﻿ 七阶Kaup-Kupershmidt方程的经典李群分析和精确解
 上海理工大学学报  2020, Vol. 42 Issue (5): 424-429, 447 PDF

Classical Lie group analysis and exact solutions of the seventh-order Kaup-Kupershmidt equation
LU Yihui, HU Hengchun
College of Science, University of Shanghai for Science and Technology, Shanghai 200093, China
Abstract: The classical Lie group analysis was used to obtain the corresponding infinitesimals of the seventh-order Kaup-Kupershmidt (KK) equation, and two different reduction equations were obtained. Many different forms of exact solutions of the seventh-order KK equation were achieved by solving the reduction equations, such as the rational function solution, the Jacobi elliptic function solution, the hyperbolic function solutions, the trigonometric function solutions and the power series solution, meanwhile, the convergence of the power series in the solution was proved.
Key words: seventh-order Kaup-Kupershmidt equation     classical Lie group analysis     exact solutions
1 问题的提出

KdV型方程是最经典的可积模型之一，它可以描述浅水波、等离子体中的离子声波和大气或海洋中的内波。同时，KdV方程族中描述表面波和内部波、重力–毛细管波以及水波与浮冰间相互作用的高阶可积元等问题也得到了广泛的研究[11]。这里，研究高阶KdV型方程之一的七阶Kaup-Kupershmidt（KK）方程[12-13]，KK方程形式如下：

 $\begin{split}{u_t} + &2\;016{u^3}{u_x} + 630u_x^3 + 2\;268u{u_x}{u_{2x}} + 504{u^2}{u_{3x}} +\\ &252{u_{2x}}{u_{3x}} + 147{u_x}{u_{4x}} + 42u{u_{5x}} + {u_{7x}} = 0\\[-10pt]\end{split}$ (1)

 $\begin{split}{u_t} +& a{u^3}{u_x} + bu_x^3 + cu{u_x}{u_{2x}} + d{u^2}{u_{3x}} + e{u_{2x}}{u_{3x}} + \\ &f{u_x}{u_{4x}} + gu{u_{5x}} + {u_{7x}} = 0\end{split}$ (2)

 $\begin{split}&a = 2\;016,\;b = 630,\;c = 2\;268,\;d = 504,\;e = 252,\\ &f = 147,\;g = 42\end{split}$

2 七阶KK方程的对称和对称约化

 $\begin{split}&x \to x + \varepsilon \xi (x,t,u),t \to t + \varepsilon \tau (x,t,u),\\ & u \to u + \varepsilon \eta (x,t,u)\end{split}$ (3)

 ${{V}} = \xi (x,t,u)\partial x + \tau (x,t,u)\partial t + \eta (x,t,u)\partial u$ (4)

 $\Pr {{{V}}^{(7)}}(\varDelta )\left| {_{\varDelta = 0} = 0} \right.$

 $\begin{split}\varDelta = &{u_t} + 2\;016{u^3}{u_x} + 630u_x^3 + 2\;268u{u_x}{u_{2x}} + 504{u^2}{u_{3x}} +\\ &252{u_{2x}}{u_{3x}} + 147{u_x}{u_{4x}} + 42u{u_{5x}} + {u_{7x}}\end{split}$

 $\Pr {{{V}}^{(7)}} = \eta \partial u + {\eta ^x}\partial {u_x} + {\eta ^{2x}}\partial {u_{2x}} + \cdots + {\eta ^{7x}}\partial {u_{7x}}$

 ${\eta ^{kx}} = {\rm D}_x^k(\eta - \tau {u_t} - \xi {u_x}) + \tau {u_{kxt}} + \xi {u_{(k + 1)x}},\;k = 1,2, \cdots ,7$

 ${{\rm D}_x} = \partial x + {u_x}\partial u + {u_{tx}}\partial {u_t} + {u_{2x}}\partial {u_x} + \cdots$

 $\begin{array}{l} {\xi _t} = {\xi _u} = {\xi _{xx}} = {\tau _x} = {\tau _u} = {\eta _x} = {\eta _t} = {\eta _{uu}} = 0,\\ \eta + u{\eta _u} + u{\tau _t} - 3u{\xi _x} = 0,{\rm{ }}{\eta _u} - 5{\xi _x} + {\tau _t} = 0,\\ u{\tau _t} - u{\xi _x} + 3\eta = 0,{\rm{ }}u{\tau _t} + 2\eta - 3u{\xi _x} = 0,\\ {\tau _t} - 7{\xi _x} = 0,{\rm{ }}{\tau _t} + 2{\eta _u} - 3{\xi _x} = 0 \end{array}$

 ${{V}} = ({c_1}x + {c_3})\frac{\partial }{{\partial x}} + (7{c_1}t + {c_2})\frac{\partial }{{\partial t}} + ( - 2{c_1}u)\frac{\partial }{{\partial u}}$ (5)

 ${{{V}}_1} = \frac{\partial }{{\partial x}},{{{V}}_2} = \frac{\partial }{{\partial t}},{{{V}}_3} = x\frac{\partial }{{\partial x}} + 7t\frac{\partial }{{\partial t}} - 2u\frac{\partial }{{\partial u}} \\$

 $\begin{split} &\left[ {{{{V}}_1},{{{V}}_1}} \right] = \left[ {{{{V}}_2},{{{V}}_2}} \right] = \left[ {{{{V}}_3},{{{V}}_3}} \right] = 0,\;\\ &\left[ {{{{V}}_1},{{{V}}_2}} \right] =- \left[ {{{{V}}_2},{{{V}}_1}} \right] = 0, \\ &\left[ {{{{V}}_1},{{{V}}_3}} \right] = - \left[ {{{{V}}_3},{{{V}}_1}} \right] ={{{V}}_1},\;\\ &\left[ {{{{V}}_2},{{{V}}_3}} \right] = - \left[ {{{{V}}_3},{{{V}}_2}} \right] = 7{{{V}}_2} \end{split}$

 $\begin{split}&{g_1}:(x + \varepsilon ,t,u),\;{g_2}:(x,t + \varepsilon ,u),\;\\ &{g_3}:(x{{\rm{e}}^\varepsilon },t{{\rm{e}}^{7\varepsilon }},u{{\rm{e}}^{ - 2\varepsilon }})\end{split}$ (6)

 ${u_1} = f(x - \varepsilon ,t),\;{u_2} = f(x,t - \varepsilon ),\;{u_3} = {{\rm{e}}^{ - 2\varepsilon }}f(x{{\rm{e}}^{ - \varepsilon }},t{{\rm{e}}^{ - 7\varepsilon }})$

 $\dfrac{{{\rm{d}}x}}{{{c_1}x + {c_3}}} = \dfrac{{{\rm{d}}t}}{{7{c_1}t + {c_2}}} = \dfrac{{{\rm{d}}u}}{{ - 2{c_1}u}}$

a. 当 ${c_1} = 0$ 时，得方程（1）对应的相似变量和相似变换

 $z = x - kt,\;u(x,t) = f(z),\;(k = {c_3}/{c_2},{c_2} \ne 0)$

$f(z)$ 满足常微分方程

 $\begin{split} &- kf' + 2\;016{f^3}f' + 630{f'^3} + 2\;268ff'f'' + \\ &\quad 504{f^2}{f^{(3)}} +252f''{f^{(3)}} + 147f'{f^{(4)}} +\\ & \quad42f{f^{(5)}} + {f^{(7)}} = 0 \end{split}$ (7)

b. 当 ${c_1} \ne 0$ 时，得到对应的相似变量和相似变换

 $z = \dfrac{{x + \dfrac{c_3}{c_1}}}{{{{\left(t + \dfrac{c_2}{7{c_1}}\right)}^{\frac{1}{7}}}}},\;u(x,t) = \dfrac{{f(z)}}{{{{\left(t +\dfrac {c_2}{7{c_1}}\right)}^{\frac{2}{7}}}}}$

$f(z)$ 满足常微分方程

 $\begin{split}& -\frac{2}{7}f - \dfrac{1}{7}zf' + 2\;016{f^3}f' + 630{f'^3} + \\ &\quad2\;268ff'f'' +504{f^2}{f^{(3)}}+252f''{f^{(3)}}+\\ &\quad147f'{f^{(4)}}+42f{f^{(5)}} + {f^{(7)}}=0 \end{split}$ (8)
3 约化方程的求解

3.1 利用扩展的tanh方法来求解约化方程（7）

 $f(z) = \sum\limits_{i = - n}^n {{A_i}H{{(z)}^i}}$

 $H'(z) = 1 - {H^2}(z)$

 $H(z) = \tanh z$

 $\begin{split}f(z) =& {A_{ - 2}}{\tanh ^{ - 2}}\ z + {A_{ - 1}}{\tanh ^{ - 1}}\ z+\\ &{A_0} + {A_1}\tanh \ z +{A_2}{\tanh ^2}\ z\end{split}$ (9)

 $\begin{array}{l} {\bf{a}} .\; u(x,t) = \dfrac{1}{3} - \dfrac{1}{2}{\tanh ^2}\left(x + \dfrac{4}{3}t\right)\\ {\bf{b}}.\; u(x,t) = - \dfrac{1}{2}{\tanh ^{ - 2}}\left(x + \dfrac{{256}}{3}t\right) + \dfrac{1}{3} - \dfrac{1}{2}{\tanh ^2}\left(x + \dfrac{{256}}{3}t\right)\\ {\bf{c}}.\; u(x,t) = \dfrac{1}{3} - \dfrac{1}{2}{\tanh ^{ - 2}}\left(x + \dfrac{4}{3}t\right) \end{array}$
3.2 利用微分方程展开法来求解方程（7）的精确解

 $f(z) = \sum\limits_{i = 1}^n {{P_i}{\phi ^i}(z)} + {P_0}$

$\phi '$ 是下面一阶常微分方程的解

 $\phi '(z) = \varepsilon \sqrt {\sum\limits_{j = 0}^r {{h_j}{\phi ^j}} }$ (10)

 $r = n + 2$

 $f = {P_2}{\phi ^2} + {P_1}\phi + {P_0}$ (11)

$\varepsilon = 1$ ，将方程（11）和（10）代入方程（7）并令 ${\phi ^i}{\phi ^{'}}$ 的系数为0，可以得到相应的代数方程，解之得：

 $\begin{split} {\bf{a}} .\; k =& - \frac{1}{{48}}h_2^3 - \dfrac{9}{{32}}h_3^2{h_0} + \frac{3}{{32}}{h_1}{h_2}{h_3},{P_0} = \frac{1}{{24}}{h_2},\\{P_1} =& - \frac{1}{8}{h_3},{P_2} = {h_4} = 0 \end{split}$

 $\begin{split} {\bf{b}}. \;k =& - \dfrac{{72{h_0}{h_2}P_2^4 + 63P_2^2P_1^2h_2^2 + 216P_2^3P_1^2{h_0} + 189{P_2}P_1^4{h_2} + 18P_2^3 + {P_1}{h_1}{h_2} + 8P_2^3h_2^3 - 27P_2^4 + h_1^2 + 189h_1^6}}{{{\rm{6}}P_2^3}}\\ {h_3} =& - 4{P_1},{h_4} = - 2{P_2},{P_0} = - \dfrac{1}{{12}}\dfrac{{3P_1^2 + 2{P_2}{h_2}}}{{{P_2}}} \end{split}$

a. $\;{h_0} = {h_1} = {h_4} = 0,{h_2} > 0$

 $f = - \frac{1}{{24}}{h_2} + \frac{1}{8}{h_2}\sec {{\rm{h}}^2}\left(\frac{{\sqrt {{h_2}} }}{2}z\right)$

b. $\;{h_0} = {h_1} = {h_4} = 0,{h_2} < 0$

 $f = - \frac{1}{{24}}{h_2} + \frac{1}{8}{h_2}{\rm{sec}}{{\rm{h}}^2}\left(\frac{{\sqrt { - {h_2}} }}{2}z\right)$

c. $\;{h_0} = {h_1} = {h_4} = {h_2} = 0$

 $f = - \frac{1}{{8{z^2}}}$

d. $\;{h_4} = 0,{h_3} > 0$

 $f = - \dfrac{1}{{24}}{h_2} + \frac{1}{8}{h_3}\wp \left(\dfrac{{\sqrt {{h_3}} }}{2}z,{g_2},{g_3}\right)$

a. $\;{h_2} = 2{m^2} - 1,{h_0} = 1 - {m^2},{h_4} = - {m^2}$

 $f = - \frac{1}{6}{h_2} - \frac{1}{2}{h_4}{({\rm{cn}}(z))^2}$

b. $\;{h_2} = 2{m^2} - 1,{h_0} = - {m^2},{h_4} = 1 - {m^2}$

 $f = - \frac{1}{6}{h_2} - \frac{1}{2}{h_4}{({\rm{nc}}(z))^2}$

c. $\;{h_2} = 2{m^2} - 1,{h_0} = 1,{h_4} = ({m^2} - 1){m^2}$

 $f = - \frac{1}{6}{h_2} - \frac{1}{2}{h_4}{({\rm{sd}}(z))^2}$

d. $\;{h_2} = 2{m^2} - 1,{h_0} = ({m^2} - 1){m^2},{h_4} = 1$

 $f = - \frac{1}{6}{h_2} - \frac{1}{2}{h_4}{({\rm{ds}}(z))^2}$

e. $\;{h_2} = \dfrac{{1 - 2{m^2}}}{2},{h_0} = {h_4} = \dfrac{1}{4}$

 $f = - \frac{1}{6}{h_2} - \frac{1}{2}{h_4}{({\rm{ns}}(z) \pm {\rm{cs}}(z))^2}$

f. $\;{h_2} = \dfrac{{1 - 2{m^2}}}{2},{h_0} = {h_4} = \dfrac{{1 - {m^2}}}{4}$

 $f = - \frac{1}{6}{h_2} - \frac{1}{2}{h_4}{({\rm{nc}}(z) \pm {\rm{sc}}(z))^2}$

g. $\;{h_2} = \dfrac{{1 - 2{m^2}}}{2},{h_0} = \dfrac{{{m^2}}}{4},{h_4} = \dfrac{1}{4}$

 $f = - \frac{1}{6}{h_2} - \frac{1}{2}{h_4}{({\rm{ns}}(z) \pm {\rm{ds}}(z))^2}$

3.3 方程（8）的幂级数解

 $f(z) = \sum\limits_{n = 0}^\infty {{C_n}{z^n}}$ (12)

 $\begin{split} {C_{n + 7}} =& - \frac{1}{{(n + 1)(n + 2)(n + 3)(n + 4)(n + 5)(n + 6)(n + 7)}}\left[ { - \frac{2}{7}{C_n} - \frac{1}{7}n{C_n}}+ \right. \\ & 2\;016\sum\limits_{j = 0}^i {\sum\limits_{i = 0}^k {\sum\limits_{k = 0}^n {{C_j}{C_{i - j}}{C_{k - i}}{C_{n - k + 1}}} } } (n - k + 1)+ \\ & 630\sum\limits_{i = 0}^k {\sum\limits_{k = 0}^n {{C_{i + 1}}{C_{k - i + 1}}{C_{n - k + 1}}(i + 1)(k - i + 1)(n - k + 1)} } +\\ & 2\;268\sum\limits_{i = 0}^k {\sum\limits_{k = 0}^n {{C_{i + 1}}{C_{k - i + 1}}{C_{n - k + 2}}(k - i + 1)(n - k + 2)(n - k + 1)} } +\\ & 504\sum\limits_{i = 0}^k {\sum\limits_{k = 0}^n {{C_i}{C_{k - i}}{C_{n - k + 3}}(n - k + 3)(n - k + 2)(n - k + 1)} } +\\ & 252\sum\limits_{k = 0}^n {{C_{k + 2}}{C_{n - k + 3}}(k + 2)(k + 1)(n - k + 3)(n - k + 2)(n - k + 1)} +\\ & 147\sum\limits_{k = 0}^n {{C_{k + 1}}{C_{n - k + 4}}(k + 1)(n - k + 4)(n - k + 3)(n - k + 2)(n - k + 1)} +\\ &\left. { 42\sum\limits_{k = 0}^n {{C_k}{C_{n - k + 5}}(n - k + 5)(n - k + 4)(n - k + 3)(n - k + 2)(n - k + 1)} } \right] \\ \end{split}$

 $\begin{array}{l} | {{C_{n + 7}}} | \leqslant 2\;268\left( {\displaystyle\sum\limits_{j = 0}^i {\displaystyle\sum\limits_{i = 0}^k {\displaystyle\sum\limits_{k = 0}^n {| {{C_j}} || {{C_{i - j}}} || {{C_{k - i}}} || {{C_{n - k + 1}}} |} } } + \displaystyle\sum\limits_{k = 0}^n {\left| {{C_k}} \right|\left| {{C_{n - k + 5}}} \right|} }+\displaystyle\sum\limits_{i = 0}^k {\displaystyle\sum\limits_{k = 0}^n {\left| {{C_{i + 1}}} \right|\left| {{C_{k - i + 1}}} \right|\left| {{C_{n - k + 1}}} \right| + } } \right. \\ \;\;\;\;\;\;\;\; \displaystyle\sum\limits_{i = 0}^k {\displaystyle\sum\limits_{k = 0}^n {\left| {{C_{i + 1}}} \right|\left| {{C_{k - i + 1}}} \right|\left| {{C_{n - k + 2}}} \right|} } +\displaystyle\sum\limits_{i = 0}^k {\displaystyle\sum\limits_{k = 0}^n {\left| {{C_i}} \right|\left| {{C_{k - i}}} \right|\left| {{C_{n - k + 3}}} \right|} } + \displaystyle\sum\limits_{k = 0}^n {\left| {{C_{k + 2}}} \right|\left| {{C_{n - k + 3}}} \right|}+\\ \;\;\;\;\;\;\;\; \left. { \displaystyle\sum\limits_{k = 0}^n {\left| {{C_{k + 1}}} \right|} \left| {{C_{n - k + 4}}} \right| + \left| {{C_n}} \right|} \right){\rm{, }}\;n = 0,1,2, \cdots \\ \end{array}$

 $\delta = R(z) = \sum\limits_{n = 0}^\infty {{r^n}{z^n}}$

$\left| {{r_0}} \right| = \left| {{C_0}} \right|,...,\left| {{r_6}} \right| = \left| {{C_6}} \right|$ ，且

 $\begin{split} {r_{n + 7}} = &2\;268\left( \displaystyle\sum\limits_{j = 0}^i \displaystyle\sum\limits_{i = 0}^k \displaystyle\sum\limits_{k = 0}^n {r_j}{r_{i - j}}{r_{k - i}}{r_{n - k + 1}} +\right. \\ &\displaystyle\sum\limits_{k = 0}^n {{r_k}{r_{n - k + 5}}} +\displaystyle\sum\limits_{i = 0}^k {\displaystyle\sum\limits_{k = 0}^n {{r_{i + 1}}{r_{k - i + 1}}} } {r_{n - k + 1}} +\\ &\displaystyle\sum\limits_{i = 0}^k {\displaystyle\sum\limits_{k = 0}^n {{r_{i + 1}}{r_{k - i + 1}}} } {r_{n - k + 2}} +\\ &\displaystyle\sum\limits_{i = 0}^k {\displaystyle\sum\limits_{k = 0}^n {{r_i}{r_{k - i}}} } {r_{n - k + 3}} + \displaystyle\sum\limits_{k = 0}^n {{r_{k + 2}}{r_{n - k + 3}}}+ \\ &\left.{\displaystyle\sum\limits_{k = 0}^n {{r_{k + 1}}{r_{n - k + 4}}} + {r_n}} \right),\;n = 0,1,2, \cdots \end{split}$

 $\begin{split} R(z) &= {r_0} + {r_1}z + {r_2}{z^2} + {r_3}{z^3} + {r_4}{z^4} + {r_5}{z^5} + {r_6}{z^6} + \\ &\displaystyle\sum\limits_{n = 0}^\infty {{r_{n + 7}}{z^{n + 7}}}{\rm{ }} = {r_0} + {r_1}z + {r_2}{z^2} + {r_3}{z^3} + {r_4}{z^4} + {r_5}{z^5} +\\ &{r_6}{z^6}+2\;268\left( {\displaystyle\sum\limits_{n = 0}^\infty {\displaystyle\sum\limits_{j = 0}^i {\displaystyle\sum\limits_{i = 0}^k {\displaystyle\sum\limits_{k = 0}^n {{r_j}{r_{i - j}}{r_{k - i}}{r_{n - k + 1}} + } } } } } \right.\\ &\displaystyle\sum\limits_{n = 0}^\infty {\displaystyle\sum\limits_{k = 0}^n {{r_k}{r_{n - k + 5}}} }+ \displaystyle\sum\limits_{n = 0}^\infty {\displaystyle\sum\limits_{i = 0}^k {\displaystyle\sum\limits_{k = 0}^n {{r_{i + 1}}{r_{k - i + 1}}} } {r_{n - k + 1}} }+\\ & \displaystyle\sum\limits_{n = 0}^\infty {\displaystyle\sum\limits_{i = 0}^k {\displaystyle\sum\limits_{k = 0}^n {{r_{i + 1}}{r_{k - i + 1}}} } {r_{n - k + 2}}} + \displaystyle\sum\limits_{n = 0}^\infty {\displaystyle\sum\limits_{i = 0}^k {\displaystyle\sum\limits_{k = 0}^n {{r_i}{r_{k - i}}} } {r_{n - k + 3}} + }\\ &\left.\displaystyle\sum\limits_{n = 0}^\infty {\displaystyle\sum\limits_{k = 0}^n {{r_{k + 2}}{r_{n - k + 3}}} }{ + \displaystyle\sum\limits_{n = 0}^\infty {\displaystyle\sum\limits_{k = 0}^n {{r_{k + 1}}{r_{n - k + 4}}} + {r_n}} } \right){z^{n + 7}} \end{split}$

 $\begin{split} F(z,&\delta ) = \delta - {r_0} - {r_1}z - {r_2}{z^2} - {r_3}{z^3} - {r_4}{z^4} - {r_5}{z^5} - {r_6}{z^6} -\\ &2\;268[{z^6}{R^3}(R - {r_0}) + {z^4}{(R - {r_0})^3} + {z^3}{(R - {r_0})^2}. \\ &(R - {r_0} - {r_1}z)+ {z^4}{R^2}(R - {r_0} - {r_1}z - {r_2}{z^2}) + \\ &{z^2}(R - {r_0} - {r_1}z)(R - {r_0} - {r_1}z - {r_2}{z^2})+\\ &{z^2}(R - {r_0}) (R - {r_0} - {r_1}z - {r_2}{z^2} - {r_3}{z^3})+\\ &{z^2}R(R - {r_0} - {r_1}z - {r_2}{z^2} - {r_3}{z^3} - {r_4}{z^4}) + {z^7}R] = 0 \end{split}$

 $\begin{array}{l} u = \dfrac{{{C_0}}}{A} + \dfrac{{{C_1}z}}{A} + \dfrac{{{C_2}{z^2}}}{A} + \dfrac{{{C_3}{z^3}}}{A} + \dfrac{{{C_4}{z^4}}}{A} + \dfrac{{{C_5}{z^5}}}{A} + \dfrac{{{C_6}{z^6}}}{A} -\\ \;\;\;\;\;\;\;\; \displaystyle\sum\limits_{n = 0}^\infty {\frac{1}{{(n + 1)(n + 2)(n + 3)(n + 4)(n + 5)(n + 6)(n + 7)}}\left[ { - \frac{2}{7}{C_n} - \frac{1}{7}n{C_n}} \right.} +\\ \;\;\;\;\;\;\;\; 630\displaystyle\sum\limits_{i = 0}^k {\displaystyle\sum\limits_{k = 0}^n {{C_{i + 1}}{C_{k - i + 1}}{C_{n - k + 1}}(i + 1)(k - i + 1)(n - k + 1)} } +\\ \;\;\;\;\;\;\;\; 2\;268\displaystyle\sum\limits_{i = 0}^k {\displaystyle\sum\limits_{k = 0}^n {{C_{i + 1}}{C_{k - i + 1}}{C_{n - k + 2}}(k - i + 1)(n - k + 2)(n - k + 1)} } {\rm{ + }}\\ \;\;\;\;\;\;\;\;504\displaystyle\sum\limits_{i = 0}^k {\displaystyle\sum\limits_{k = 0}^n {{C_i}{C_{k - i}}{C_{n - k + 3}}(n - k + 3)(n - k + 2)(n - k + 1)} } +\\ \;\;\;\;\;\;\;\; 252\displaystyle\sum\limits_{k = 0}^n {{C_{k + 2}}{C_{n - k + 3}}(k + 2)(k + 1)(n - k + 3)(n - k + 2)(n - k + 1)} +\\ \;\;\;\;\;\;\;\; 147\displaystyle\sum\limits_{k = 0}^n {{C_{k + 1}}{C_{n - k + 4}}(k + 1)(n - k + 4)(n - k + 3)(n - k + 2)(n - k + 1)} +\\ \left. {\;\;\;\;\;\;\;\; 42\displaystyle\sum\limits_{k = 0}^n {{C_k}{C_{n - k + 5}}(n - k + 5)(n - k + 4)(n - k + 3)(n - k + 2)(n - k + 1)} } \right]\dfrac{{{z^{n + 7}}}}{A} \\ \end{array}$

4 结　论

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