**等幂求和**，即**法烏爾哈貝爾公式**（英語：），是指求幂数相同的变数之和{\displaystyle \sum \_{i=1}^{n}x\_{i}^{m}}。

常见公式
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* 三角形數： \sum^{n}\_{i=1} i = \frac{n(n+1)}{2}
* 正方形數： \sum\_{i=1}^{n} (2i - 1) = n^2
* 調和級數：\sum\_{n=1}^k\,\frac{1}{n} \;=\; \ln k + \gamma + \varepsilon\_k

一般数列的等幂和
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### 自然数等幂和

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

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

\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

\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}

\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

\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}

\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

{\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}

{\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}}，其中B\_{i}是伯努利数。

\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)

### 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

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| --- | --- |
|  | 維基教科書中的相關電子：**組合數求和** |

伯努利数也通用於等差数列的等幂和。

\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)}


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}}

多项式根的等幂和
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条目：对称多项式
\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

s\_{m}+a\_{1}s\_{{m-1}}+a\_{2}s\_{{m-2}}+...+a\_{{m-1}}s\_{1}+ma\_{m}=0

### 组合公式

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m=n=3

\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})
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})
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m=n=3

\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 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
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* 有形數
* 斯特灵数
* 隙积术
* 欧拉-麦克劳林求和公式

参考资料
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1. 1 2 3 4 5 6  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.. 科学. 1999, (4)  [2014-04-18]. （原始内容存档于2019-06-10）.
3. ↑  罗见今.. 内蒙古师范大学学报(自然科学汉文版). 1982, (1)  [2015-03-29]. （原始内容存档于2019-06-08）.
4. ↑  金晶 杨婷娜 朱伟义.. 渭南师范学院学报. 2012, (2)  [2015-09-20]. （原始内容存档于2019-06-09）.
5. ↑  陶家元.. 成都大学学报(自然科学版). 1999, (1)  [2016-05-18]. （原始内容存档于2020-01-15）.
6. ↑  黄婷 车茂林 彭杰 张莉.. 内江师范学院学报. 2011, (8)  [2014-03-30]. （原始内容存档于2020-02-12）.
7. ↑  黄嘉威.. 数学学习与研究. 2016, (7)  [2016-05-17]. （原始内容存档于2020-01-15）.
8. ↑  田达武.. 数学教学通讯. 2009, (36)  [2014-05-24]. （原始内容存档于2020-01-15）.
9. ↑  沈南山.. 数学通报. 2005, (3)  [2014-03-20]. （原始内容存档于2019-06-08）.

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基本乘法公式及恆等式 （因式分解） |
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| 分配律 | (a+b)(c+d)= ac+ad+bc+bd\,\! |
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| 二項式定理 | 

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| 和差平方 | {\displaystyle (a\pm b)^{2}=a^{2}\pm 2ab+b^{2}} |
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| 和差立方 | {\displaystyle (a\pm b)^{3}=a^{3}\pm 3a^{2}b+3ab^{2}\pm b^{3}} |

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| 多項式定理 | 

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| 三數和平方 |  (a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca \,\! |

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| 等冪和差 | 

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| 平方和 | a^2+b^2=(a+bi)(a-bi) |
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| 平方差 | {\displaystyle a^{2}-b^{2}=\left(a+b\right)\left(a-b\right)} |
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| 立方和差 | a^{3}\pm b^{3}=(a\pm b)(a^{2}\mp ab+b^{2})\,\! |

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| 等冪和差逆定理 | a^4+a^2b^2+b^4 = (a^2+ab+b^2)(a^2-ab+b^2)\,\! |
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| 對稱多項式 | a^3+b^3+c^3= (a+b+c)^3+3(a+b+c)(-ab-bc-ca)+3abc\,\! |

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