Application and Popularization of Formal Calculation

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.

Share and Cite:

Peng, J. (2022) Application and Popularization of Formal Calculation. Open Access Library Journal, 9, 1-21. doi: 10.4236/oalib.1109483.

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: $Seri{e}_{i}=\left\{{K}_{i},{K}_{i}+{D}_{i},{K}_{i}+2{D}_{i},\cdots ,{K}_{i}+\left(N-1\right){D}_{i}\right\}$, $i\in \left[1,M\right]$

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:

$PS=\left[{K}_{1}:{D}_{1},{K}_{2}:{D}_{2},\cdots ,{K}_{M}:{D}_{M}\right],PT=\left[{T}_{1},{T}_{2},\cdots ,{T}_{M}\right]$

$PSA=\left[{K}_{1}:{D}_{1},\cdots ,{K}_{M}:{D}_{M},{K}_{M+1}:{D}_{M+1}\right]=\left[PS,\text{}{K}_{M+1}:{D}_{M+1}\right],$

$PTA=\left[PT,\text{}{T}_{M+1}=\text{}{T}_{M}+2-p\right]$

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

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

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

$SUM\left(N,\left[{K}_{1}:{D}_{1}\right],\left[{T}_{1}=1\right]\right)={\sum }_{n=0}^{N-1}\left({K}_{1}+n×{D}_{1}\right)$

$SUM\left(N,PSA,PTA\right)={\sum }_{n=0}^{N-1}\left({K}_{M+1}+n×{D}_{M+1}\right)×{\nabla }^{p}SUM\left(n+1\right)$

For example:

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

$\begin{array}{l}SUM\left(N,PS,\left[1,3,\cdots ,2M-1\right]\right)\\ ={\sum }_{{n}_{M}=0}^{N-1}\left({K}_{M}+{n}_{M}{D}_{M}\right)\cdots {\sum }_{{n}_{1}=0}^{{n}_{2}}\left({K}_{2}+{n}_{2}{D}_{2}\right){\sum }_{{n}_{1}=0}^{{n}_{2}}\left({K}_{1}+{n}_{1}{D}_{1}\right)\end{array}$

$SUM\left(N,PS,\left[1,2,4\right]\right)={\sum }_{{n}_{3}=0}^{N-1}\left({K}_{3}+{n}_{3}{D}_{3}\right){\sum }_{n=0}^{{n}_{3}}\left({K}_{2}+n{D}_{1}\right)\left({K}_{1}+n{D}_{1}\right)$

$SUM\left(N,PS,\left[1,3,4\right]\right)={\sum }_{n=0}^{N-1}\left({K}_{3}+n{D}_{3}\right)\left({K}_{2}+n{D}_{2}\right){\sum }_{{n}_{1}=0}^{n}\left({K}_{1}+{n}_{1}{D}_{1}\right)$

$SUM\left(N,PS,\left[1,4\right]\right)={\sum }_{n=0}^{N-1}\left({K}_{2}+n{D}_{2}\right){\sum }_{{n}_{2}=0}^{n}{\sum }_{{n}_{1}=0}^{{n}_{2}}\left({K}_{1}+{n}_{1}{D}_{1}\right)$

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: $\left({T}_{1}+{K}_{1}\right)\left({T}_{2}+{K}_{2}\right)\cdots \left({T}_{M}+{K}_{M}\right)=\sum {\prod }_{i=1}^{M}{X}_{i}$, ${X}_{i}={T}_{i}$ or ${K}_{i}$

$X\left(T\right)=\text{Countof}\text{\hspace{0.17em}}\left\{{X}_{1},\cdots ,{X}_{M}\right\}\in T$,

${X}_{T-1}=\text{Countof}\text{\hspace{0.17em}}\left\{{X}_{1},\cdots ,{X}_{i-1}\right\}\in T$, ${X}_{K-1}=\text{Countof}\text{\hspace{0.17em}}\left\{{X}_{1},\cdots ,{X}_{i-1}\right\}\in K$.

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

1.1) $H\left(q\right)={\sum }_{\prod {X}_{i}\text{\hspace{0.17em}}\text{with}\text{\hspace{0.17em}}X\left(T\right)=q}{\prod }_{i=1}^{M}{B}_{i}$, $q=X\left(T\right)$, $SUM\left(N\right)=$

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

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

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

$\stackrel{{\text{Form}}_{\text{1}}}{\to }{B}_{i}=\left\{\begin{array}{l}\left({T}_{i}-{X}_{K-1}\right){D}_{i};{X}_{i}={T}_{i}\\ {K}_{i}+{X}_{T-1}{D}_{i};{X}_{i}={K}_{i}\end{array}$

$\stackrel{{\text{Form}}_{\text{2}}}{\to }{B}_{i}=\left\{\begin{array}{l}\left({T}_{i}-{X}_{K-1}\right){D}_{i};{X}_{i}={T}_{i}\\ {K}_{i}+\left({X}_{K-1}-{T}_{i}\right){D}_{i};{X}_{i}={K}_{i}\end{array}$

$\stackrel{{\text{Form}}_{\text{3}}}{\to }{B}_{i}=\left\{\begin{array}{l}-{K}_{i}+\left({T}_{i}-{X}_{T-1}\right){D}_{i};{X}_{i}={T}_{i}\\ {K}_{i}+{X}_{T-1}{D}_{i};{X}_{i}={K}_{i}\end{array}$

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 $f\left(n\right)=\sum {A}_{i}\left(\begin{array}{c}{N}_{i}\\ {m}_{i}\end{array}\right)$, ${m}_{i}$ is not changed with n, then ${\nabla }^{p}f\left(n\right)=\sum {A}_{i}\left(\begin{array}{c}{N}_{i}-p\\ {m}_{i}-p\end{array}\right)$

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

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

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

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

1.9) ${\sum }_{q=0}^{M}{H}_{1}\left(q\right)\left(\begin{array}{c}A\\ B-q\end{array}\right)={\sum }_{q=0}^{M}{H}_{2}\left(q\right)\left(\begin{array}{c}A+q\\ B\end{array}\right)={\sum }_{q=0}^{M}{H}_{3}\left(q\right)\left(\begin{array}{c}A+M-q\\ B-q\end{array}\right)$

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

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

Sometimes simple expressions can be obtained.

Define

${F}_{q}^{K=\left\{{K}_{1},{K}_{2},\cdots \right\}}=\sum {\prod }_{i=1}^{q}{I}_{i}$, ${I}_{i}\in K$ and ${I}_{i}\ne {I}_{j}$. ${F}_{q}^{\left\{1,2,\cdots ,N\right\}}$ abbreviated as ${F}_{q}^{N}$

${E}_{q}^{K=\left\{{K}_{1},{K}_{2},\cdots \right\}}=\sum {\prod }_{i=1}^{q}{I}_{i}$, ${I}_{i}\in K$, ${E}_{q}^{\left\{1,2,\cdots ,N\right\}}$ abbreviated as ${E}_{q}^{N}$

${F}_{q}^{N}=\underset{1\le {\lambda }_{1}<{\lambda }_{2}<\cdots <{\lambda }_{q}\le N}{\sum }\prod \lambda ={S}_{1}\left(N+q,N\right)$

${E}_{q}^{N}=\underset{1\le {\lambda }_{1}<{\lambda }_{2}<\cdots <{\lambda }_{q}\le N}{\sum }\prod \lambda =\underset{{\lambda }_{1}+{\lambda }_{2}+\cdots +{\lambda }_{N}=q}{\sum }{1}^{{\lambda }_{1}}{2}^{{\lambda }_{2}}\cdots {N}^{{\lambda }_{N}}={S}_{2}\left(N+q,N\right)$

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

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

$H\left(q,T\right)=\left(\prod X\in T\right)\text{\hspace{0.17em}}\text{of}\text{\hspace{0.17em}}H\left(q\right),\text{\hspace{0.17em}}H\left(q,K\right)=\left(\prod X\in K\right)\text{\hspace{0.17em}}\text{of}\text{\hspace{0.17em}}H\left( q \right)$

$H\left(q,\sum T\right)=\sum H\left(q,T\right),\text{\hspace{0.17em}}H\left(q,\sum K\right)=\sum H\left(q,K\right)$

$PT=\left[{T}_{1}=T+1,\cdots ,{T}_{M}=T+M\right]\to {H}_{1}\left(q,T\right)={H}_{2}\left(q,T\right)={D}^{q}{T}_{1}\cdots {T}_{q}$

$PT=\left[1,2,\cdots ,M\right],D=1\to {H}_{1}\left(q,\sum K\right)={F}_{M-q}^{K}{E}_{0}^{q}+{F}_{M-q-1}^{K}{E}_{1}^{q}+\cdots +{F}_{0}^{K}{E}_{M-q}^{q}$

$f\left(n\right)={a}_{0}+{a}_{1}n+{a}_{2}{n}^{2}+\cdots +{a}_{M}{n}^{M}={a}_{M}\left({K}_{1}+n\right)\left({K}_{2}+n\right)\cdots \left({K}_{M}+n\right)$

$F\left(N\right)={\sum }_{n=0}^{N-1}f\left(n\right)={a}_{M}SUM\left(N,\left[{K}_{1},{K}_{2},\cdots ,{K}_{M}\right],\left[1,2,\cdots ,M\right]\right)$

According to Vieta’s formulas, ${F}_{i}^{K}=\frac{{a}_{M-i}}{{a}_{M}}$

${H}_{1}\left(q\right)=q!{\sum }_{i=0}^{M-q}{F}_{M-q-i}^{K}{E}_{i}^{q}=q!{\sum }_{i=0}^{M-q}\frac{{a}_{q+i}}{{a}_{M}}{E}_{i}^{q}$

${a}_{M}{H}_{1}\left(0\right)=0!{a}_{0}$

${a}_{M}{H}_{1}\left(1\right)=1!\left({a}_{1}+{a}_{2}+\cdots +{a}_{M}\right)$

$\begin{array}{c}{a}_{M}{H}_{1}\left(2\right)=2!\left({a}_{2}+3{a}_{3}+7{a}_{4}+\cdots \right)\\ =2!\left({S}_{2}\left(2,2\right){a}_{2}+{S}_{2}\left(3,2\right){a}_{3}+{S}_{2}\left(4,2\right){a}_{4}+\cdots \right)\end{array}$

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

${a}_{M}\left[\begin{array}{c}{H}_{1}\left(0\right)\\ ⋮\\ {H}_{1}\left(M\right)\end{array}\right]=\left[\begin{array}{c}0!\\ ⋮\\ M!\end{array}\right]\left[\begin{array}{ccc}{S}_{2}\left(0,0\right)& \cdots & {S}_{2}\left(M,0\right)\\ ⋮& \ddots & ⋮\\ {S}_{2}\left(0,M\right)& \cdots & {S}_{2}\left(M,M\right)\end{array}\right]\left[\begin{array}{c}{a}_{0}\\ ⋮\\ {a}_{M}\end{array}\right]$

It’s the same as: $F\left(N\right)={\sum }_{q=0}^{M}{a}_{q}{\sum }_{n=0}^{N-1}{n}^{q}={\sum }_{q=0}^{M}{a}_{q}{\sum }_{i=1}^{q}i!{S}_{2}\left(q,i\right)\left(\begin{array}{c}N\\ i+1\end{array}\right)$

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

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

${H}_{1}\left(q,K\right)={\left[P:D\right]}_{M-q}$,

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

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

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

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

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

[Proof]

$\begin{array}{l}=\frac{1}{A!M!}\nabla SUM\left(N,\left[1,2,\cdots ,A,B+1,B+2,\cdots ,B+M\right],\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left[1,2,\cdots ,A,A+1,\cdots ,A+M\right]\right)\\ =\frac{1}{M!}\nabla SUM\left(N,\left[B+1,B+2,\cdots ,B+M\right],\left[A+1,A+2,\cdots ,A+M\right]\right)\\ =\frac{1}{M!}\nabla SUM\left(N,\left[B+M,\cdots ,B+2,B+1\right],\left[A+1,A+2,\cdots ,A+M\right]\right)\end{array}$

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

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

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

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

q.e.d.

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

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

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

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

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

${H}_{1}\left(q\right)\ne 0\to {X}_{1},{X}_{2},\cdots ,{X}_{B}\in T\to {H}_{1}\left(q

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

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

[Proof]

$\begin{array}{l}=\frac{1}{A!M!}\nabla SUM\left(N,\left[X,X-1,\cdots ,X-A+1,Y,Y-1,\cdots ,Y-M+1\right],\left[1,2,\cdots \right]\right)\\ =\frac{1}{A!M!}\nabla SUM\left(N,\left[1,2,\cdots ,Y,0,-1,\cdots ,-\left(M-Y\right)+1,X,\cdots ,X-A+1\right],\left[1,2,\cdots \right]\right)\\ =\frac{Y!}{A!M!}SUM\left(N,\left[0,-1,\cdots ,-\left(M-Y\right)+1,X,\cdots ,X-A+1\right],\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left[Y+1,Y+2,\cdots ,M+A\right]\right)\end{array}$

${H}_{1}\left(q\ge M-Y,\sum K\right)\stackrel{Z=M+A-Y}{\to }\left(\begin{array}{c}Z-\left(M-Y\right)\\ Z-q\end{array}\right){\left[X+\left(M-Y\right)\right]}_{Z-q}$

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

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

q.e.d.

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

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

[Proof]

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

$\sum {H}_{1}\left(q,K\right)=\left\{\begin{array}{l}0,q

q.e.d.

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

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

[Proof]

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

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

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

q.e.d.

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

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

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

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

q.e.d.

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

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

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

2.2.1)

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

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

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

Change M to M − 1 and q! to (q + 1)!à

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

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

${H}_{2}\left(q,K\right)={\left[P:D\right]}^{M-q}$

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

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

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

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

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

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]

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

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

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

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

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

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

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

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

${H}_{3}\left(q\ge A\right)=A!\left(q-A\right)!\left(\begin{array}{c}B\\ q-A\end{array}\right)\left(\begin{array}{c}B\\ q-A\end{array}\right){\left[A\right]}_{A+B-q}$

$\left(\begin{array}{c}n\\ A\end{array}\right)\left(\begin{array}{c}n\\ B\end{array}\right)=\frac{1}{A!B!}{\sum }_{q=A}^{A+B}{H}_{3}\left(q\right)\left(\begin{array}{c}n+A+B-q\\ A+B\end{array}\right)$

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

q.e.d.

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

[1] has obtained: ${H}_{1}\left(q\right)={\sum }_{x=q}^{M}{H}_{2}\left(x\right)\left(\begin{array}{c}x\\ q\end{array}\right)={\sum }_{x=0}^{q}{H}_{3}\left(x\right)\left(\begin{array}{c}M-x\\ M-q\end{array}\right)$

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

[Proof]

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

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

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

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

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

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

q.e.d.

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

2.5.2)

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

2.5.3)

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

2.6. Summary

3. r-Flod Sum

Define ${\sum }_{\left(r\right)}^{N}f\left(k\right)={\sum }_{{k}_{r}=1}^{N}\cdots {\sum }_{{k}_{2}=1}^{{k}_{3}}{\sum }_{k=1}^{{k}_{2}}f\left(k\right)={\sum }_{{k}_{r}=0}^{N-1}\cdots {\sum }_{{k}_{2}=0}^{{k}_{3}}{\sum }_{k=0}^{{k}_{2}}f\left(k+1\right)$

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

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

[Proof]

$PS1=\left[PS,1:0,1:0,\cdots \right],PT1=\left[PT,{T}_{M}+2,{T}_{M}+4,\cdots ,{T}_{M}+2\left(r-1\right)\right]$

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

In ${H}_{1}\left(q,PS1,PT1\right)$, $i>M$, $\left\{\begin{array}{l}\left({T}_{i}-{X}_{K-1}\right){D}_{i}=0;{X}_{i}={T}_{i}\\ {K}_{i}+{X}_{T-1}{D}_{i}=1;{X}_{i}={K}_{i}\end{array}$

${H}_{1}\left(q>M,PS1,PT1\right)=0,{H}_{1}\left(q\le M,PS1,PT1\right)={H}_{1}\left(q,PS,PT\right)$

q.e.d.

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

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

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

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

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

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

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

3.3) ${\sum }_{{k}_{r}=0}^{N-1}\cdots {\sum }_{{k}_{2}=0}^{{k}_{3}}{\sum }_{k=0}^{{k}_{2}}\left(\begin{array}{c}{k}_{i}\\ p\end{array}\right)=\left(\begin{array}{c}p+i-1\\ p\end{array}\right)\left(\begin{array}{c}N+r-1\\ r+p\end{array}\right)\stackrel{p=1}{\to }i\left(\begin{array}{c}N+r-1\\ r+1\end{array}\right)$

[Proof]

${\sum }_{{k}_{r}=0}^{N-1}\cdots {\sum }_{{k}_{2}=0}^{{k}_{3}}{\sum }_{k=0}^{{k}_{2}}\left(\begin{array}{c}{k}_{i}\\ p\end{array}\right)={\sum }_{{k}_{r}=0}^{N-1}\cdots {\sum }_{{k}_{i}=0}^{{k}_{i+1}}\left(\begin{array}{c}{k}_{i}\\ p\end{array}\right)\left(\begin{array}{c}{k}_{i}+i-1\\ i-1\end{array}\right)$

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

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

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

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

q.e.d.

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

4. ${\sum }_{n=0}^{N-1}{\nabla }^{{P}_{1}}SUM\left(n+1,\left[{K}_{1}\right],\left[{T}_{1}\right]\right){\nabla }^{{P}_{2}}SUM\left(n+1,\left[{K}_{2}\right],\left[{T}_{2}\right]\right)\cdots$

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.

4.2) ${P}_{1}\le {T}_{1},{P}_{2}\le {T}_{2},{D}_{1}\ne 0,{D}_{2}\ne 0$

$PS=\left[1,2,\cdots ,{T}_{1}-{P}_{1},\frac{{K}_{1}\left({T}_{1}+1-{P}_{1}\right)}{{T}_{1}{D}_{1}},1,2,\cdots ,{T}_{2}-{P}_{2},\frac{{K}_{2}\left({T}_{2}+1-{P}_{2}\right)}{{T}_{2}{D}_{2}}\right]$

$PT=\left[1,2,3,\cdots ,{T}_{1}+{T}_{2}+2-{P}_{1}-{P}_{2}\right]$

$\begin{array}{l}{\sum }_{n=0}^{N-1}{\nabla }^{{P}_{1}}SUM\left(n+1,\left[{K}_{1}:{D}_{1}\right],\left[{T}_{1}\right]\right){\nabla }^{{P}_{2}}SUM\left(n+1,\left[{K}_{2}:{D}_{2}\right],\left[{T}_{2}\right]\right)\\ =\frac{{T}_{1}{T}_{2}{D}_{1}{D}_{2}}{\left({T}_{1}+1-{P}_{1}\right)!\left({T}_{2}+1-{P}_{2}\right)!}SUM\left(N,PS,PT\right)\end{array}$

4.3)

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

${T}_{1}={T}_{2}=1\to {\sum }_{n=0}^{N-1}\left(n+{K}_{1}\right)\left(n+{K}_{2}\right)=SUM\left(N,\left[{K}_{1},{K}_{2}\right],\left[1,2\right]\right)$

4.4)

$\begin{array}{l}{\sum }_{n=0}^{N-1}SUM\left(n+1,\left[{K}_{1}\right],\left[{T}_{1}\right]\right)SUM\left(n+1,\left[{K}_{2}\right],\left[{T}_{2}\right]\right)\\ =\frac{{T}_{1}{T}_{2}}{\left({T}_{1}+1\right)!\left({T}_{2}+1\right)!}SUM\left(N,\left[1,\cdots ,{T}_{1},\frac{{K}_{1}\left({T}_{1}+1\right)}{{T}_{1}},1,\cdots ,{T}_{2},\frac{{K}_{2}\left({T}_{2}+1\right)}{{T}_{2}}\right],\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left[1,2,\cdots ,{T}_{1}+{T}_{2}+2\right]\right)\end{array}$

The following calculation problems have been solved:

${\sum }_{n=0}^{N-1}{\nabla }^{{P}_{1}}SUM\left(n+1,\left[{K}_{1}:{D}_{1}\right],\left[{T}_{1}\right]\right){\nabla }^{{P}_{2}}\left(\dots \right){\nabla }^{{P}_{3}}\left(\dots \right){\nabla }^{{P}_{4}}\left(\dots \right)\cdots$

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

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

$\begin{array}{l}\nabla {\sum }^{\text{​}}\left(K+{n}_{1}+\cdots +{n}_{M}\right)\stackrel{\text{Repeatableselection}M-1\text{\hspace{0.17em}}\text{in}\text{\hspace{0.17em}}\left[0,n\right],\text{countofitems}=\left(\begin{array}{c}n+1+M-1-1\\ M-1\end{array}\right)}{\to }\\ =\left(K+n\right)\left(\begin{array}{c}n+M-1\\ M-1\end{array}\right)+{\sum }_{{n}_{M-1}=0}^{n}\cdots {\sum }_{{n}_{1}=0}^{{n}_{2}}\left({n}_{1}+{n}_{2}+\cdots +{n}_{M-1}\right)\end{array}$

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

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

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

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

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

4.6) $\begin{array}{l}{\sum }_{{n}_{1},{n}_{2},\cdots ,{n}_{M}=0,{n}_{1}\le {n}_{2}\le \cdots \le {n}_{M}}^{N-1}\left(K+{n}_{1}{d}_{1}+\cdots +{n}_{M}{d}_{M}\right)\\ ={\nabla }^{\left(\begin{array}{c}M\\ 2\end{array}\right)}SUM\left(N,\left[K:\left({d}_{1}+2{d}_{2}+\cdots +M{d}_{M}\right){\left(\begin{array}{c}M+1\\ 2\end{array}\right)}^{-1}\right],\left[\left(\begin{array}{c}M+1\\ 2\end{array}\right)\right]\right)\end{array}$

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

${\sum }_{0}^{N-1}{\nabla }^{{P}_{1}}\left({\sum }^{\text{​}}\left(K+{n}_{1}{d}_{1,1}+\cdots +{n}_{{M}_{1}}{d}_{1,M}\right)\right){\nabla }^{{P}_{2}}\left({\sum }^{\text{​}}\left(K+{n}_{1}{d}_{2,1}+\cdots \right)\right){▽}^{{P}_{3}}\left(\dots \right)\cdots$

Investigation

${\sum }_{{n}_{1},{n}_{2},\cdots ,{n}_{M}=0,{n}_{1}\le {n}_{2}\le \cdots \le {n}_{M}}^{N-1}\left({K}_{1}+{n}_{1}{d}_{1,1}+\cdots +{n}_{M}{d}_{M,1}\right)\left({K}_{2}+{n}_{1}{d}_{1,2}+\cdots +{n}_{M}{d}_{M,2}\right)$

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

Suppose

$\begin{array}{c}F\left(N,M\right)={\sum }_{{n}_{1},{n}_{2},\cdots ,{n}_{M}=0,{n}_{1}\le {n}_{2}\le \cdots \le {n}_{M}}^{N-1}\left({K}_{1}+{n}_{1}{d}_{1,1}+\cdots \right)\left({K}_{2}+{n}_{1}{d}_{1,2}+\cdots \right)\\ =A\left(M\right)\left(\begin{array}{c}N+M-1\\ M+2\end{array}\right)+B\left(M\right)\left(\begin{array}{c}N+M-1\\ M+1\end{array}\right)+{K}_{1}{K}_{2}\left(\begin{array}{c}N+M-1\\ M\end{array}\right)\end{array}$

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

$\begin{array}{l}F\left(N,M+1\right)={\sum }_{{n}_{M+1}=0}^{N-1}F\left(N,M\right)\\ \text{ }+{\sum }_{{n}_{M+1}=0}^{N-1}{n}_{M+1}{d}_{M+1,2}{\sum }_{{n}_{1},{n}_{2},\cdots ,{n}_{M}=0,{n}_{1}\le {n}_{2}\le \cdots \le {n}_{M}}^{N-1}\left({K}_{1}+{n}_{1}{d}_{1,1}+\cdots \right)\\ \text{ }+{\sum }_{{n}_{M+1}=0}^{N-1}{n}_{M+1}{d}_{M+1,1}{\sum }_{{n}_{1},{n}_{2},\cdots ,{n}_{M}=0,{n}_{1}\le {n}_{2}\le \cdots \le {n}_{M}}^{N-1}\left({K}_{2}+{n}_{1}{d}_{2,1}+\cdots \right)\\ \text{ }+{\sum }_{{n}_{M+1}=0}^{N-1}{n}_{M+1}{d}_{M+1,1}{d}_{M+1,2}{\sum }_{{n}_{1},{n}_{2},\cdots ,{n}_{M}=0,{n}_{1}\le {n}_{2}\le \cdots \le {n}_{M}}^{N-1}1\end{array}$

$\begin{array}{l}=A\left(M\right)\left(\begin{array}{c}N+M\\ \left(M+1\right)+2\end{array}\right)+B\left(M\right)\left(\begin{array}{c}N+M\\ \left(M+1\right)+1\end{array}\right)+{K}_{1}{K}_{2}\left(\begin{array}{c}N+M\\ M+1\end{array}\right)\\ \text{ }+SUM\left(N,\left[{K}_{1}:{D}_{1}{\left(\begin{array}{c}M+1\\ 2\end{array}\right)}^{-1},0:{d}_{M+1,2}\right]\left[\left(\begin{array}{c}M+1\\ 2\end{array}\right),\left(\begin{array}{c}M+1\\ 2\end{array}\right)+2-\left(\begin{array}{c}M\\ 2\end{array}\right)\right]\right)\\ \text{ }+SUM\left(N,\left[{K}_{2}:{D}_{2}{\left(\begin{array}{c}M+1\\ 2\end{array}\right)}^{-1},0:{d}_{M+1,1}\right]\left[\left(\begin{array}{c}M+1\\ 2\end{array}\right),M+2\right]\right)\\ \text{ }+SUM\left(N,\left[1:0,1:0,\cdots ,1:0,0:{d}_{M+1,1},0:{d}_{M+1,2}\right]\left[1,3,\cdots ,2M-1,2M\right]\right)\end{array}$

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

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

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

$\begin{array}{c}A\left(M\right)=A\left(M-1\right)+M\left(M+1\right){d}_{M,1}{d}_{M,2}+\left(M+1\right)\left({D}_{i}^{M}{d}_{M,2}+{D}_{i}^{M}{d}_{M,1}\right)\\ =A\left(M-1\right)+\left(M+1\right)\left({D}_{1}{d}_{M,2}+{D}_{2}{d}_{M,1}-M{d}_{M,1}{d}_{M,2}\right)\end{array}$

$\begin{array}{c}B\left(M\right)=B\left(M-1\right)+M\left({d}_{M,1}{d}_{M,2}+{K}_{1}{d}_{M,2}+{K}_{2}{d}_{M,1}\right)+{D}_{i}^{M}{d}_{M,2}+{D}_{i}^{M}{d}_{M,1}\\ =B\left(M-1\right)+M\left({K}_{1}{d}_{M,2}+{K}_{2}{d}_{M,1}-{d}_{M,1}{d}_{M,2}\right)+{D}_{1}{d}_{M,2}+{D}_{2}{d}_{M,1}\end{array}$

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

${d}_{i,1}=1,{d}_{i,2}=1\to$

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

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

5. Formal Calculation of Gaussian Coefficients

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

${G}_{0}^{N}=1,{G}_{M<0\text{\hspace{0.17em}}\text{or}\text{\hspace{0.17em}}M>N}^{N}=0,{G}_{M}^{N}={G}_{N-M}^{N},{G}_{M}^{N}={q}^{M}{G}_{M}^{N-1}+{G}_{M-1}^{N-1}$

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

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

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

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

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

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

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

[1] has obtained the formal formula of ${\sum }_{n=0}^{N-1}{q}^{n}{\prod }_{i=1}^{M}\left({K}_{i}+{D}_{i}{q}^{n}\right)$, but it cannot be generalized to ${\sum }_{{n}_{M}=0}^{N-1}\cdots {\sum }_{{n}_{2}=0}^{{n}_{3}}\cdots {\sum }_{{n}_{1}=0}^{{n}_{2}}\cdots$.

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 ${\nabla }_{q}^{P}$ :

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

$SU{M}_{q}\left(N,\left[{K}_{1}:{D}_{1}\right],\left[{T}_{1}=1\right]\right)={\sum }_{n=0}^{N-1}{q}^{n}\left({D}_{1}{G}_{1}^{n}+{K}_{1}\right)$

$\begin{array}{l}SU{M}_{q}\left(N,\left[{K}_{1}:{D}_{1},{K}_{2}:{D}_{2}\right],\left[{T}_{1}=1,{T}_{2}={T}_{1}+2-P\right]\right)\\ ={\sum }_{n=0}^{N-1}{q}^{n}{\nabla }^{P}SU{M}_{q}\left(n+1,\left[{K}_{1}:{D}_{1}\right],\left[1\right]\right)\left({D}_{2}{G}_{2}^{n}+{K}_{2}\right)\end{array}$

Recursive definie $SU{M}_{q}\left(N,\left[{K}_{1}:{D}_{1},{K}_{2}:{D}_{2},\cdots \right],PT\right)$.

5.5) ${\nabla }^{1}SU{M}_{q}\left(N,\left[{K}_{1}:{D}_{1},{K}_{2}:{D}_{2},\cdots \right],\left[1,2,\cdots ,M\right]\right)={\prod }_{i=1}^{M}\left({D}_{i}{G}_{1}^{n}+{K}_{i}\right)$

5.6) $\begin{array}{l}SU{M}_{q}\left(N,\left[{K}_{1}:{D}_{1},{K}_{2}:{D}_{2},\cdots \right],\left[1,3,\cdots ,2M-1\right]\right)\\ ={\sum }_{{n}_{M}=0}^{N-1}\cdots {\sum }_{{n}_{2}=0}^{{n}_{3}}{q}^{{n}_{2}}\left({D}_{2}{G}_{1}^{{n}_{2}}+{K}_{2}\right){\sum }_{{n}_{1}=0}^{{n}_{2}}{q}^{{n}_{1}}\left({D}_{1}{G}_{1}^{{n}_{1}}+{K}_{1}\right)\end{array}$

The Form: $\left({T}_{1}+{K}_{1}\right)\left({T}_{2}+{K}_{2}\right)\cdots \left({T}_{M}+{K}_{M}\right)=\sum {\prod }_{i=1}^{M}{X}_{i}$

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

$\left\{\begin{array}{l}\stackrel{{\text{Form}}_{1}}{\to }{\sum }_{g=0}^{M}{H}_{1}^{q}\left(g\right){G}_{N-1-\text{g}}^{N+{T}_{M}-M}={\sum }_{g=0}^{M}{H}_{1}^{q}\left(g\right){G}_{{T}_{M}-M+1+\text{g}}^{N+{T}_{M}-M}\\ \stackrel{{\text{Form}}_{2}}{\to }{\sum }_{g=0}^{M}{H}_{2}^{q}\left(g\right){G}_{N-1}^{N+{T}_{M}-M+g}={\sum }_{g=0}^{M}{H}_{2}^{q}\left(g\right){G}_{{T}_{M}-M+1+\text{g}}^{N+{T}_{M}-M+g}\\ \stackrel{{\text{Form}}_{3}}{\to }{\sum }_{g=0}^{M}{H}_{3}^{q}\left(g\right){G}_{N-1-g}^{N+{T}_{M}-g}={\sum }_{g=0}^{M}{H}_{3}^{q}\left(g\right){G}_{{T}_{M}+1}^{N+{T}_{M}+g}\end{array}$

$\left\{\begin{array}{l}{B}_{i}\stackrel{{\text{Form}}_{1}}{\to }\left\{\begin{array}{l}{q}^{\left\{{T}_{i}-{T}_{i-1}\right\}{X}_{T-1}+1}{G}_{1}^{{T}_{i}-{X}_{K-1}}{D}_{i};{X}_{i}={T}_{i}\\ {q}^{\left({T}_{i}-{T}_{i-1}-1\right){X}_{T-1}}\left({K}_{i}+{G}_{1}^{{X}_{T-1}}{D}_{i}\right);{X}_{i}={K}_{i}\end{array}\\ {B}_{i}\stackrel{{\text{Form}}_{2}}{\to }\left\{\begin{array}{l}{q}^{-\left({T}_{i}-{X}_{K-1}\right)}{G}_{1}^{{T}_{i}-{X}_{K-1}}{D}_{i};{X}_{i}={T}_{i}\\ {K}_{i}-{q}^{-\left({T}_{i}-{X}_{K-1}\right)}{G}_{1}^{{T}_{i}-{X}_{K-1}}{D}_{i};{X}_{i}={K}_{i}\end{array}\\ {B}_{i}\stackrel{{\text{Form}}_{3}}{\to }\left\{\begin{array}{l}{q}^{\left({T}_{i}-{T}_{i-1}-1\right){X}_{T-1+1}}\left\{\left({q}^{{X}_{T-1}}{G}_{1}^{{T}_{i}}-{q}^{{T}_{i}}{G}_{1}^{{X}_{T-1}}\right){D}_{i}-{K}_{i}{q}^{{T}_{i}}\right\};{X}_{i}={T}_{i}\\ {q}^{\left({T}_{i}-{T}_{i-1}-1\right){X}_{T-1}}\left({K}_{i}+{G}_{1}^{{X}_{T-1}}{D}_{i}\right);{X}_{i}={K}_{i}\end{array}\end{array}$

[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 $f\left(N\right)=\sum {A}_{i}{G}_{{M}_{i}+1}^{N+{M}_{i}}$, ${\nabla }_{q}^{P}f\left(N\right)=\sum {A}_{i}{G}_{{M}_{i}+1-P}^{N+{M}_{i}-P}$, Form2 is simplest.

Assume $SU{M}_{q}\left(N,PS,PT\right)={\sum }_{g=0}^{M}{H}_{2}^{q}\left(g\right){G}_{{T}_{M}-M+1+g}^{N+{T}_{M}-M+g},X={T}_{M}-M+1-P$

${\nabla }_{q}^{P}SU{M}_{q}\left(n+1,PS,PT\right)={\sum }_{g=0}^{M}{H}_{2}^{q}\left(g\right){G}_{{T}_{M}-M+1+g-P}^{n+1+{T}_{M}-M+g-P}={\sum }_{g=0}^{M}{H}_{2}^{q}\left(g\right){G}_{X+g}^{n+X+g}$

$\begin{array}{l}SU{M}_{q}\left(N,PSA,PTA\right)\\ ={\sum }_{n=0}^{N-1}{q}^{n}{\sum }_{g=0}^{M}{H}_{2}^{q}\left(g\right){G}_{X+g}^{n+X+g}\left({D}_{M+1}{G}_{1}^{n}+{K}_{M+1}\right)\\ ={\sum }_{g=0}^{M}{D}_{M+1}{q}^{-X-g-1}{G}_{1}^{X+g+1}{H}_{2}^{q}\left(g\right){G}_{X+g+2}^{N+X+g+1}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+{\sum }_{g=0}^{M}\left({K}_{M+1}-{q}^{-X-g-1}{G}_{1}^{X+g+1}{D}_{M+1}\right){H}_{2}^{q}\left(g\right){G}_{X+g+1}^{N+X+g}\end{array}$

$\begin{array}{l}={\sum }_{g=0}^{M}{D}_{M+1}{q}^{-\left({T}_{M+1}-\left[M-g\right]\right)}{G}_{1}^{{T}_{M+1}-\left[M-g\right]}{H}_{2}^{q}\left(g\right){G}_{{T}_{M+1}-\left(M+1\right)+g+2}^{N+{T}_{M+1}-\left(M+1\right)+g+1}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+{\sum }_{g=0}^{M}\left({K}_{M+1}-{q}^{-\left({T}_{M+1}-\left[M-g\right]\right)}{G}_{1}^{{T}_{M+1}-\left[M-g\right]}{D}_{M+1}\right){H}_{2}^{q}\left(g\right){G}_{{T}_{M+1}-\left(M+1\right)+g+1}^{N+{T}_{M+1}-\left(M+1\right)+g}\\ ={\sum }_{g=0}^{M+1}{H}_{2}^{q}\left(PSA,PTA,g\right){G}_{N-1}^{N+{T}_{M+1}-\left(M+1\right)+g}\to {\text{Form}}_{2}\end{array}$

If $f\left(N\right)=\sum {A}_{i}{G}_{{M}_{i}}^{N+K},{\nabla }_{q}^{P}f\left(N\right)=\sum {A}_{i}{q}^{-\left({M}_{i}-K\right)P}{G}_{{M}_{i}-P}^{N+K-P}$

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

${\nabla }_{q}^{P}SUM\left(n+1\right)={\sum }_{g=0}^{M}{q}^{-Pg}{H}_{1}^{q}\left(g\right){G}_{{T}_{M}-M+1+g-P}^{n+1+{T}_{M}-M-P}={\sum }_{g=0}^{M}{q}^{-Pg}{H}_{1}^{q}\left(g\right){G}_{X+g}^{n+X}$

$\begin{array}{l}SU{M}_{q}\left(N,PSA,PTA\right)\\ ={\sum }_{n=0}^{N-1}{q}^{n}{\sum }_{g=0}^{M}{q}^{-Pg}{H}_{1}^{q}\left(g\right){G}_{X+g}^{n+X}×\left({D}_{M+1}{G}_{1}^{n}+{K}_{M+1}\right)\\ ={\sum }_{g=0}^{M}{q}^{-Pg}{D}_{M+1}{q}^{2g+1}{G}_{1}^{X+g+1}{H}_{1}^{q}\left(g\right){G}_{X+g+2}^{N+X}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+{\sum }_{g=0}^{M}{q}^{-Pg}\left({D}_{M+1}{q}^{g}{G}_{1}^{g}+{K}_{M+1}{q}^{g}\right){H}_{1}^{q}\left(g\right){G}_{X+g+1}^{N+X}\end{array}$

$\begin{array}{l}={\sum }_{g=0}^{M}{D}_{M+1}{q}^{\left(2-P\right)g+1}{G}_{1}^{{T}_{M+1}-\left[M-g\right]}{H}_{1}^{q}\left(g\right){G}_{N-1-\left(g+1\right)}^{N+{T}_{M+1}-\left(M+1\right)}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+{\sum }_{g=0}^{M}{q}^{g\left(1-P\right)}\left({G}_{1}^{g}{D}_{M+1}+{K}_{M+1}\right){H}_{1}^{q}\left(g\right){G}_{N-1-g}^{N+{T}_{M+1}-\left(M+1\right)}\\ ={\sum }_{g=0}^{M+1}{H}_{1}^{q}\left(PSA,PTA,g\right){G}_{N-1-g}^{N+{T}_{M+1}-\left(M+1\right)}\to {\text{Form}}_{1}\end{array}$

If $f\left(N\right)=\sum {A}_{i}{G}_{M}^{N+{K}_{i}},{▽}_{q}^{P}f\left(N\right)=\sum {A}_{i}{q}^{-\left(M-{K}_{i}\right)P}{G}_{M-P}^{N+{K}_{i}-P}\to {\text{Form}}_{3}$

q.e.d.

From the proof processà

5.8) ${\sum }_{g=0}^{M}{H}_{1}^{q}\left(g\right){G}_{B-g}^{A}={\sum }_{g=0}^{M}{H}_{2}^{q}\left(g\right){G}_{B}^{A+g}={\sum }_{g=0}^{M}{H}_{3}^{q}\left(g\right){G}_{B-g}^{A+M-g}$

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

PT’s domain can be extended to $ℂ$.

$\begin{array}{l}SU{M}_{q}\left(N,\left[{K}_{1},{K}_{2}\right],\left[1,2\right]\right)\\ \stackrel{{\text{Form}}_{\text{2}}}{\to }{q}^{-1}{q}^{-2}{G}_{1}^{1}{G}_{1}^{2}{G}_{3}^{N+2}+{q}^{-1}\left[\left({K}_{1}-{q}^{-1}{G}_{1}^{1}\right)+\left({K}_{2}-{q}^{-2}{G}_{1}^{2}\right)\right]{G}_{2}^{N+1}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}+\left({K}_{1}-{q}^{-1}{G}_{1}^{1}\right)\left({K}_{2}-{q}^{-1}{G}_{1}^{1}\right){G}_{1}^{N}\end{array}$

$\begin{array}{l}SU{M}_{q}\left(N,\left[{K}_{1},{K}_{2}\right],\left[1,3\right]\right)\\ \stackrel{{\text{Form}}_{\text{2}}}{\to }{q}^{-1}{q}^{-3}{G}_{1}^{1}{G}_{1}^{3}{G}_{4}^{N+3}+\left[\left({K}_{1}-{q}^{-1}{G}_{1}^{1}\right){q}^{-2}{G}_{1}^{2}+{q}^{-1}\left({K}_{2}-{q}^{-3}{G}_{1}^{3}\right)\right]{G}_{3}^{N+2}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}+\left({K}_{1}-{q}^{-1}{G}_{1}^{1}\right)\left({K}_{2}-{q}^{-2}{G}_{1}^{2}\right){G}_{2}^{N+1}\end{array}$

$\begin{array}{l}SU{M}_{q}\left(N,\left[{K}_{1},{K}_{2},{K}_{3}\right],\left[1,2,3\right]\right)\\ \stackrel{{\text{Form}}_{\text{2}}}{\to }{q}^{-1}{q}^{-2}{q}^{-3}{G}_{1}^{1}{G}_{1}^{2}{G}_{1}^{3}{G}_{4}^{N+3}+{q}^{-1}{q}^{-2}{G}_{1}^{1}{G}_{1}^{2}\left[\left({K}_{1}-{q}^{-1}{G}_{1}^{1}\right)+\left({K}_{2}-{q}^{-2}{G}_{1}^{2}\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}+\left({K}_{3}-{q}^{-3}{G}_{1}^{3}\right)\right]{G}_{3}^{N+2}+{q}^{-1}{G}_{1}^{1}{G}_{2}^{N+1}\left[\left({K}_{1}-{q}^{-1}{G}_{1}^{1}\right)\left({K}_{2}-{q}^{-1}{G}_{1}^{1}\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}+\left({K}_{1}-{q}^{-1}{G}_{1}^{1}\right)\left({K}_{3}-{q}^{-2}{G}_{1}^{2}\right)+\left({K}_{2}-{q}^{-2}{G}_{1}^{2}\right)\left({K}_{3}-{q}^{-2}{G}_{1}^{2}\right)\right]\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}+\left({K}_{1}-{q}^{-1}{G}_{1}^{1}\right)\left({K}_{2}-{q}^{-1}{G}_{1}^{1}\right)\left({K}_{3}-{q}^{-1}{G}_{1}^{1}\right){G}_{1}^{N}\end{array}$

Form2à

5.9) $SU{M}_{q}\left(N,\left[{q}^{-{T}_{1}}{G}_{1}^{{T}_{1}},{q}^{-{T}_{2}}{G}_{1}^{{T}_{2}},\cdots \right],PT\right)={\prod }_{i=1}^{M}{q}^{-{T}_{i}}{G}_{1}^{{T}_{i}}{G}_{{T}_{M}+1}^{N+{T}_{M}}$

5.10) $\begin{array}{l}SU{M}_{q}\left(N,\left[{q}^{-{L}_{1}}{G}_{1}^{{L}_{1}},\cdots ,{q}^{-{L}_{p}}{G}_{1}^{{L}_{p}},PS\right],\left[{L}_{1},\cdots ,{L}_{P},PT\right]\right)\\ ={\prod }_{i=1}^{P}{q}^{-{L}_{i}}{G}_{1}^{{L}_{i}}SU{M}_{q}\left(N,PS,PT\right)\to \text{extendsdomainof}\text{\hspace{0.17em}}{T}_{1}\text{\hspace{0.17em}}\text{from}\text{\hspace{0.17em}}1\text{\hspace{0.17em}}\text{to}\text{\hspace{0.17em}}ℕ\end{array}$

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

[Proof]

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

${H}_{2}^{q}\left(g,T\right)={\left(q-1\right)}^{g}{\prod }_{i=1}^{g}{q}^{-i}{G}_{1}^{i}$

$1-\frac{{q}^{x}-1}{{q}^{x}\left(q-1\right)}\left(q-1\right)=\frac{1}{{q}^{x}}\to {H}_{2}^{q}\left(g,\sum K\right)={q}^{-M+g}{\sum }_{i=0}^{\left(\begin{array}{c}M\\ g\end{array}\right)-1}{q}^{-i}$

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

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) ${\sum }_{n=0}^{N-1}{q}^{n}{\prod }_{i=1}^{M}\left({K}_{i}+{D}_{i}{q}^{n}\right)={\sum }_{g=0}^{M}H\left(g\right){G}_{g+1}^{N}\to$ Conclusion of [1]

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

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

[Proof]

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

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

$1+\frac{{q}^{x}-1}{q-1}\left(q-1\right)={q}^{x}\to {H}_{1}^{q}\left(g,\sum K\right)=\sum \prod {q}^{{X}_{T-1}}$

In 1916 MacMahon [5] observed that ${G}_{K}^{M}={\sum }_{w\in \Omega \left(M,K\right)}{q}^{inv\left(w\right)}$ $\Omega \left(M,K\right)$ denotes all permutations of the multiset {0M−K, 1K} 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 $inv\left({w}_{1},\cdots ,{w}_{n}\right)=|\left\{\left(i,\text{}j\right):1\le i{w}_{j}\right\}|$.

So $\sum {q}^{{X}_{T-1}}={G}_{M-g}^{M},{H}_{1}^{q}\left(g\right)={\sum }_{g=0}^{M}{\left(q-1\right)}^{g}\left({\prod }_{i=1}^{g}{q}^{i}{G}_{1}^{i}\right){G}_{M-g}^{M}$

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

q.e.d.

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

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

5.16) ${P}_{1}<{T}_{1},{P}_{2}<{T}_{2},{T}_{1}>0,{T}_{2}>0$

$PS=\left[\frac{{G}_{1}^{1}}{{q}^{1}},\frac{{G}_{1}^{2}}{{q}^{2}},\cdots ,\frac{{G}_{1}^{{T}_{1}-{P}_{1}}}{{q}^{{T}_{1}-{P}_{1}}},\frac{{K}_{1}{G}_{1}^{{T}_{1}+1-{P}_{1}}}{{q}^{1-{P}_{1}}{G}_{1}^{{T}_{1}}},\frac{{G}_{1}^{1}}{{q}^{1}},\frac{{G}_{1}^{2}}{{q}^{2}},\cdots ,\frac{{G}_{1}^{{T}_{2}-{P}_{2}}}{{q}^{{T}_{2}-{P}_{2}}},\frac{{K}_{2}{G}_{1}^{{T}_{2}+1-{P}_{2}}}{{q}^{1-{P}_{2}}{G}_{1}^{{T}_{2}}}\right]$

$PT=\left[1,2,3,\cdots ,{T}_{1}+{T}_{2}+2-{P}_{1}-{P}_{2}\right]$

$\begin{array}{l}{\sum }_{n=0}^{N-1}{q}^{n}{\nabla }^{{P}_{1}}SU{M}_{q}\left(n+1,\left[{K}_{1}\right],\left[{T}_{1}\right]\right){\nabla }^{{P}_{2}}SU{M}_{q}\left(n+1,\left[{K}_{2}\right],\left[{T}_{2}\right]\right)\\ =\frac{{q}^{1}}{{G}_{1}^{1}}\frac{{q}^{2}}{{G}_{1}^{2}}\cdots \frac{{q}^{{T}_{1}-{P}_{1}}}{{G}_{1}^{{T}_{1}-{P}_{1}}}\frac{{q}^{1-{P}_{1}}{G}_{1}^{{T}_{1}}}{{G}_{1}^{{T}_{1}-{P}_{1}+1}}\frac{{q}^{1}}{{G}_{1}^{1}}\frac{{q}^{2}}{{G}_{1}^{2}}\cdots \frac{{q}^{{T}_{2}-{P}_{2}}}{{G}_{1}^{{T}_{2}-{P}_{2}}}\frac{{q}^{1-{P}_{2}}{G}_{1}^{{T}_{2}}}{{G}_{1}^{{T}_{2}-{P}_{2}+1}}SUM\left(N,PS,PT\right)\end{array}$

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

[1] has a conclusion:

$\begin{array}{l}{H}_{1}\left(q\right)={\sum }_{x=q}^{M}{H}_{2}\left(x\right)\left(\begin{array}{c}x\\ q\end{array}\right)={\sum }_{x=0}^{q}{H}_{3}\left(x\right)\left(\begin{array}{c}M-x\\ M-q\end{array}\right)\\ {H}_{2}\left(q\right)={\sum }_{x=q}^{M}{\left(-1\right)}^{x+q}{H}_{1}\left(x\right)\left(\begin{array}{c}x\\ q\end{array}\right)\\ {H}_{3}\left(q\right)={\sum }_{x=0}^{q}{\left(-1\right)}^{x+q}{H}_{1}\left(x\right)\left(\begin{array}{c}M-x\\ M-q\end{array}\right)\end{array}$

Generally, Hq(g) has no such attribute; things get complicated.

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

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

[Proof]

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

$\to {H}_{1}^{q}\left(g\right){\left({\prod }_{i=1}^{g}{q}^{\left\{{T}_{i}-{T}_{i-1}\right\}{X}_{T-1}+1}\right)}^{-1}=\left({\sum }_{x=g}^{M}{H}_{2}^{q}\left(x\right){G}_{g}^{x}\right){\left({\prod }_{i=1}^{g}{q}^{-\left({T}_{i}-{X}_{K-1}\right)}\right)}^{-1}$

$PS1=\left[PS,{K}_{M+1}:{D}_{M+1}\right],PT1=\left[PT,M+1\right]$

Suppose it is true at M, ${H}_{1}^{q}\left(g\right)=\left({\sum }_{x=g}^{M}\dots \right)=\left({\sum }_{x=0}^{M}\dots \right)$

$\begin{array}{l}{H}_{1}^{q}\left(PS1,PT1,g\right)\\ ={q}^{{X}_{T-1}+1}{G}_{1}^{{T}_{M+1}-{X}_{K-1},\text{choice}\text{\hspace{0.17em}}T}{D}_{M+1}{H}_{1}^{q}\left(g-1\right)+\left({K}_{M+1}+{G}_{1}^{{X}_{T-1},\text{choice}\text{\hspace{0.17em}}K}{D}_{M+1}\right){H}_{1}^{q}\left(g\right)\\ ={q}^{g}{G}_{1}^{g}{D}_{M+1}{H}_{1}^{q}\left(g-1\right)+\left({K}_{M+1}+{G}_{1}^{g}{D}_{M+1}\right){H}_{1}^{q}\left( g \right)\end{array}$

$\begin{array}{l}={q}^{g}{G}_{1}^{g}{D}_{M+1}{\sum }_{x=0}^{M}{H}_{2}^{q}\left(x\right)\left\{{G}_{g-1}^{x}={G}_{g}^{x+1}-{q}^{g}{G}_{g}^{x}\right\}{q}^{2\left(\begin{array}{c}g\\ 2\end{array}\right)=2\left(\begin{array}{c}g+1\\ 2\end{array}\right)-2g}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+\left({K}_{M+1}+{G}_{1}^{g}{D}_{M+1}\right){\sum }_{x=0}^{M}{H}_{2}^{q}\left(x\right){G}_{g}^{x}{q}^{2\left(\begin{array}{c}g+1\\ 2\end{array}\right)}\\ ={q}^{-g}{G}_{1}^{g}{D}_{M+1}{\sum }_{x=0}^{M}{H}_{2}^{q}\left(x\right){G}_{g}^{x+1}{q}^{2\left(\begin{array}{c}g+1\\ 2\end{array}\right)}+{K}_{M+1}{\sum }_{x=0}^{M}{H}_{2}^{q}\left(x\right){G}_{g}^{x}{q}^{2\left(\begin{array}{c}g+1\\ 2\end{array}\right)}\end{array}$

$\begin{array}{l}{q}^{2\left(\begin{array}{c}g+1\\ 2\end{array}\right)}{\sum }_{x=0}^{M+1}{H}_{2}^{q}\left(PS1,PT1,x\right){G}_{g}^{x}\\ ={q}^{2\left(\begin{array}{c}g+1\\ 2\end{array}\right)}\left\{{\sum }_{x=1}^{M+1}{H}_{2}^{q}\left(x-1\right){q}^{-x}{G}_{1}^{x}{D}_{M+1}{G}_{g}^{x}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+{\sum }_{x=0}^{M}{H}_{2}^{q}\left(x\right)\left({K}_{M+1}-{q}^{-x-1}{G}_{1}^{x+1}{D}_{M+1}\right){G}_{g}^{x}\right\}\end{array}$

$\begin{array}{l}={q}^{2\left(\begin{array}{c}g+1\\ 2\end{array}\right)}\left\{{\sum }_{x=0}^{M}{H}_{2}^{q}\left(x\right){q}^{-x-1}{G}_{1}^{x+1}{D}_{M+1}{G}_{g}^{x+1}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+{\sum }_{x=0}^{M}{H}_{2}^{q}\left(x\right)\left({K}_{M+1}-{q}^{-x-1}{G}_{1}^{x+1}{D}_{M+1}\right){G}_{g}^{x}\right\}\end{array}$

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

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 Di 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

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

$\begin{array}{c}SUM\left(N\right)=SUM\left(N,PS,PT{A}_{1},PT{A}_{2}\right)\\ ={\sum }_{n=0}^{N-1}\left({K}_{M+1}+n×{D}_{M+1,1}+\left(\begin{array}{c}n\\ 2\end{array}\right)×{D}_{M+1,1}\right)×{\nabla }^{p}SUM\left(n+1\right)\end{array}$

$\left({K}_{1}+{T}_{1,1}+{T}_{1,2}\right)\left({K}_{2}+{T}_{2,1}+{T}_{2,2}\right)\cdots \left({K}_{M}+{T}_{M,1}+{T}_{M,2}\right)=\sum {\prod }_{i=1}^{M}{X}_{i}$

$X\left(P{T}_{1}\right),X\left(P{T}_{2}\right)=\text{countof}\text{\hspace{0.17em}}X\in P{T}_{1},P{T}_{2};X\left(PT\right)=X\left(P{T}_{1}\right)+2X\left(P{T}_{2}\right)$

${X}_{P{T}_{1}},{X}_{P{T}_{2}}=\text{countof}\text{\hspace{0.17em}}\left\{{X}_{1},{X}_{2},\cdots ,{X}_{i}\right\}\in P{T}_{1},P{T}_{2};{X}_{PT}={X}_{P{T}_{1}}+2{X}_{P{T}_{2}}$

6.1) $\begin{array}{l}H\left(q\right)={\sum }_{\prod {X}_{i}\text{\hspace{0.17em}}\text{with}\text{\hspace{0.17em}}X\left(PT\right)=q}{\prod }_{i=1}^{M}{B}_{i},\\ SUM\left(N\right)\stackrel{{\text{Form}}_{\text{1}}}{\to }{\sum }_{q=0}^{2M}{H}_{1}\left(q\right)\left(\begin{array}{c}N+{T}_{M}-M\\ {T}_{M}-M+1+q\end{array}\right)\end{array}$

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

[Proof]

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

(**) $\begin{array}{c}{\sum }_{n=0}^{N-1}\left(\begin{array}{c}n\\ 2\end{array}\right)\left(\begin{array}{c}n+K\\ M\end{array}\right)=\left(\begin{array}{c}M+2\\ 2\end{array}\right)\left(\begin{array}{c}N+K\\ M+3\end{array}\right)+\left(M-k\right)\left(M+1\right)\left(\begin{array}{c}N+K\\ M+2\end{array}\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+\left(\begin{array}{c}M\end{array}\end{array}$