查看“︁等幂求和”︁的源代码

'''等幂求和'''，即'''法烏爾哈貝爾公式'''（{{lang-en|Faulhaber's formula}}），是指求幂数相同的变数之和<math>\sum_{i=1}^{n} x_i^m</math>。

==常见公式==
*[[三角形數]]：<math> \sum^{n}_{i=1} i = \frac{n(n+1)}{2}</math>
*[[正方形數]]：<math> \sum_{i=1}^{n} (2i - 1) = n^2 </math>
*[[調和級數]]：<math>\sum_{n=1}^k\,\frac{1}{n} \;=\; \ln k + \gamma + \varepsilon_k </math>
==一般数列的等幂和==
===自然数等幂和===
<math> \sum_{i=1}^{n} i^{0} =n</math>

<math> \sum_{i=1}^{n} i^{1} = \frac{n(n+1)}{2} = \frac{1}{2}n^2 +\frac{1}{2} n </math>

<math> \sum_{i=1}^{n} i^{2} = \frac{n(n+1)(2n+1)}{6} = \frac{1}{3}n^3 + \frac{1}{2}n^2 + \frac{1}{6}n</math><ref name="數學傳播2002年6月">{{cite journal |title=連續整數冪次和公式的另類思考 |author=李政豐 |url=https://web.math.sinica.edu.tw/math_media/d262/26210.pdf |format=PDF |archiveurl=https://web.archive.org/web/20211216144002/https://web.math.sinica.edu.tw/math_media/d262/26210.pdf |archivedate=2021-12-16 |language=zh-tw |date=2002-06 |volume=26 |issue=2 |journal=《[[數學傳播]]》 |page=頁73–74,76 |publisher=中央研究院數學研究所 |location=臺北市 |accessdate=2021-12-16}}</ref>

<math> \sum_{i=1}^{n} i^{3} = \left[\frac{n(n+1)}{2}\right]^{2} = \frac{1}{4}n^4 + \frac{1}{2}n^3 + \frac{1}{4}n^2</math><ref name="數學傳播2002年6月"/>

<math> \sum_{i=1}^{n} i^{4} = \frac{n(n+1)(2n+1)(3n^2+3n-1)}{30} = \frac{1}{5}n^5 + \frac{1}{2}n^4 + \frac{1}{3}n^3 - \frac{1}{30}n</math><ref name="數學傳播2002年6月"/>

<math> \sum_{i=1}^{n} i^{5} = \frac{n^{2}(n+1)^{2}(2n^2+2n-1)}{12} = \frac{1}{6}n^6 + \frac{1}{2}n^5 + \frac{5}{12}n^4 - \frac{1}{12}n^2</math><ref name="數學傳播2002年6月"/>

<math> \sum_{i=1}^{n} i^{6} = \frac{n(n+1)(2n+1)(3n^4+6n^3-3n+1)}{42} = \frac{1}{7}n^7 + \frac{1}{2}n^6 + \frac{1}{2}n^5 - \frac{1}{6}n^3 + \frac{1}{42}n</math><ref name="數學傳播2002年6月"/>

<math> \sum_{i=1}^{n} i^{7} = \frac{n^2(n+1)^2(3n^4+6n^3-n^2-4n+2)}{24} = \frac{1}{8}n^8 + \frac{1}{2}n^7 + \frac{7}{12}n^6 - \frac{7}{24}n^4 + \frac{1}{12}n^2</math>

<math> \sum_{i=1}^{n} i^{8} = \frac{n(n+1)(2n+1)(5n^6+15n^5+5n^4-15n^3-n^2+9n-3)}{90}=\frac{1}{9}n^9 + \frac{1}{2}n^8 + \frac{2}{3}n^7 - \frac{7}{15}n^5 + \frac{2}{9}n^3 - \frac{1}{30}n</math>

<math> \sum_{i=1}^{n} i^{9} = \frac{n^2(n+1)^2(n^2+n-1)(2n^4+4n^3-n^2-3n+3)}{20}=\frac{1}{10}n^{10}+\frac{1}{2}n^9+\frac{3}{4}n^8-\frac{7}{10}n^6+\frac{1}{2}n^4-\frac{3}{20}n^2</math>

<math> \sum_{i=1}^{n} i^{10} = \frac{n(n+1)(2n+1)(n^2+n-1)(3n^6+9n^5+2n^4-11n^3+3n^2+10n-5)}{66}=\frac{1}{11}n^{11}+\frac{1}{2}n^{10}+\frac{5}{6}n^9-n^7+n^5-\frac{1}{2}n^3+\frac{5}{66}n</math><ref name="數學傳播2002年6月"/>

<math>\sum_{i=0}^{n} i^{m-1} = \sum_{k=0}^m  S_k^m n^k</math>，其中<math>S_0^m = 0</math>，<math>S_m^m = \frac{1}{m}</math>，當m−k為大於1的奇數時，<math>S_k^m = 0</math>。

<math>\sum_{i=0}^{n} i^m = {1\over{m+1}}\sum_{i=0}^m{m+1\choose{i}} B_i\, n^{m+1-i}</math><ref>{{cite journal|author=谈祥柏|year=1999|title=伯努利数|journal=科学|issue=4|url=http://www.cnki.com.cn/Article/CJFDTotal-KXZZ199904024.htm|access-date=2014-04-18|archive-date=2019-06-10|archive-url=https://web.archive.org/web/20190610084130/http://www.cnki.com.cn/Article/CJFDTotal-KXZZ199904024.htm|dead-url=no}}</ref>，其中<math>B_i</math>是[[伯努利数]]。

<math>\displaystyle \sum_{i=1}^n i^{m+1} = \sum_{k=0}^m  L_k^m \binom{n+k+1}{m+2},\left(L_k^m = \sum_{r=0}^k (-1)^r \binom{m+2}{r} (k+1-r)^{m+1}\right)</math><ref>{{cite journal|author=罗见今|year=1982|title=《垛积比类》内容分析|journal=内蒙古师范大学学报(自然科学汉文版)|issue=1|url=http://www.cnki.com.cn/Article/CJFDTotal-NMSB198201010.htm|access-date=2015-03-29|archive-date=2019-06-08|archive-url=https://web.archive.org/web/20190608035337/http://www.cnki.com.cn/Article/CJFDTOTAL-NMSB198201010.htm|dead-url=no}}</ref>

===奇數等冪和與偶數等幂和===
<math> \sum_{i=1}^{n} (2i-1)^{0} =n</math>

<math> \sum_{i=1}^{n} (2i)^{0} =n</math>

<math> \sum_{i=1}^{n} (2i-1)^{1} =n^2 </math>

<math> \sum_{i=1}^{n} (2i)^{1} = n(n+1) </math>

<math> \sum_{i=1}^{n} (2i-1)^{2} = \frac{n(2n-1)(2n+1)}{3}</math>

<math> \sum_{i=1}^{n} (2i)^{2} = \frac{2n(n+1)(2n+1)}{3}</math>

<math> \sum_{i=1}^{n} (2i-1)^{3} = n^2(2n^2-1)</math>

<math> \sum_{i=1}^{n} (2i)^{3} = 2 \left[ n(n+1) \right]^2</math>

===多项式求和===
{{Wikibooks|代數/本書課文/求和/組合數求和|組合數求和}}

伯努利数也通用於[[等差数列]]的等幂和。<ref>{{cite journal|author=金晶 杨婷娜 朱伟义|year=2012|title=等差数列前n项等幂和计算公式及算法实现|journal=渭南师范学院学报|issue=2|url=http://www.cnki.com.cn/Article/CJFDTotal-WOLF201202010.htm|access-date=2015-09-20|archive-date=2019-06-09|archive-url=https://web.archive.org/web/20190609060638/http://www.cnki.com.cn/Article/CJFDTotal-WOLF201202010.htm|dead-url=no}}</ref>

<math> \sum_{i=1}^n (a_1+(i-1)d)^m = \frac{1}{m+1} \sum_{i=0}^m B_i d^{i-1} {m+1\choose i}(a_{n+1}^{m+1-i}-a_1^{m+1-i}) </math>

也可以利用[[帕斯卡矩阵]]，把多项式的和写成矩阵相乘。

<math>\sum_{k=1}^n p(k)=
\begin{pmatrix}C_n^1 & C_n^2 & \cdots & C_n^{m+1}\end{pmatrix}
\begin{pmatrix}
C_0^0 & 0 & \cdots & 0\\
-C_1^0 & C_1^1 & \cdots & 0\\
\vdots & \vdots & \ddots & \vdots\\
(-1)^mC_m^0 & (-1)^{m-1}C_m^1 & \cdots & C_m^m\\
\end{pmatrix}
\begin{pmatrix}p(1)\\p(2)\\\vdots\\p(m+1)\end{pmatrix}
=\sum_{j=1}^{m+1} C_{n}^{j}\Delta^{j-1}p(1)</math>
<ref>{{cite journal|author=陶家元|year=1999|title=高阶等差数列的前n项求和|journal=成都大学学报(自然科学版)|issue=1|url=http://www.cnki.com.cn/Article/CJFDTOTAL-CDDD199901006.htm|access-date=2016-05-18|archive-date=2020-01-15|archive-url=https://web.archive.org/web/20200115072151/http://www.cnki.com.cn/Article/CJFDTOTAL-CDDD199901006.htm|dead-url=no}}</ref>
<ref>{{cite journal|author=黄婷 车茂林 彭杰 张莉|year=2011|title=自然数幂和通项公式证明的新方法|journal=内江师范学院学报|issue=8|url=http://www.cnki.com.cn/Article/CJFDTotal-NJSG201108006.htm|access-date=2014-03-30|archive-date=2020-02-12|archive-url=https://web.archive.org/web/20200212225516/http://www.cnki.com.cn/Article/CJFDTOTAL-NJSG201108006.htm|dead-url=no}}</ref>
<ref>{{cite journal|author=黄嘉威|year=2016|title=方幂和及其推广和式|journal=数学学习与研究|issue=7|url=http://www.cnki.com.cn/Article/CJFDTOTAL-SXYG201607113.htm|access-date=2016-05-17|archive-date=2020-01-15|archive-url=https://web.archive.org/web/20200115072155/http://www.cnki.com.cn/Article/CJFDTOTAL-SXYG201607113.htm|dead-url=no}}</ref>
:其中<math>\Delta p(n)=p(n+1)-p(n)</math>

也可以将数列表达成组合数然后利用[[朱世杰恒等式]]求和。

:<math>\sum_{i=1}^n \left(i^2-i\right)=2\sum_{i=1}^n C_i^2=2C_{n+1}^3</math><ref>{{cite journal|author=田达武|year=2009|title=朱世杰恒等式及其应用|journal=数学教学通讯|issue=36|url=http://www.cnki.com.cn/Article/CJFDTotal-SXUJ200936022.htm|access-date=2014-05-24|archive-date=2020-01-15|archive-url=https://web.archive.org/web/20200115072156/http://www.cnki.com.cn/Article/CJFDTotal-SXUJ200936022.htm|dead-url=no}}</ref>
==多项式根的等幂和==
{{main|对称多项式}}
<math>\prod_{r=1}^n (x-x_r)=\sum_{r=0}^n a_r x^r=0,s_m=\sum_{r=1}^n x_r^m</math>
===牛顿公式===
<math>s_m+a_1s_{m-1}+a_2s_{m-2}+...+a_{m-1}s_1+ma_m=0</math><ref>{{cite journal|author=沈南山|year=2005|title=牛顿(Newton)公式的一个注记及其应用|journal=数学通报|issue=3|url=http://www.cnki.com.cn/Article/CJFDTotal-SXTB20050300H.htm|access-date=2014-03-20|archive-date=2019-06-08|archive-url=https://web.archive.org/web/20190608132237/http://www.cnki.com.cn/Article/CJFDTotal-SXTB20050300H.htm|dead-url=no}}</ref>
===组合公式===
<math>s_m=\sum_{r_i=0}^{\lfloor \frac{m}{i} \rfloor} \frac{m(r_1+r_2+...+r_n -1)!}{r_1!r_2!...r_n!} \prod_{i=1}^n (-a_{n-i})^{r_i}</math>

取<math>m=n=3</math>
:<math>\displaystyle x_1^3+x_2^3+x_3^3=\frac{3(3-1)!}{3!}(x_1+x_2+x_3)^3+\frac{3(1+1-1)!}{1!1!}(x_1+x_2+x_3)(-x_1x_2-x_1x_3-x_2x_3)+\frac{3(1-1)!}{1!}(x_1x_2x_3)</math>
:<math>x_1^3+x_2^3+x_3^3=(x_1+x_2+x_3)^3-3(x_1+x_2+x_3)(x_1x_2+x_1x_3+x_2x_3)+3(x_1x_2x_3)</math>

<math>a_{n-m}=\sum_{r_i=0}^{\lfloor \frac{m}{i} \rfloor} \prod_{i=1}^m \frac{(-s_i)^{r_i}}{i^{r_i} r_i !}</math>

取<math>m=n=3</math>
:<math>\displaystyle -x_1x_2x_3=\frac{1}{1^3 3!}(-x_1-x_2-x_3)^3
+\frac{1}{1^1 1! 2^1 1!}(-x_1-x_2-x_3)(-x_1^2-x_2^2-x_3^2)
+\frac{1}{3^1 1!}(-x_1^3-x_2^3-x_3^3)</math>
:<math>\displaystyle x_1x_2x_3=\frac{1}{6}(x_1+x_2+x_3)^3-\frac{1}{2}(x_1+x_2+x_3)(x_1^2+x_2^2+x_3^2)+\frac{1}{3}(x_1^3+x_2^3+x_3^3)</math>

==参见==
*[[有形數]]
*[[斯特灵数]]
*[[隙积术]]
*[[欧拉-麦克劳林求和公式]]

==参考资料==
{{reflist}}

{{Basic identity}}
[[Category:有限差分]]