{1,i}+2d_{2,i}+\cdots +M{d}_{M,i},D_i^M=d_{1,i}+2d_{2,i}+\cdots +\left(M-1\right)d_{M-1,i}$

$\begin{array}{l}F\left(N,M+1\right)={\displaystyle {\sum }_{n_{M+1}=0}^{N-1}F\left(N,M\right)}\\ +{\displaystyle {\sum }_{n_{M+1}=0}^{N-1}{n}_{M+1}{d}_{M+1,2}}{\displaystyle {\sum }_{n_1,n_2,\cdots ,n_M=0,n_1\le n_2\le \cdots \le n_M}^{N-1}\left(K_1+n_1{d}_{1,1}+\cdots \right)}\\ +{\displaystyle {\sum }_{n_{M+1}=0}^{N-1}{n}_{M+1}{d}_{M+1,1}}{\displaystyle {\sum }_{n_1,n_2,\cdots ,n_M=0,n_1\le n_2\le \cdots \le n_M}^{N-1}\left(K_2+n_1{d}_{2,1}+\cdots \right)}\\ +{\displaystyle {\sum }_{n_{M+1}=0}^{N-1}{n}_{M+1}{d}_{M+1,1}{d}_{M+1,2}}{\displaystyle {\sum }_{n_1,n_2,\cdots ,n_M=0,n_1\le n_2\le \cdots \le n_M}^{N-1}1}\end{array}$

$\begin{array}{l}=A\left(M\right)\left(\begin{array}{c}N+M\\ \left(M+1\right)+2\end{array}\right)+B\left(M\right)\left(\begin{array}{c}N+M\\ \left(M+1\right)+1\end{array}\right)+K_1{K}_2\left(\begin{array}{c}N+M\\ M+1\end{array}\right)\\ +SUM\left(N,\left[K_1:D_1{\left(\begin{array}{c}M+1\\ 2\end{array}\right)}^{-1},0:d_{M+1,2}\right]\left[\left(\begin{array}{c}M+1\\ 2\end{array}\right),\left(\begin{array}{c}M+1\\ 2\end{array}\right)+2-\left(\begin{array}{c}M\\ 2\end{array}\right)\right]\right)\\ +SUM\left(N,\left[K_2:D_2{\left(\begin{array}{c}M+1\\ 2\end{array}\right)}^{-1},0:d_{M+1,1}\right]\left[\left(\begin{array}{c}M+1\\ 2\end{array}\right),M+2\right]\right)\\ +SUM\left(N,\left[1:0,1:0,\cdots ,1:0,0:d_{M+1,1},0:d_{M+1,2}\right]\left[1,3,\cdots ,2M-1,2M\right]\right)\end{array}$

$\begin{array}{l}=A\left(M\right)\left(\begin{array}{c}N+M\\ \left(M+1\right)+2\end{array}\right)+B\left(M\right)\left(\begin{array}{c}N+M\\ \left(M+1\right)+1\end{array}\right)+K_1{K}_2\left(\begin{array}{c}N+M\\ M+1\end{array}\right)\\ +D_1\left(M+2\right)d_{M+1,2}\left(\begin{array}{c}N+M\\ \left(M+1\right)+2\end{array}\right)+\left\{K_1\left(M+1\right)d_{M+1,2}+D_1{d}_{M+1,2}\right\}\left(\begin{array}{c}N+M\\ \left(M+1\right)+1\end{array}\right)\end{array}$

$\begin{array}{l}+D_2\left(M+2\right)d_{M+1,1}\left(\begin{array}{c}N+M\\ \left(M+1\right)+2\end{array}\right)+\left\{K_2{\left(M+1\right)}_{M+1,1}+D_2{d}_{M+1,1}\right\}\left(\begin{array}{c}N+M\\ \left(M+1\right)+1\end{array}\right)\\ +\left(M+1\right)\left(M+2\right)d_{M+1,1}{d}_{M+1,2}\left(\begin{array}{c}N+M\\ \left(M+1\right)+2\end{array}\right)+\left(M+1\right)d_{M+1,1}{d}_{M+1,2}\left(\begin{array}{c}N+M\\ \left(M+1\right)+1\end{array}\right)\end{array}$

$A\left(0\right)=0,B\left(0\right)=0$

$\begin{array}{c}A\left(M\right)=A\left(M-1\right)+M\left(M+1\right)d_{M,1}{d}_{M,2}+\left(M+1\right)\left(D_i^M{d}_{M,2}+D_i^M{d}_{M,1}\right)\\ =A\left(M-1\right)+\left(M+1\right)\left(D_1{d}_{M,2}+D_2{d}_{M,1}-M{d}_{M,1}{d}_{M,2}\right)\end{array}$

$\begin{array}{c}B\left(M\right)=B\left(M-1\right)+M(d_{M,1}{d}_{M,2}+K_1{d}_{M,2}+K_2{d}_{M,1})+D_i^M{d}_{M,2}+D_i^M{d}_{M,1}\\ =B\left(M-1\right)+M\left(K_1{d}_{M,2}+K_2{d}_{M,1}-d_{M,1}{d}_{M,2}\right)+D_1{d}_{M,2}+D_2{d}_{M,1}\end{array}$

4.7) $F\left(N,M\right)=A\left(M\right)\left(\begin{array}{c}N+M-1\\ M+2\end{array}\right)+B\left(M\right)\left(\begin{array}{c}N+M-1\\ M+1\end{array}\right)+K_1{K}_2\left(\begin{array}{c}N+M-1\\ M\end{array}\right)$

$d_{i,1}=1,d_{i,2}=1\to$

$A\left(M\right)={\displaystyle {\sum }_{n=0}^M\left\{n\left(n+1\right)+2\left(n+1\right)\left(\begin{array}{c}n\\ 2\end{array}\right)\right\}}=2\left(\begin{array}{c}M+2\\ 3\end{array}\right)+6\left(\begin{array}{c}M+2\\ 4\end{array}\right)$

$B\left(M\right)={\displaystyle {\sum }_{n=0}^M\left\{n\left(K_1+K_2+1\right)+2\left(\begin{array}{c}n\\ 2\end{array}\right)\right\}}=\left(\begin{array}{c}M+1\\ 2\end{array}\right)\left(K_1+K_2+1\right)+2\left(\begin{array}{c}M+1\\ 3\end{array}\right)$

5. Formal Calculation of Gaussian Coefficients

Define: $G_M^N={\left[\begin{array}{c}N\\ M\end{array}\right]}_q=\frac{\left(q^N-1\right)\left(q^{N-1}-1\right)\cdots \left(q^{N-M+1}-1\right)}{\left(q^M-1\right)\left(q^{M-1}-1\right)\cdots \left(q-1\right)}$

$G_0^N=1,G_{M<0\text{or}M>N}^N=0,G_M^N=G_{N-M}^N,G_M^N=q^M{G}_M^{N-1}+G_{M-1}^{N-1}$

5.1) $G_M^N=G_M^{N-1}+q^{N-M}{G}_{M-1}^{N-1}$

5.2) ${\displaystyle {\sum }_{n=0}^{N-1}{q}^n{G}_M^{n+M}}=G_{M+1}^{N+M}$

5.3) ${\displaystyle {\sum }_{n=0}^{N-1}{q}^n{G}_M^{n+K}}=q^{M-K}{G}_{M+1}^{N+K};{\displaystyle {\sum }_{n=0}^{N-1}{q}^n{G}_M^n}=q^M{G}_{M+1}^N$

5.4) ${\displaystyle {\sum }_{n=0}^{N-1}{q}^n{G}_1^n{G}_M^{n+K}},M>0,M\ge K$

① $=q^{2\left(M-K\right)+1}{G}_1^{M+1}{G}_{M+2}^{N+K}+q^{M-K}{G}_1^{M-K}{G}_{M+1}^{N+K}$,

② $=q^{M-2K-1}{G}_1^{M+1}{G}_{M+2}^{N+K+1}+q^{M-K}\left(G_1^{M-K}-q^{-K-1}{G}_1^{M+1}\right)G_{M+1}^{N+K}$,

③ $=\left(q^{2\left(M-K\right)+1}{G}_1^{M+1}-q^{2M-K+2}{G}_1^{M-K}\right)G_{M+2}^{N+K}+q^{M-K}{G}_1^{M-K}{G}_{M+2}^{N+K+1}$.

[1] has obtained the formal formula of ${\displaystyle {\sum }_{n=0}^{N-1}{q}^n}{\displaystyle {\prod }_{i=1}^M\left(K_i+D_i{q}^n\right)}$, but it cannot be generalized to ${\displaystyle {\sum }_{n_M=0}^{N-1}\cdots {\displaystyle {\sum }_{n_2=0}^{n_3}\cdots }{\displaystyle {\sum }_{n_1=0}^{n_2}\cdots }}$.

Notice $K_i+D_{i}n=K_i+D_i\left(\begin{array}{c}n\\ 1\end{array}\right)$, this inspired use $G_1^n$ instead of $q^n$.

The difficulty lies in the definition of ${\nabla }_q^P$.

Recursive define operator ${\nabla }_q^P$ :

${\nabla }_q^{0}f\left(N\right)=f\left(N\right),{\displaystyle {\sum }_{n=0}^{N-1}{q}^n{\nabla }_q^{1}f\left(n+1\right)}=f\left(N\right),{\displaystyle {\sum }_{n=0}^{N-1}{q}^{n}f\left(n+1\right)}={\nabla }_q^{-1}f(\; N\; )$

$SU{M}_q\left(N,\left[K_1:D_1\right],\left[T_1=1\right]\right)={\displaystyle {\sum }_{n=0}^{N-1}{q}^n}\left(D_1{G}_1^n+K_1\right)$

$\begin{array}{l}SU{M}_q\left(N,\left[K_1:D_1,K_2:D_2\right],\left[T_1=1,T_2=T_1+2-P\right]\right)\\ ={\displaystyle {\sum }_{n=0}^{N-1}{q}^n{\nabla }^{P}SU{M}_q\left(n+1,\left[K_1:D_1\right],\left[1\right]\right)\left(D_2{G}_2^n+K_2\right)}\end{array}$

Recursive definie $SU{M}_q\left(N,\left[K_1:D_1,K_2:D_2,\cdots \right],PT\right)$.

5.5) ${\nabla }^{1}SU{M}_q\left(N,\left[K_1:D_1,K_2:D_2,\cdots \right],\left[1,2,\cdots ,M\right]\right)={\displaystyle {\prod }_{i=1}^M\left(D_i{G}_1^n+K_i\right)}$

5.6) $\begin{array}{l}SU{M}_q\left(N,\left[K_1:D_1,K_2:D_2,\cdots \right],\left[1,3,\cdots ,2M-1\right]\right)\\ ={\displaystyle {\sum }_{n_M=0}^{N-1}\cdots {\displaystyle {\sum }_{n_2=0}^{n_3}{q}^{n_2}}}\left(D_2{G}_1^{n_2}+K_2\right){\displaystyle {\sum }_{n_1=0}^{n_2}{q}^{n_1}}\left(D_1{G}_1^{n_1}+K_1\right)\end{array}$

The Form: $\left(T_1+K_1\right)\left(T_2+K_2\right)\cdots \left(T_M+K_M\right)={\displaystyle \sum {\displaystyle {\prod }_{i=1}^M{X}_i}}$

5.7) $H^q\left(g\right)=H^q\left(PS,PT,g\right)={\displaystyle {\sum }_{X\left(T\right)=g}{\displaystyle {\prod }_{i=1}^M{B}_i}},SU{M}_q\left(N\right)=$

$\{\begin{array}{l}\stackrel{{\text{Form}}_1}{\to }{\displaystyle {\sum }_{g=0}^M{H}_1^q\left(g\right)G_{N-1-\text{g}}^{N+T_M-M}}={\displaystyle {\sum }_{g=0}^M{H}_1^q\left(g\right)G_{T_M-M+1+\text{g}}^{N+T_M-M}}\\ \stackrel{{\text{Form}}_2}{\to }{\displaystyle {\sum }_{g=0}^M{H}_2^q\left(g\right)G_{N-1}^{N+T_M-M+g}}={\displaystyle {\sum }_{g=0}^M{H}_2^q\left(g\right)G_{T_M-M+1+\text{g}}^{N+T_M-M+g}}\\ \stackrel{{\text{Form}}_3}{\to }{\displaystyle {\sum }_{g=0}^M{H}_3^q\left(g\right)G_{N-1-g}^{N+T_M-g}}={\displaystyle {\sum }_{g=0}^M{H}_3^q\left(g\right)G_{T_M+1}^{N+T_M+g}}\end{array}$

$\{\begin{array}{l}{B}_i\stackrel{{\text{Form}}_1}{\to }\{\begin{array}{l}{q}^{\left\{T_i-T_{i-1}\right\}{X}_{T-1}+1}{G}_1^{T_i-X_{K-1}}{D}_i;X_i=T_i\\ q^{\left(T_i-T_{i-1}-1\right)X_{T-1}}\left(K_i+G_1^{X_{T-1}}{D}_i\right);X_i=K_i\end{array}\\ B_i\stackrel{{\text{Form}}_2}{\to }\{\begin{array}{l}{q}^{-\left(T_i-X_{K-1}\right)}{G}_1^{T_i-X_{K-1}}{D}_i;X_i=T_i\\ K_i-q^{-\left(T_i-X_{K-1}\right)}{G}_1^{T_i-X_{K-1}}{D}_i;X_i=K_i\end{array}\\ B_i\stackrel{{\text{Form}}_3}{\to }\{\begin{array}{l}{q}^{\left(T_i-T_{i-1}-1\right)X_{T-1+1}}\left\{\left(q^{X_{T-1}}{G}_1^{T_i}-q^{T_i}{G}_1^{X_{T-1}}\right)D_i-K_i{q}^{T_i}\right\};X_i=T_i\\ q^{\left(T_i-T_{i-1}-1\right)X_{T-1}}\left(K_i+G_1^{X_{T-1}}{D}_i\right);X_i=K_i\end{array}\end{array}$

[Proof]

$\begin{array}{l}SU{M}_q\left(N,\left[K_1:D_1\right],\left[1\right]\right)\stackrel{{\text{Form}}_1}{\to }{q}^1{D}_1{G}_2^N+K_1{G}_1^N\\ \stackrel{{\text{Form}}_2}{\to }{q}^{-1}{D}_1{G}_2^{N+1}+\left(K_1-q^{-1}{D}_1\right)G_1^N\\ \stackrel{{\text{Form}}_3}{\to }\left(q^1{D}_1-K_1{q}^2\right)G_2^N+K_1{G}_2^{N+1}\end{array}$

If $f\left(N\right)={\displaystyle \sum A_i{G}_{M_i+1}^{N+M_i}}$, ${\nabla }_q^{P}f\left(N\right)={\displaystyle \sum A_i{G}_{M_i+1-P}^{N+M_i-P}}$, Form₂ is simplest.

Assume $SU{M}_q\left(N,PS,PT\right)={\displaystyle {\sum }_{g=0}^M{H}_2^q}\left(g\right)G_{T_M-M+1+g}^{N+T_M-M+g},X=T_M-M+1-P$

${\nabla }_q^{P}SU{M}_q\left(n+1,PS,PT\right)={\displaystyle {\sum }_{g=0}^M{H}_2^q}\left(g\right)G_{T_M-M+1+g-P}^{n+1+T_M-M+g-P}={\displaystyle {\sum }_{g=0}^M{H}_2^q}\left(g\right)G_{X+g}^{n+X+g}$

$\begin{array}{l}SU{M}_q\left(N,PSA,PTA\right)\\ ={\displaystyle {\sum }_{n=0}^{N-1}{q}^n}{\displaystyle {\sum }_{g=0}^M{H}_2^q}\left(g\right)G_{X+g}^{n+X+g}\left(D_{M+1}{G}_1^n+K_{M+1}\right)\\ ={\displaystyle {\sum }_{g=0}^M{D}_{M+1}{q}^{-X-g-1}{G}_1^{X+g+1}{H}_2^q\left(g\right)G_{X+g+2}^{N+X+g+1}}\\ +{\displaystyle {\sum }_{g=0}^M\left(K_{M+1}-q^{-X-g-1}{G}_1^{X+g+1}{D}_{M+1}\right)H_2^q\left(g\right)G_{X+g+1}^{N+X+g}}\end{array}$

$\begin{array}{l}={\displaystyle {\sum }_{g=0}^M{D}_{M+1}{q}^{-\left(T_{M+1}-\left[M-g\right]\right)}{G}_1^{T_{M+1}-\left[M-g\right]}{H}_2^q\left(g\right)G_{T_{M+1}-\left(M+1\right)+g+2}^{N+T_{M+1}-\left(M+1\right)+g+1}}\\ +{\displaystyle {\sum }_{g=0}^M\left(K_{M+1}-q^{-\left(T_{M+1}-\left[M-g\right]\right)}{G}_1^{T_{M+1}-\left[M-g\right]}{D}_{M+1}\right)H_2^q\left(g\right)G_{T_{M+1}-\left(M+1\right)+g+1}^{N+T_{M+1}-\left(M+1\right)+g}}\\ ={\displaystyle {\sum }_{g=0}^{M+1}{H}_2^q}\left(PSA,PTA,g\right)G_{N-1}^{N+T_{M+1}-\left(M+1\right)+g}\to {\text{Form}}_2\end{array}$

If $f\left(N\right)={\displaystyle \sum A_i{G}_{M_i}^{N+K}},{\nabla }_q^{P}f\left(N\right)={\displaystyle \sum A_i{q}^{-\left(M_i-K\right)P}{G}_{M_i-P}^{N+K-P}}$

Assume $SU{M}_q\left(N,PS,PT\right)={\displaystyle {\sum }_{g=0}^M{H}_1^q}\left(g\right)G_{T_M-M+1+g}^{N+T_M-M},X=T_M-M+1-P$

${\nabla }_q^{P}SUM\left(n+1\right)={\displaystyle {\sum }_{g=0}^M{q}^{-Pg}{H}_1^q\left(g\right)G_{T_M-M+1+g-P}^{n+1+T_M-M-P}}={\displaystyle {\sum }_{g=0}^M{q}^{-Pg}{H}_1^q\left(g\right)G_{X+g}^{n+X}}$

$\begin{array}{l}SU{M}_q\left(N,PSA,PTA\right)\\ ={\displaystyle {\sum }_{n=0}^{N-1}{q}^n}{\displaystyle {\sum }_{g=0}^M{q}^{-Pg}{H}_1^q}\left(g\right)G_{X+g}^{n+X}\times \left(D_{M+1}{G}_1^n+K_{M+1}\right)\\ ={\displaystyle {\sum }_{g=0}^M{q}^{-Pg}{D}_{M+1}{q}^{2g+1}{G}_1^{X+g+1}{H}_1^q\left(g\right)G_{X+g+2}^{N+X}}\\ +{\displaystyle {\sum }_{g=0}^M{q}^{-Pg}\left(D_{M+1}{q}^g{G}_1^g+K_{M+1}{q}^g\right)H_1^q\left(g\right)G_{X+g+1}^{N+X}}\end{array}$

$\begin{array}{l}={\displaystyle {\sum }_{g=0}^M{D}_{M+1}{q}^{\left(2-P\right)g+1}{G}_1^{T_{M+1}-\left[M-g\right]}{H}_1^q\left(g\right)G_{N-1-\left(g+1\right)}^{N+T_{M+1}-\left(M+1\right)}}\\ +{\displaystyle {\sum }_{g=0}^M{q}^{g\left(1-P\right)}}\left(G_1^g{D}_{M+1}+K_{M+1}\right)H_1^q\left(g\right)G_{N-1-g}^{N+T_{M+1}-\left(M+1\right)}\\ ={\displaystyle {\sum }_{g=0}^{M+1}{H}_1^q\left(PSA,PTA,g\right)G_{N-1-g}^{N+T_{M+1}-\left(M+1\right)}}\to {\text{Form}}_1\end{array}$

If $f\left(N\right)={\displaystyle \sum A_i{G}_M^{N+K_i}},{▽}_q^{P}f\left(N\right)={\displaystyle \sum A_i{q}^{-\left(M-K_i\right)P}{G}_{M-P}^{N+K_i-P}}\to {\text{Form}}_3$

q.e.d.

From the proof processà

5.8) ${\displaystyle {\sum }_{g=0}^M{H}_1^q\left(g\right)G_{B-g}^A}={\displaystyle {\sum }_{g=0}^M{H}_2^q\left(g\right)G_B^{A+g}}={\displaystyle {\sum }_{g=0}^M{H}_3^q\left(g\right)G_{B-g}^{A+M-g}}$

When regardless of the actual meaning, Form₁ = Form₂ = Form₃ is still established.

PT’s domain can be extended to $ℂ$.

$\begin{array}{l}SU{M}_q\left(N,\left[K_1,K_2\right],\left[1,2\right]\right)\\ \stackrel{{\text{Form}}_{\text{2}}}{\to }{q}^{-1}{q}^{-2}{G}_1^1{G}_1^2{G}_3^{N+2}+q^{-1}\left[\left(K_1-q^{-1}{G}_1^1\right)+\left(K_2-q^{-2}{G}_1^2\right)\right]G_2^{N+1}\\ +\left(K_1-q^{-1}{G}_1^1\right)\left(K_2-q^{-1}{G}_1^1\right)G_1^N\end{array}$

$\begin{array}{l}SU{M}_q\left(N,\left[K_1,K_2\right],\left[1,3\right]\right)\\ \stackrel{{\text{Form}}_{\text{2}}}{\to }{q}^{-1}{q}^{-3}{G}_1^1{G}_1^3{G}_4^{N+3}+\left[\left(K_1-q^{-1}{G}_1^1\right)q^{-2}{G}_1^2+q^{-1}\left(K_2-q^{-3}{G}_1^3\right)\right]G_3^{N+2}\\ +\left(K_1-q^{-1}{G}_1^1\right)\left(K_2-q^{-2}{G}_1^2\right)G_2^{N+1}\end{array}$

$\begin{array}{l}SU{M}_q\left(N,\left[K_1,K_2,K_3\right],\left[1,2,3\right]\right)\\ \stackrel{{\text{Form}}_{\text{2}}}{\to }{q}^{-1}{q}^{-2}{q}^{-3}{G}_1^1{G}_1^2{G}_1^3{G}_4^{N+3}+q^{-1}{q}^{-2}{G}_1^1{G}_1^2[\left(K_1-q^{-1}{G}_1^1\right)+\left(K_2-q^{-2}{G}_1^2\right)\\ +\left(K_3-q^{-3}{G}_1^3\right)]G_3^{N+2}+q^{-1}{G}_1^1{G}_2^{N+1}[\left(K_1-q^{-1}{G}_1^1\right)\left(K_2-q^{-1}{G}_1^1\right)\\ +\left(K_1-q^{-1}{G}_1^1\right)\left(K_3-q^{-2}{G}_1^2\right)+\left(K_2-q^{-2}{G}_1^2\right)\left(K_3-q^{-2}{G}_1^2\right)]\\ +\left(K_1-q^{-1}{G}_1^1\right)\left(K_2-q^{-1}{G}_1^1\right)\left(K_3-q^{-1}{G}_1^1\right)G_1^N\end{array}$

Form₂à

5.9) $SU{M}_q\left(N,\left[q^{-T_1}{G}_1^{T_1},q^{-T_2}{G}_1^{T_2},\cdots \right],PT\right)={\displaystyle {\prod }_{i=1}^M{q}^{-T_i}{G}_1^{T_i}{G}_{T_M+1}^{N+T_M}}$

5.10) $\begin{array}{l}SU{M}_q\left(N,\left[q^{-L_1}{G}_1^{L_1},\cdots ,q^{-L_p}{G}_1^{L_p},PS\right],\left[L_1,\cdots ,L_P,PT\right]\right)\\ ={\displaystyle {\prod }_{i=1}^P{q}^{-L_i}{G}_1^{L_i}SU{M}_q\left(N,PS,PT\right)}\to \text{extendsdomainof}{T}_1\text{from}1\text{to}\mathbb{N}\end{array}$

5.11) $SU{M}_q\left(N,PS,\left[T+1,T+2,\cdots ,T+M\right]\right)$ , $K_i$ can exchange order

5.12) $q^{Mn}=\frac{q^{-M}}{q^{-1}-1}{\displaystyle {\sum }_{g=0}^M{q}^g}\left(q^{-\left(\begin{array}{c}M\\ g\end{array}\right)}-1\right){\displaystyle {\prod }_{i=1}^g\left(q^n-q^{-i}\right)}$

[Proof]

$q^{Mn}=\nabla SU{M}_q\left(N,\left[1,1,\cdots \right]:q-1,\left[1,2,\cdots ,M\right]\right)$

$H_2^q\left(g,T\right)={\left(q-1\right)}^g{\displaystyle {\prod }_{i=1}^g{q}^{-i}{G}_1^i}$

$1-\frac{q^x-1}{q^x\left(q-1\right)}\left(q-1\right)=\frac{1}{q^x}\to H_2^q\left(g,{\displaystyle \sum K}\right)=q^{-M+g}{\displaystyle {\sum }_{i=0}^{\left(\begin{array}{c}M\\ g\end{array}\right)-1}{q}^{-i}}$

$q^{Mn}={\displaystyle {\sum }_{g=0}^M{\left(q-1\right)}^g}\left({\displaystyle {\prod }_{i=1}^g{q}^{-i}{G}_1^i}\right)q^{-M+g}\frac{q^{-\left(\begin{array}{c}M\\ g\end{array}\right)}-1}{q^{-1}-1}{G}_g^{n+g}$

q.e.d.

$SUM\left(N,\left[K_1+D_1:\left(q-1\right)D_1,K_2+D_2:\left(q-1\right)D_2,\cdots \right],\left[1,2,\cdots ,M\right]\right)\stackrel{{\text{Form}}_{\text{1}}}{\to }$

5.13) ${\displaystyle {\sum }_{n=0}^{N-1}{q}^n}{\displaystyle {\prod }_{i=1}^M\left(K_i+D_i{q}^n\right)}={\displaystyle {\sum }_{g=0}^{M}H\left(g\right)G_{g+1}^N}\to$ Conclusion of [1]

$H\left(g\right)={\displaystyle {\sum }_{X\left(T\right)=g}{\displaystyle {\prod }_{i=1}^M\{\begin{array}{l}{q}^{X_T}\left(q^{X_T}-1\right)D_i;X\in T\\ K_i+q^{X_{T-1}}{D}_i;X\in K\end{array}}}$

5.14) $q^{Mn}={\displaystyle {\sum }_{g=0}^M{q}^{-g}}{\displaystyle {\prod }_{i=1}^g{q}^i}\left(q^{n-g+i}-1\right)G_g^M$

[Proof]

$q^{Mn}=\nabla SU{M}_q\left(N,\left[1,1,,\cdots \right]:q-1,\left[1,2,\cdots ,M\right]\right)$

$H_1^q\left(g,T\right)={\left(q-1\right)}^g{\displaystyle {\prod }_{i=1}^g{q}^i{G}_1^i}$

$1+\frac{q^x-1}{q-1}\left(q-1\right)=q^x\to H_1^q\left(g,{\displaystyle \sum K}\right)={\displaystyle \sum {\displaystyle \prod q^{X_{T-1}}}}$

In 1916 MacMahon [5] observed that $G_K^M={\displaystyle {\sum }_{w\in \Omega \left(M,K\right)}{q}^{inv\left(w\right)}}$ $\Omega \left(M,K\right)$ denotes all permutations of the multiset {0^M−K, 1^K} that is, all words $w=w_1,\cdots ,w_n$ with n − k zeroes and k ones, and inv(・) denotes the inversion statistic defined by $inv\left(w_1,\cdots ,w_n\right)=\left|\left\{\left(i,j\right):1\le i<j\le n,w_i>w_j\right\}\right|$.

So ${\displaystyle \sum q^{X_{T-1}}}=G_{M-g}^M,H_1^q\left(g\right)={\displaystyle {\sum }_{g=0}^M{\left(q-1\right)}^g}\left({\displaystyle {\prod }_{i=1}^g{q}^i{G}_1^i}\right)G_{M-g}^M$

$q^{Mn}=\nabla SU{M}_q\left(N\right)={\displaystyle {\sum }_{g=0}^M{\left(q-1\right)}^g}\left({\displaystyle {\prod }_{i=1}^g{q}^i{G}_1^i}\right)G_{M-g}^M{q}^{-g}{G}_g^n$

q.e.d.

$\begin{array}{l}\frac{q^{Mn}-1}{q^M-1}=\frac{{\displaystyle {\sum }_{g=1}^M{\left(q-1\right)}^g}\left({\displaystyle {\prod }_{i=1}^g{q}^i{G}_1^i}\right)G_{M-g}^M{q}^{-g}{G}_g^n}{q^M-1}\\ =\frac{{\displaystyle {\sum }_{g=1}^M\left({\displaystyle {\prod }_{i=1}^g{q}^i}\left(q^i-1\right)\right)G_g^M{q}^{-g}{G}_g^n}}{q^M-1}\to \end{array}$

5.15) $\frac{q^{Mn}-1}{q^M-1}={\displaystyle {\sum }_{g=1}^M\left({\displaystyle {\prod }_{i=1}^{g-1}{q}^i\left(q^i-1\right)}\right)G_{M-g}^{M-1}{G}_g^n}\to$ Conclusion of [1]

5.16) $P_1<T_1,P_2<T_2,T_1>0,T_2>0$

$PS=\left[\frac{G_1^1}{q^1},\frac{G_1^2}{q^2},\cdots ,\frac{G_1^{T_1-P_1}}{q^{T_1-P_1}},\frac{K_1{G}_1^{T_1+1-P_1}}{q^{1-P_1}{G}_1^{T_1}},\frac{G_1^1}{q^1},\frac{G_1^2}{q^2},\cdots ,\frac{G_1^{T_2-P_2}}{q^{T_2-P_2}},\frac{K_2{G}_1^{T_2+1-P_2}}{q^{1-P_2}{G}_1^{T_2}}\right]$

$PT=\left[1,2,3,\cdots ,T_1+T_2+2-P_1-P_2\right]$

$\begin{array}{l}{\displaystyle {\sum }_{n=0}^{N-1}{q}^n{\nabla }^{P_1}SU{M}_q\left(n+1,\left[K_1\right],\left[T_1\right]\right){\nabla }^{P_2}SU{M}_q\left(n+1,\left[K_2\right],\left[T_2\right]\right)}\\ =\frac{q^1}{G_1^1}\frac{q^2}{G_1^2}\cdots \frac{q^{T_1-P_1}}{G_1^{T_1-P_1}}\frac{q^{1-P_1}{G}_1^{T_1}}{G_1^{T_1-P_1+1}}\frac{q^1}{G_1^1}\frac{q^2}{G_1^2}\cdots \frac{q^{T_2-P_2}}{G_1^{T_2-P_2}}\frac{q^{1-P_2}{G}_1^{T_2}}{G_1^{T_2-P_2+1}}SUM\left(N,PS,PT\right)\end{array}$

[Proof]

$\begin{array}{c}\frac{q^{n+K}-1}{q^K-1}=\frac{q^{n+K}-1}{q^K-1}\frac{q^1-1}{q^1-1}=\frac{q^K\left(q^n-1\right)+q^K-1}{G_1^K{q}^1-1}\\ =\frac{q^K}{G_1^K}{G}_1^n+\frac{G_1^K}{G_1^K}=\frac{q^K}{G_1^K}\left(G_1^n+G_1^K{q}^{-K}\right)\end{array}$

$\begin{array}{l}{\nabla }^{P}SU{M}_q\left(N,\left[K\right],\left[T\right]\right)\\ \stackrel{{\text{Form}}_{\text{2}}}{\to }{q}^{-T}{G}_1^T\left(\begin{array}{c}n+T+1-P\\ T+1-P\end{array}\right)+\left(K-q^{-T}{G}_1^T\right)\left(\begin{array}{c}n+T-P\\ T-P\end{array}\right)\\ =q^{-T}{G}_1^T\frac{\left(q^{n+T+1-P}-1\right)\cdots \left(q^{n+1}-1\right)}{\left(q^{T+1-P}-1\right)\cdots \left(q-1\right)}+\left(K-q^{-T}{G}_1^T\right)\frac{\left(q^{n+T-P}-1\right)\cdots \left(q^{n+1}-1\right)}{\left(q^{T-P}-1\right)\cdots \left(q-1\right)}\end{array}$

$\begin{array}{l}=\frac{\left(q^{n+T-P}-1\right)\cdots \left(q^{n+1}-1\right)}{\left(q^{T-P}-1\right)\cdots \left(q-1\right)}\left(K-q^{-T}{G}_1^T+q^{-T}{G}_1^T\frac{q^{n+T+1-P}-1}{q^{T+1-P}-1}\right)\\ =\frac{\left(q^{n+T-P}-1\right)\cdots \left(q^{n+1}-1\right)}{\left(q^{T-P}-1\right)\cdots \left(q-1\right)}\left(K-q^{-T}{G}_1^T+q^{-T}{G}_1^T\frac{1}{G_1^{T+1-P}}\left(q^{T+1-P}{G}_1^n+G_1^{T+1-P}\right)\right)\\ =\frac{\left(q^{n+T-P}-1\right)\cdots \left(q^{n+1}-1\right)}{\left(q^{T-P}-1\right)\cdots \left(q-1\right)}\frac{q^{1-P}{G}_1^T}{G_1^{T+1-P}}\left(\frac{K{G}_1^{T+1-P}}{q^{1-P}{G}_1^T}+G_1^n\right)\end{array}$

q.e.d.

5.17) ${\lim }_{q\to 1}{H}^q\left(g\right)=H\left(g\right),{\lim }_{q\to 1}SU{M}_q\left(N\right)=SUM(\; N\; )$

[1] has a conclusion:

$\begin{array}{l}{H}_1\left(q\right)={\displaystyle {\sum }_{x=q}^M{H}_2\left(x\right)}\left(\begin{array}{c}x\\ q\end{array}\right)={\displaystyle {\sum }_{x=0}^q{H}_3\left(x\right)}\left(\begin{array}{c}M-x\\ M-q\end{array}\right)\\ H_2\left(q\right)={\displaystyle {\sum }_{x=q}^M{\left(-1\right)}^{x+q}{H}_1\left(x\right)}\left(\begin{array}{c}x\\ q\end{array}\right)\\ H_3\left(q\right)={\displaystyle {\sum }_{x=0}^q{\left(-1\right)}^{x+q}{H}_1\left(x\right)}\left(\begin{array}{c}M-x\\ M-q\end{array}\right)\end{array}$

Generally, H^q(g) has no such attribute; things get complicated.

When $PT=\left[1,2,\cdots ,M\right]$, the situation is relatively simple.

5.18) $PT=\left[1,2,\cdots ,M\right],H_1^q\left(g\right)={\displaystyle {\sum }_{x=g}^M{H}_2^q\left(x\right)G_g^x{q}^{2\left(\begin{array}{c}g+1\\ 2\end{array}\right)}}$

[Proof]

This is to prove: $H_1^q\left(g\right)q^{-\left(\begin{array}{c}g+1\\ 2\end{array}\right)}=\left({\displaystyle {\sum }_{x=g}^M{H}_2^q\left(x\right)G_g^x}\right)q^{\left(\begin{array}{c}g+1\\ 2\end{array}\right)}$

$\to H_1^q\left(g\right){\left({\displaystyle {\prod }_{i=1}^g{q}^{\left\{T_i-T_{i-1}\right\}{X}_{T-1}+1}}\right)}^{-1}=\left({\displaystyle {\sum }_{x=g}^M{H}_2^q\left(x\right)G_g^x}\right){\left({\displaystyle {\prod }_{i=1}^g{q}^{-\left(T_i-X_{K-1}\right)}}\right)}^{-1}$

$PS1=\left[PS,K_{M+1}:D_{M+1}\right],PT1=\left[PT,M+1\right]$

Suppose it is true at M, $H_1^q\left(g\right)=\left({\displaystyle {\sum }_{x=g}^M\dots }\right)=\left({\displaystyle {\sum }_{x=0}^M\dots }\right)$

$\begin{array}{l}{H}_1^q\left(PS1,PT1,g\right)\\ =q^{X_{T-1}+1}{G}_1^{T_{M+1}-X_{K-1},\text{choice}T}{D}_{M+1}{H}_1^q\left(g-1\right)+\left(K_{M+1}+G_1^{X_{T-1},\text{choice}K}{D}_{M+1}\right)H_1^q\left(g\right)\\ =q^g{G}_1^g{D}_{M+1}{H}_1^q\left(g-1\right)+\left(K_{M+1}+G_1^g{D}_{M+1}\right)H_1^q(\; g\; )\end{array}$

$\begin{array}{l}=q^g{G}_1^g{D}_{M+1}{\displaystyle {\sum }_{x=0}^M{H}_2^q}\left(x\right)\left\{G_{g-1}^x=G_g^{x+1}-q^g{G}_g^x\right\}{q}^{2\left(\begin{array}{c}g\\ 2\end{array}\right)=2\left(\begin{array}{c}g+1\\ 2\end{array}\right)-2g}\\ +\left(K_{M+1}+G_1^g{D}_{M+1}\right){\displaystyle {\sum }_{x=0}^M{H}_2^q}\left(x\right)G_g^x{q}^{2\left(\begin{array}{c}g+1\\ 2\end{array}\right)}\\ =q^{-g}{G}_1^g{D}_{M+1}{\displaystyle {\sum }_{x=0}^M{H}_2^q}\left(x\right)G_g^{x+1}{q}^{2\left(\begin{array}{c}g+1\\ 2\end{array}\right)}+K_{M+1}{\displaystyle {\sum }_{x=0}^M{H}_2^q}\left(x\right)G_g^x{q}^{2\left(\begin{array}{c}g+1\\ 2\end{array}\right)}\end{array}$

$\begin{array}{l}{q}^{2\left(\begin{array}{c}g+1\\ 2\end{array}\right)}{\displaystyle {\sum }_{x=0}^{M+1}{H}_2^q}\left(PS1,PT1,x\right)G_g^x\\ =q^{2\left(\begin{array}{c}g+1\\ 2\end{array}\right)}\{{\displaystyle {\sum }_{x=1}^{M+1}{H}_2^q}\left(x-1\right)q^{-x}{G}_1^x{D}_{M+1}{G}_g^x\\ +{\displaystyle {\sum }_{x=0}^M{H}_2^q}\left(x\right)\left(K_{M+1}-q^{-x-1}{G}_1^{x+1}{D}_{M+1}\right)G_g^x\}\end{array}$

$\begin{array}{l}=q^{2\left(\begin{array}{c}g+1\\ 2\end{array}\right)}\{{\displaystyle {\sum }_{x=0}^M{H}_2^q}\left(x\right)q^{-x-1}{G}_1^{x+1}{D}_{M+1}{G}_g^{x+1}\\ +{\displaystyle {\sum }_{x=0}^M{H}_2^q}\left(x\right)\left(K_{M+1}-q^{-x-1}{G}_1^{x+1}{D}_{M+1}\right)G_g^x\}\end{array}$

$\begin{array}{l}{H}_1^q\left(PS1,PT1,g\right)-q^{2\left(\begin{array}{c}g+1\\ 2\end{array}\right)}{\displaystyle {\sum }_{x=0}^{M+1}{H}_2^q}\left(PS1,PT1,x\right)G_g^x\\ =q^{2\left(\begin{array}{c}g+1\\ 2\end{array}\right)}{D}_{M+1}{\displaystyle {\sum }_{x=0}^M{H}_2}\left(x\right)\left\{\left(q^{-g}{G}_1^g-q^{-x-1}{G}_1^{x+1}\right)G_g^{x+1}+q^{-x-1}{G}_1^{x+1}{G}_g^x\right\}\\ \stackrel{\left\{\mathrm{..}\right\}=0}{\to }=0\end{array}$

q.e.d.

6. Multiparameter Forms

(1.1) and (5.7) Use the Form: $\left(T_1+K_1\right)\left(T_2+K_2\right)\cdots \left(T_M+K_M\right)$

The form has T and K parameters, and more parameters will be used in this section. This section moves the D_i to PT.

$PS=\left[K_1,K_2,\cdots ,K_M\right],PSA=\left[PS,K_{M+1}\right]$

$P{T}_1=\left[T_{1,1}:D_{1,1},T_{2,1}:D_{2,1},\cdots ,T_{M,1}:D_{M,1}\right]=\left[T_1:D_{1,1},T_2:D_{2,1},\cdots ,T_M:D_{M,1}\right]$

$P{T}_2=\left[T_{1,2}:D_{1,2},T_{2,2}:D_{2,2},\cdots ,T_{M,2}:D_{M,2}\right]=\left[T_1:D_{1,2},T_2:D_{1,2},\cdots ,T_M:D_{M,2}\right]$

$PT{A}_1=\left[P{T}_1,T_{M+1}=T_M+2-p:D_{M+1,1}\right],PT{A}_2=\left[P{T}_2,T_{M+1}:D_{M+1,2}\right]$

Define

$SUM\left(N,\left[K_1\right],\left[T_{1,1}=1:D_{1,1}\right],\left[T_{1,2}=1:D_{1,2}\right]\right)={\displaystyle {\sum }_{n=0}^{N-1}\left(K_1+n{D}_{1,1}+\left(\begin{array}{c}n\\ 2\end{array}\right)D_{1,2}\right)}$

$\begin{array}{c}SUM\left(N\right)=SUM\left(N,PS,PT{A}_1,PT{A}_2\right)\\ ={\displaystyle {\sum }_{n=0}^{N-1}\left(K_{M+1}+n\times D_{M+1,1}+\left(\begin{array}{c}n\\ 2\end{array}\right)\times D_{M+1,1}\right)\times {\nabla }^{p}SUM\left(n+1\right)}\end{array}$

$\left(K_1+T_{1,1}+T_{1,2}\right)\left(K_2+T_{2,1}+T_{2,2}\right)\cdots \left(K_M+T_{M,1}+T_{M,2}\right)={\displaystyle \sum {\displaystyle {\prod }_{i=1}^M{X}_i}}$

$X\left(P{T}_1\right),X\left(P{T}_2\right)=\text{countof}X\in P{T}_1,P{T}_2;X\left(PT\right)=X\left(P{T}_1\right)+2X\left(P{T}_2\right)$

$X_{P{T}_1},X_{P{T}_2}=\text{countof}\left\{X_1,X_2,\cdots ,X_i\right\}\in P{T}_1,P{T}_2;X_{PT}=X_{P{T}_1}+2X_{P{T}_2}$

6.1) $\begin{array}{l}H\left(q\right)={\displaystyle {\sum }_{{\displaystyle \prod X_i}\text{with}X\left(PT\right)=q}{\displaystyle {\prod }_{i=1}^M{B}_i}},\\ SUM\left(N\right)\stackrel{{\text{Form}}_{\text{1}}}{\to }{\displaystyle {\sum }_{q=0}^{2M}{H}_1}\left(q\right)\left(\begin{array}{c}N+T_M-M\\ T_M-M+1+q\end{array}\right)\end{array}$

$\stackrel{{\text{Form}}_1}{\to }{B}_i=\{\begin{array}{l}\left(\begin{array}{c}{X}_{PT}\\ 0\end{array}\right)K_i+\left(\begin{array}{c}{X}_{PT}\\ 1\end{array}\right)D_{i,1}+\left(\begin{array}{c}{X}_{PT}\\ 2\end{array}\right)D_{i,2};X_i=K_i\\ \left(\begin{array}{c}{T}_i+X_{PT}-i\\ 1\end{array}\right)D_{i,1}+\left(\begin{array}{c}{T}_i+X_{PT}-i\\ 1\end{array}\right)\left(\begin{array}{c}{X}_{PT}-1\\ 1\end{array}\right)D_{i,2};X_i=P{T}_{i,1}\\ \left(\begin{array}{c}{T}_i+X_{PT}-i\\ 2\end{array}\right)D_{i,2};X_i=P{T}_{i,2}\end{array}$

[Proof]

(*) ${\displaystyle {\sum }_{n=0}^{N-1}n}\left(\begin{array}{c}n+K\\ M\end{array}\right)=\left(M+1\right)\left(\begin{array}{c}N+K\\ M+2\end{array}\right)+\left(M-K\right)\left(\begin{array}{c}N+K\\ M+1\end{array}\right)$

(**) $\begin{array}{c}{\displaystyle {\sum }_{n=0}^{N-1}\left(\begin{array}{c}n\\ 2\end{array}\right)\left(\begin{array}{c}n+K\\ M\end{array}\right)}=\left(\begin{array}{c}M+2\\ 2\end{array}\right)\left(\begin{array}{c}N+K\\ M+3\end{array}\right)+\left(M-k\right)\left(M+1\right)\left(\begin{array}{c}N+K\\ M+2\end{array}\right)\\ +(\begin{array}{c}M\end{array}\end{array}$