等幂求和，即法乌尔哈贝尔公式（英语：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

• 三角形数：${\displaystyle \sum _{i=1}^{n}i={\frac {n(n+1)}{2))}$
• 正方形数：${\displaystyle \sum _{i=1}^{n}(2i-1)=n^{2))$
• 调和级数：${\displaystyle \sum _{n=1}^{k}\,{\frac {1}{n))\;=\;\ln k+\gamma +\varepsilon _{k))$

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${\displaystyle S_{0}^{m}=0}$，${\displaystyle S_{m}^{m}={\frac {1}{m))}$，当m−k为大于1的奇数时，${\displaystyle S_{k}^{m}=0}$。

${\displaystyle \sum _{i=0}^{n}i^{m}={1 \over {m+1))\sum _{i=0}^{m}{m+1 \choose {i))B_{i}\,n^{m+1-i))$^[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]

奇数等幂和与偶数等幂和

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

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

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

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

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

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

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

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

多项式求和

维基教科书中的相关电子教程：组合数求和

伯努利数也通用于等差数列的等幂和。^[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

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

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

• 有形数
• 斯特灵数
• 隙积术
• 欧拉-麦克劳林求和公式

参考资料

查 论 编 基本乘法公式及恒等式 （因式分解） 分配律 ${\displaystyle (a+b)(c+d)=ac+ad+bc+bd\,\!}$ 二项式定理 和与差的平方 ${\displaystyle (a\pm b)^{2}=a^{2}\pm 2ab+b^{2))$ 和与差的立方 ${\displaystyle (a\pm b)^{3}=a^{3}\pm 3a^{2}b+3ab^{2}\pm b^{3))$ 多项式定理 三数和平方 ${\displaystyle (a+b+c)^{2}=a^{2}+b^{2}+c^{2}+2ab+2bc+2ca\,\!}$ 等幂和差 平方和 ${\displaystyle a^{2}+b^{2}=(a+bi)(a-bi)}$ 平方差 ${\displaystyle a^{2}-b^{2}=\left(a+b\right)\left(a-b\right)}$ 立方和和立方差 ${\displaystyle a^{3}\pm b^{3}=(a\pm b)(a^{2}\mp ab+b^{2})\,\!}$ 等幂和差逆定理 ${\displaystyle a^{4}+a^{2}b^{2}+b^{4}=(a^{2}+ab+b^{2})(a^{2}-ab+b^{2})\,\!}$ 对称多项式 ${\displaystyle a^{3}+b^{3}+c^{3}=(a+b+c)^{3}+3(a+b+c)(-ab-bc-ca)+3abc\,\!}$