等幂求和

等幂求和，即法烏爾哈貝爾公式（Template:Lang-en），是指求幂数相同的变数之和$\sum _{i=1}^n{x}_i^m$。

常见公式

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

一般数列的等幂和

自然数等幂和

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

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

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

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

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

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

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

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

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

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

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

$\sum _{i=0}^n{i}^{m-1}=\sum _{k=0}^m{S}_k^m{n}^k$，其中$S_0^m=0$，$S_m^m=\frac{1}{m}$，當m−k為大於1的奇數時，$S_k^m=0$。

$\sum _{i=0}^n{i}^m=\frac{1}{m+1}\sum _{i=0}^m(\genfrac{}{}{0ex}{}{m+1}{i})B_i{n}^{m+1-i}$，其中$B_i$是伯努利数。

$\sum _{i=1}^n{i}^{m+1}=\sum _{k=0}^m{L}_k^m(\genfrac{}{}{0ex}{}{n+k+1}{m+2}),(L_k^m=\sum _{r=0}^k(-1{)}^r(\genfrac{}{}{0ex}{}{m+2}{r})(k+1-r{)}^{m+1})$

奇數等冪和與偶數等幂和

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

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

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

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

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

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

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

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

多项式求和

伯努利数也通用於等差数列的等幂和。

$\sum _{i=1}^n(a_1+(i-1)d{)}^m=\frac{1}{m+1}\sum _{i=0}^m{B}_i{d}^{i-1}(\genfrac{}{}{0ex}{}{m+1}{i})(a_{n+1}^{m+1-i}-a_1^{m+1-i})$

也可以利用帕斯卡矩阵，把多项式的和写成矩阵相乘。

$\sum _{k=1}^{n}p(k)=\left(\begin{array}{cccc}{C}_n^1& C_n^2& \cdots & C_n^{m+1}\end{array}\right)\left(\begin{array}{cccc}{C}_0^0& 0& \cdots & 0\\ -C_1^0& C_1^1& \cdots & 0\\ ⋮& ⋮& \ddots & ⋮\\ (-1{)}^m{C}_m^0& (-1{)}^{m-1}{C}_m^1& \cdots & C_m^m\end{array}\right)\left(\begin{array}{c}p(1)\\ p(2)\\ ⋮\\ p(m+1)\end{array}\right)=\sum _{j=1}^{m+1}{C}_n^j{\mathrm{\Delta }}^{j-1}p(1)$

其中$\mathrm{\Delta }p(n)=p(n+1)-p(n)$

也可以将数列表达成组合数然后利用朱世杰恒等式求和。

$\sum _{i=1}^n(i^2-i)=2\sum _{i=1}^n{C}_i^2=2C_{n+1}^3$

多项式根的等幂和

Template:Main $\prod _{r=1}^n(x-x_r)=\sum _{r=0}^n{a}_r{x}^r=0,s_m=\sum _{r=1}^n{x}_r^m$

牛顿公式

$s_m+a_1{s}_{m-1}+a_2{s}_{m-2}+...+a_{m-1}{s}_1+m{a}_m=0$

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$m=n=3$

$x_1^3+x_2^3+x_3^3=\frac{3(3-1)!}{3!}(x_1+x_2+x_3{)}^3+\frac{3(1+1-1)!}{1!1!}(x_1+x_2+x_3)(-x_1{x}_2-x_1{x}_3-x_2{x}_3)+\frac{3(1-1)!}{1!}(x_1{x}_2{x}_3)$

$x_1^3+x_2^3+x_3^3=(x_1+x_2+x_3{)}^3-3(x_1+x_2+x_3)(x_1{x}_2+x_1{x}_3+x_2{x}_3)+3(x_1{x}_2{x}_3)$

$a_{n-m}=\sum _{r_i=0}^{\lfloor \frac{m}{i}\rfloor }\prod _{i=1}^m\frac{(-s_i{)}^{r_i}}{i^{r_i}{r}_i!}$

取$m=n=3$

$-x_1{x}_2{x}_3=\frac{1}{1^{3}3!}(-x_1-x_2-x_3{)}^3+\frac{1}{1^{1}1!2^{1}1!}(-x_1-x_2-x_3)(-x_1^2-x_2^2-x_3^2)+\frac{1}{3^{1}1!}(-x_1^3-x_2^3-x_3^3)$

$x_1{x}_2{x}_3=\frac{1}{6}(x_1+x_2+x_3{)}^3-\frac{1}{2}(x_1+x_2+x_3)(x_1^2+x_2^2+x_3^2)+\frac{1}{3}(x_1^3+x_2^3+x_3^3)$

参见

• 有形數
• 斯特灵数
• 隙积术
• 欧拉-麦克劳林求和公式