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 ∑ i = 1 n x i m {\displaystyle \sum _{i=1}^{n}x_{i}^{m}} 。 三角形數： ∑ i = 1 n i = n ( n + 1 ) 2 {\displaystyle \sum _{i=1}^{n}i={\frac {n(n+1)}{2}}} 正方形數： ∑ i = 1 n ( 2 i − 1 ) = n 2 {\displaystyle \sum _{i=1}^{n}(2i-1)=n^{2}} 調和級數： ∑ n = 1 k 1 n = ln ⁡ k + γ + ε k {\displaystyle \sum _{n=1}^{k}\,{\frac {1}{n}}\;=\;\ln k+\gamma +\varepsilon _{k}} ∑ i = 1 n i 0 = n {\displaystyle \sum _{i=1}^{n}i^{0}=n} ∑ i = 1 n i 1 = n ( n + 1 ) 2 = 1 2 n 2 + 1 2 n {\displaystyle \sum _{i=1}^{n}i^{1}={\frac {n(n+1)}{2}}={\frac {1}{2}}n^{2}+{\frac {1}{2}}n} ∑ i = 1 n i 2 = n ( n + 1 ) ( 2 n + 1 ) 6 = 1 3 n 3 + 1 2 n 2 + 1 6 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} ∑ i = 1 n i 3 = [ n ( n + 1 ) 2 ] 2 = 1 4 n 4 + 1 2 n 3 + 1 4 n 2 {\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}} ∑ i = 1 n i 4 = n ( n + 1 ) ( 2 n + 1 ) ( 3 n 2 + 3 n − 1 ) 30 = 1 5 n 5 + 1 2 n 4 + 1 3 n 3 − 1 30 n {\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} ∑ i = 1 n i 5 = n 2 ( n + 1 ) 2 ( 2 n 2 + 2 n − 1 ) 12 = 1 6 n 6 + 1 2 n 5 + 5 12 n 4 − 1 12 n 2 {\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}} ∑ i = 1 n i 6 = n ( n + 1 ) ( 2 n + 1 ) ( 3 n 4 + 6 n 3 − 3 n + 1 ) 42 = 1 7 n 7 + 1 2 n 6 + 1 2 n 5 − 1 6 n 3 + 1 42 n {\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} ∑ i = 1 n i 7 = n 2 ( n + 1 ) 2 ( 3 n 4 + 6 n 3 − n 2 − 4 n + 2 ) 24 = 1 8 n 8 + 1 2 n 7 + 7 12 n 6 − 7 24 n 4 + 1 12 n 2 {\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}} ∑ i = 1 n i 8 = n ( n + 1 ) ( 2 n + 1 ) ( 5 n 6 + 15 n 5 + 5 n 4 − 15 n 3 − n 2 + 9 n − 3 ) 90 = 1 9 n 9 + 1 2 n 8 + 2 3 n 7 − 7 15 n 5 + 2 9 n 3 − 1 30 n {\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} ∑ i = 1 n i 9 = n 2 ( n + 1 ) 2 ( n 2 + n − 1 ) ( 2 n 4 + 4 n 3 − n 2 − 3 n + 3 ) 20 = 1 10 n 10 + 1 2 n 9 + 3 4 n 8 − 7 10 n 6 + 1 2 n 4 − 3 20 n 2 {\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}} ∑ i = 1 n i 10 = n ( n + 1 ) ( 2 n + 1 ) ( n 2 + n − 1 ) ( 3 n 6 + 9 n 5 + 2 n 4 − 11 n 3 + 3 n 2 + 10 n − 5 ) 66 = 1 11 n 11 + 1 2 n 10 + 5 6 n 9 − n 7 + n 5 − 1 2 n 3 + 5 66 n {\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} ∑ i = 0 n i m − 1 = ∑ k = 0 m S k m n k {\displaystyle \sum _{i=0}^{n}i^{m-1}=\sum _{k=0}^{m}S_{k}^{m}n^{k}} ，其中 S 0 m = 0 {\displaystyle S_{0}^{m}=0} ， S m m = 1 m {\displaystyle S_{m}^{m}={\frac {1}{m}}} ，當m−k為大於1的奇數時， S k m = 0 {\displaystyle S_{k}^{m}=0} 。 ∑ i = 0 n i m = 1 m + 1 ∑ i = 0 m ( m + 1 i ) B i n m + 1 − i {\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 {\displaystyle B_{i}} 是伯努利数。 ∑ i = 1 n i m + 1 = ∑ k = 0 m L k m ( n + k + 1 m + 2 ) , ( L k m = ∑ r = 0 k ( − 1 ) r ( m + 2 r ) ( k + 1 − r ) m + 1 ) {\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)} ∑ i = 1 n ( 2 i − 1 ) 0 = n {\displaystyle \sum _{i=1}^{n}(2i-1)^{0}=n} ∑ i = 1 n ( 2 i ) 0 = n {\displaystyle \sum _{i=1}^{n}(2i)^{0}=n} ∑ i = 1 n ( 2 i − 1 ) 1 = n 2 {\displaystyle \sum _{i=1}^{n}(2i-1)^{1}=n^{2}} ∑ i = 1 n ( 2 i ) 1 = n ( n + 1 ) {\displaystyle \sum _{i=1}^{n}(2i)^{1}=n(n+1)} ∑ i = 1 n ( 2 i − 1 ) 2 = n ( 2 n − 1 ) ( 2 n + 1 ) 3 {\displaystyle \sum _{i=1}^{n}(2i-1)^{2}={\frac {n(2n-1)(2n+1)}{3}}} ∑ i = 1 n ( 2 i ) 2 = 2 n ( n + 1 ) ( 2 n + 1 ) 3 {\displaystyle \sum _{i=1}^{n}(2i)^{2}={\frac {2n(n+1)(2n+1)}{3}}} ∑ i = 1 n ( 2 i − 1 ) 3 = n 2 ( 2 n 2 − 1 ) {\displaystyle \sum _{i=1}^{n}(2i-1)^{3}=n^{2}(2n^{2}-1)} ∑ i = 1 n ( 2 i ) 3 = 2 [ n ( n + 1 ) ] 2 {\displaystyle \sum _{i=1}^{n}(2i)^{3}=2\left[n(n+1)\right]^{2}} 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。 ∑ i = 1 n ( a 1 + ( i − 1 ) d ) m = 1 m + 1 ∑ i = 0 m B i d i − 1 ( m + 1 i ) ( a n + 1 m + 1 − i − a 1 m + 1 − i ) {\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。 ∑ k = 1 n p ( k ) = ( C n 1 C n 2 ⋯ C n m + 1 ) ( C 0 0 0 ⋯ 0 − C 1 0 C 1 1 ⋯ 0 ⋮ ⋮ ⋱ ⋮ ( − 1 ) m C m 0 ( − 1 ) m − 1 C m 1 ⋯ C m m ) ( p ( 1 ) p ( 2 ) ⋮ p ( m + 1 ) ) = ∑ j = 1 m + 1 C n j Δ j − 1 p ( 1 ) {\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 Δ p ( n ) = p ( n + 1 ) − p ( n ) {\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。 ∑ i = 1 n ( i 2 − i ) = 2 ∑ i = 1 n C i 2 = 2 C n + 1 3 {\displaystyle \sum _{i=1}^{n}\left(i^{2}-i\right)=2\sum _{i=1}^{n}C_{i}^{2}=2C_{n+1}^{3}} ∏ r = 1 n ( x − x r ) = ∑ r = 0 n a r x r = 0 , s m = ∑ r = 1 n x r m {\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}} s m + a 1 s m − 1 + a 2 s m − 2 + . . . + a m − 1 s 1 + m a m = 0 {\displaystyle s_{m}+a_{1}s_{m-1}+a_{2}s_{m-2}+...+a_{m-1}s_{1}+ma_{m}=0} s m = ∑ r i = 0 ⌊ m i ⌋ m ( r 1 + r 2 + . . . + r n − 1 ) ! r 1 ! r 2 ! . . . r n ! ∏ i = 1 n ( − a n − i ) r i {\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 m = n = 3 {\displaystyle m=n=3} x 1 3 + x 2 3 + x 3 3 = 3 ( 3 − 1 ) ! 3 ! ( x 1 + x 2 + x 3 ) 3 + 3 ( 1 + 1 − 1 ) ! 1 ! 1 ! ( x 1 + x 2 + x 3 ) ( − x 1 x 2 − x 1 x 3 − x 2 x 3 ) + 3 ( 1 − 1 ) ! 1 ! ( x 1 x 2 x 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})} 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 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 = ∑ r i = 0 ⌊ m i ⌋ ∏ i = 1 m ( − s i ) r i i r i r i ! {\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 m = n = 3 {\displaystyle m=n=3} − x 1 x 2 x 3 = 1 1 3 3 ! ( − x 1 − x 2 − x 3 ) 3 + 1 1 1 1 ! 2 1 1 ! ( − x 1 − x 2 − x 3 ) ( − x 1 2 − x 2 2 − x 3 2 ) + 1 3 1 1 ! ( − x 1 3 − x 2 3 − x 3 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})} x 1 x 2 x 3 = 1 6 ( x 1 + x 2 + x 3 ) 3 − 1 2 ( x 1 + x 2 + x 3 ) ( x 1 2 + x 2 2 + x 3 2 ) + 1 3 ( 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})} 有形數 斯特灵数 隙积术 欧拉-麦克劳林求和公式 李政豐. 連續整數冪次和公式的另類思考 (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））.  谈祥柏. 伯努利数. 科学. 1999, (4) [2014-04-18]. （原始内容存档于2019-06-10）.  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]. （原始内容存档于2019-06-08）.  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]. （原始内容存档于2019-06-09）.  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]. （原始内容存档于2020-01-15）.  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]. （原始内容存档于2020-02-12）.  黄嘉威. 方幂和及其推广和式. 数学学习与研究. 2016, (7) [2016-05-17]. （原始内容存档于2020-01-15）.  田达武. 朱世杰恒等式及其应用. 数学教学通讯. 2009, (36) [2014-05-24]. （原始内容存档于2020-01-15）.  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]. （原始内容存档于2019-06-08）.