等幂求和

等幂求和，即法烏爾哈貝爾公式（英語：Faulhaber's formula），是指求幂数相同的变数之和${\displaystyle \sum _{i=1}^{n}x_{i}^{m}}$。

常见公式

• 三角形數：${\displaystyle \sum _{i=1}^{n}i={\frac {n(n+1)}{2}}}$ 
• 正方形數：${\displaystyle \sum _{i=1}^{n}(2i-1)=n^{2}}$ 
• 調和級數：${\displaystyle \sum _{n=1}^{k}\,{\frac {1}{n}}\;=\;\ln k+\gamma +\varepsilon _{k}}$ 

一般数列的等幂和

自然数等幂和

${\displaystyle \sum _{i=1}^{n}i^{0}=n}$ 

${\displaystyle \sum _{i=1}^{n}i^{1}={\frac {n(n+1)}{2}}={\frac {1}{2}}n^{2}+{\frac {1}{2}}n}$ 

${\displaystyle \sum _{i=1}^{n}i^{2}={\frac {n(n+1)(2n+1)}{6}}={\frac {1}{3}}n^{3}+{\frac {1}{2}}n^{2}+{\frac {1}{6}}n}$ ^[1]

${\displaystyle \sum _{i=1}^{n}i^{3}=\left[{\frac {n(n+1)}{2}}\right]^{2}={\frac {1}{4}}n^{4}+{\frac {1}{2}}n^{3}+{\frac {1}{4}}n^{2}}$ ^[1]

${\displaystyle \sum _{i=1}^{n}i^{4}={\frac {n(n+1)(2n+1)(3n^{2}+3n-1)}{30}}={\frac {1}{5}}n^{5}+{\frac {1}{2}}n^{4}+{\frac {1}{3}}n^{3}-{\frac {1}{30}}n}$ ^[1]

${\displaystyle \sum _{i=1}^{n}i^{5}={\frac {n^{2}(n+1)^{2}(2n^{2}+2n-1)}{12}}={\frac {1}{6}}n^{6}+{\frac {1}{2}}n^{5}+{\frac {5}{12}}n^{4}-{\frac {1}{12}}n^{2}}$ ^[1]

${\displaystyle \sum _{i=1}^{n}i^{6}={\frac {n(n+1)(2n+1)(3n^{4}+6n^{3}-3n+1)}{42}}={\frac {1}{7}}n^{7}+{\frac {1}{2}}n^{6}+{\frac {1}{2}}n^{5}-{\frac {1}{6}}n^{3}+{\frac {1}{42}}n}$ ^[1]

${\displaystyle \sum _{i=1}^{n}i^{7}={\frac {n^{2}(n+1)^{2}(3n^{4}+6n^{3}-n^{2}-4n+2)}{24}}={\frac {1}{8}}n^{8}+{\frac {1}{2}}n^{7}+{\frac {7}{12}}n^{6}-{\frac {7}{24}}n^{4}+{\frac {1}{12}}n^{2}}$ 

${\displaystyle \sum _{i=1}^{n}i^{8}={\frac {n(n+1)(2n+1)(5n^{6}+15n^{5}+5n^{4}-15n^{3}-n^{2}+9n-3)}{90}}={\frac {1}{9}}n^{9}+{\frac {1}{2}}n^{8}+{\frac {2}{3}}n^{7}-{\frac {7}{15}}n^{5}+{\frac {2}{9}}n^{3}-{\frac {1}{30}}n}$ 

${\displaystyle \sum _{i=1}^{n}i^{9}={\frac {n^{2}(n+1)^{2}(n^{2}+n-1)(2n^{4}+4n^{3}-n^{2}-3n+3)}{20}}={\frac {1}{10}}n^{10}+{\frac {1}{2}}n^{9}+{\frac {3}{4}}n^{8}-{\frac {7}{10}}n^{6}+{\frac {1}{2}}n^{4}-{\frac {3}{20}}n^{2}}$ 

${\displaystyle \sum _{i=1}^{n}i^{10}={\frac {n(n+1)(2n+1)(n^{2}+n-1)(3n^{6}+9n^{5}+2n^{4}-11n^{3}+3n^{2}+10n-5)}{66}}={\frac {1}{11}}n^{11}+{\frac {1}{2}}n^{10}+{\frac {5}{6}}n^{9}-n^{7}+n^{5}-{\frac {1}{2}}n^{3}+{\frac {5}{66}}n}$ ^[1]

${\displaystyle \sum _{i=0}^{n}i^{m-1}=\sum _{k=0}^{m}S_{k}^{m}n^{k}}$ ，其中${\displaystyle S_{0}^{m}=0}$ ，${\displaystyle S_{m}^{m}={\frac {1}{m}}}$ ，當m−k為大於1的奇數時，${\displaystyle S_{k}^{m}=0}$ 。

${\displaystyle \sum _{i=0}^{n}i^{m}={1 \over {m+1}}\sum _{i=0}^{m}{m+1 \choose {i}}B_{i}\,n^{m+1-i}}$ ^[2]，其中${\displaystyle B_{i}}$ 是伯努利数。

${\displaystyle \displaystyle \sum _{i=1}^{n}i^{m+1}=\sum _{k=0}^{m}L_{k}^{m}{\binom {n+k+1}{m+2}},\left(L_{k}^{m}=\sum _{r=0}^{k}(-1)^{r}{\binom {m+2}{r}}(k+1-r)^{m+1}\right)}$ ^[3]

奇數等冪和與偶數等幂和

${\displaystyle \sum _{i=1}^{n}(2i-1)^{0}=n}$ 

${\displaystyle \sum _{i=1}^{n}(2i)^{0}=n}$ 

${\displaystyle \sum _{i=1}^{n}(2i-1)^{1}=n^{2}}$ 

${\displaystyle \sum _{i=1}^{n}(2i)^{1}=n(n+1)}$ 

${\displaystyle \sum _{i=1}^{n}(2i-1)^{2}={\frac {n(2n-1)(2n+1)}{3}}}$ 

${\displaystyle \sum _{i=1}^{n}(2i)^{2}={\frac {2n(n+1)(2n+1)}{3}}}$ 

${\displaystyle \sum _{i=1}^{n}(2i-1)^{3}=n^{2}(2n^{2}-1)}$ 

${\displaystyle \sum _{i=1}^{n}(2i)^{3}=2\left[n(n+1)\right]^{2}}$ 

多项式求和

維基教科書中的相關電子教程：組合數求和

伯努利数也通用於等差数列的等幂和。^[4]

${\displaystyle \sum _{i=1}^{n}(a_{1}+(i-1)d)^{m}={\frac {1}{m+1}}\sum _{i=0}^{m}B_{i}d^{i-1}{m+1 \choose i}(a_{n+1}^{m+1-i}-a_{1}^{m+1-i})}$ 

也可以利用帕斯卡矩阵，把多项式的和写成矩阵相乘。

${\displaystyle \sum _{k=1}^{n}p(k)={\begin{pmatrix}C_{n}^{1}&C_{n}^{2}&\cdots &C_{n}^{m+1}\end{pmatrix}}{\begin{pmatrix}C_{0}^{0}&0&\cdots &0\\-C_{1}^{0}&C_{1}^{1}&\cdots &0\\\vdots &\vdots &\ddots &\vdots \\(-1)^{m}C_{m}^{0}&(-1)^{m-1}C_{m}^{1}&\cdots &C_{m}^{m}\\\end{pmatrix}}{\begin{pmatrix}p(1)\\p(2)\\\vdots \\p(m+1)\end{pmatrix}}=\sum _{j=1}^{m+1}C_{n}^{j}\Delta ^{j-1}p(1)}$  ^[5][6][7]

其中${\displaystyle \Delta p(n)=p(n+1)-p(n)}$ 

也可以将数列表达成组合数然后利用朱世杰恒等式求和。

${\displaystyle \sum _{i=1}^{n}\left(i^{2}-i\right)=2\sum _{i=1}^{n}C_{i}^{2}=2C_{n+1}^{3}}$ ^[8]

多项式根的等幂和

主条目：对称多项式

${\displaystyle \prod _{r=1}^{n}(x-x_{r})=\sum _{r=0}^{n}a_{r}x^{r}=0,s_{m}=\sum _{r=1}^{n}x_{r}^{m}}$ 

牛顿公式

${\displaystyle s_{m}+a_{1}s_{m-1}+a_{2}s_{m-2}+...+a_{m-1}s_{1}+ma_{m}=0}$ ^[9]

组合公式

${\displaystyle s_{m}=\sum _{r_{i}=0}^{\lfloor {\frac {m}{i}}\rfloor }{\frac {m(r_{1}+r_{2}+...+r_{n}-1)!}{r_{1}!r_{2}!...r_{n}!}}\prod _{i=1}^{n}(-a_{n-i})^{r_{i}}}$ 

取${\displaystyle m=n=3}$ 

${\displaystyle \displaystyle x_{1}^{3}+x_{2}^{3}+x_{3}^{3}={\frac {3(3-1)!}{3!}}(x_{1}+x_{2}+x_{3})^{3}+{\frac {3(1+1-1)!}{1!1!}}(x_{1}+x_{2}+x_{3})(-x_{1}x_{2}-x_{1}x_{3}-x_{2}x_{3})+{\frac {3(1-1)!}{1!}}(x_{1}x_{2}x_{3})}$ 

${\displaystyle x_{1}^{3}+x_{2}^{3}+x_{3}^{3}=(x_{1}+x_{2}+x_{3})^{3}-3(x_{1}+x_{2}+x_{3})(x_{1}x_{2}+x_{1}x_{3}+x_{2}x_{3})+3(x_{1}x_{2}x_{3})}$ 

${\displaystyle a_{n-m}=\sum _{r_{i}=0}^{\lfloor {\frac {m}{i}}\rfloor }\prod _{i=1}^{m}{\frac {(-s_{i})^{r_{i}}}{i^{r_{i}}r_{i}!}}}$ 

取${\displaystyle m=n=3}$ 

${\displaystyle \displaystyle -x_{1}x_{2}x_{3}={\frac {1}{1^{3}3!}}(-x_{1}-x_{2}-x_{3})^{3}+{\frac {1}{1^{1}1!2^{1}1!}}(-x_{1}-x_{2}-x_{3})(-x_{1}^{2}-x_{2}^{2}-x_{3}^{2})+{\frac {1}{3^{1}1!}}(-x_{1}^{3}-x_{2}^{3}-x_{3}^{3})}$ 

${\displaystyle \displaystyle x_{1}x_{2}x_{3}={\frac {1}{6}}(x_{1}+x_{2}+x_{3})^{3}-{\frac {1}{2}}(x_{1}+x_{2}+x_{3})(x_{1}^{2}+x_{2}^{2}+x_{3}^{2})+{\frac {1}{3}}(x_{1}^{3}+x_{2}^{3}+x_{3}^{3})}$ 

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

1. 李政豐. (PDF). 《數學傳播》 (臺北市: 中央研究院數學研究所). 2002-06, 26 (2): 頁73–74,76 [2021-12-16]. （原始内容 (PDF)存档于2021-12-16） （中文（臺灣））. 
2. 谈祥柏. 伯努利数. 科学. 1999, (4) [2014-04-18]. （原始内容于2019-06-10）. 
3. 罗见今. 《垛积比类》内容分析. 内蒙古师范大学学报(自然科学汉文版). 1982, (1) [2015-03-29]. （原始内容于2019-06-08）. 
4. 金晶 杨婷娜 朱伟义. 等差数列前n项等幂和计算公式及算法实现. 渭南师范学院学报. 2012, (2) [2015-09-20]. （原始内容于2019-06-09）. 
5. 陶家元. 高阶等差数列的前n项求和. 成都大学学报(自然科学版). 1999, (1) [2016-05-18]. （原始内容于2020-01-15）. 
6. 黄婷 车茂林 彭杰 张莉. 自然数幂和通项公式证明的新方法. 内江师范学院学报. 2011, (8) [2014-03-30]. （原始内容于2020-02-12）. 
7. 黄嘉威. 方幂和及其推广和式. 数学学习与研究. 2016, (7) [2016-05-17]. （原始内容于2020-01-15）. 
8. 田达武. 朱世杰恒等式及其应用. 数学教学通讯. 2009, (36) [2014-05-24]. （原始内容于2020-01-15）. 
9. 沈南山. 牛顿(Newton)公式的一个注记及其应用. 数学通报. 2005, (3) [2014-03-20]. （原始内容于2019-06-08）.