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=1nxim{\displaystyle \sum _{i=1}^{n}x_{i}^{m}}。
∑i=1ni0=n{\displaystyle \sum _{i=1}^{n}i^{0}=n}
∑i=1ni1=n(n+1)2=12n2+12n{\displaystyle \sum _{i=1}^{n}i^{1}={\frac {n(n+1)}{2}}={\frac {1}{2}}n^{2}+{\frac {1}{2}}n}
∑i=1ni2=n(n+1)(2n+1)6=13n3+12n2+16n{\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]
∑i=1ni3=[n(n+1)2]2=14n4+12n3+14n2{\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]
∑i=1ni4=n(n+1)(2n+1)(3n2+3n−1)30=15n5+12n4+13n3−130n{\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]
∑i=1ni5=n2(n+1)2(2n2+2n−1)12=16n6+12n5+512n4−112n2{\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]
∑i=1ni6=n(n+1)(2n+1)(3n4+6n3−3n+1)42=17n7+12n6+12n5−16n3+142n{\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]
∑i=1ni7=n2(n+1)2(3n4+6n3−n2−4n+2)24=18n8+12n7+712n6−724n4+112n2{\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=1ni8=n(n+1)(2n+1)(5n6+15n5+5n4−15n3−n2+9n−3)90=19n9+12n8+23n7−715n5+29n3−130n{\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=1ni9=n2(n+1)2(n2+n−1)(2n4+4n3−n2−3n+3)20=110n10+12n9+34n8−710n6+12n4−320n2{\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=1ni10=n(n+1)(2n+1)(n2+n−1)(3n6+9n5+2n4−11n3+3n2+10n−5)66=111n11+12n10+56n9−n7+n5−12n3+566n{\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]
∑i=0nim−1=∑k=0mSkmnk{\displaystyle \sum _{i=0}^{n}i^{m-1}=\sum _{k=0}^{m}S_{k}^{m}n^{k}} ,其中S0m=0{\displaystyle S_{0}^{m}=0} ,Smm=1m{\displaystyle S_{m}^{m}={\frac {1}{m}}} ,當m−k為大於1的奇數時,Skm=0{\displaystyle S_{k}^{m}=0} 。
∑i=0nim=1m+1∑i=0m(m+1i)Binm+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}} [2],其中Bi{\displaystyle B_{i}} 是伯努利数。
∑i=1nim+1=∑k=0mLkm(n+k+1m+2),(Lkm=∑r=0k(−1)r(m+2r)(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)} [3]
∑i=1n(2i−1)0=n{\displaystyle \sum _{i=1}^{n}(2i-1)^{0}=n}
∑i=1n(2i)0=n{\displaystyle \sum _{i=1}^{n}(2i)^{0}=n}
∑i=1n(2i−1)1=n2{\displaystyle \sum _{i=1}^{n}(2i-1)^{1}=n^{2}}
∑i=1n(2i)1=n(n+1){\displaystyle \sum _{i=1}^{n}(2i)^{1}=n(n+1)}
∑i=1n(2i−1)2=n(2n−1)(2n+1)3{\displaystyle \sum _{i=1}^{n}(2i-1)^{2}={\frac {n(2n-1)(2n+1)}{3}}}
∑i=1n(2i)2=2n(n+1)(2n+1)3{\displaystyle \sum _{i=1}^{n}(2i)^{2}={\frac {2n(n+1)(2n+1)}{3}}}
∑i=1n(2i−1)3=n2(2n2−1){\displaystyle \sum _{i=1}^{n}(2i-1)^{3}=n^{2}(2n^{2}-1)}
∑i=1n(2i)3=2[n(n+1)]2{\displaystyle \sum _{i=1}^{n}(2i)^{3}=2\left[n(n+1)\right]^{2}}
伯努利数也通用於等差数列的等幂和。[4]
∑i=1n(a1+(i−1)d)m=1m+1∑i=0mBidi−1(m+1i)(an+1m+1−i−a1m+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=1np(k)=(Cn1Cn2⋯Cnm+1)(C000⋯0−C10C11⋯0⋮⋮⋱⋮(−1)mCm0(−1)m−1Cm1⋯Cmm)(p(1)p(2)⋮p(m+1))=∑j=1m+1CnjΔj−1p(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)} [5][6][7]
也可以将数列表达成组合数然后利用朱世杰恒等式求和。
∏r=1n(x−xr)=∑r=0narxr=0,sm=∑r=1nxrm{\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}}
sm+a1sm−1+a2sm−2+...+am−1s1+mam=0{\displaystyle s_{m}+a_{1}s_{m-1}+a_{2}s_{m-2}+...+a_{m-1}s_{1}+ma_{m}=0} [9]
sm=∑ri=0⌊mi⌋m(r1+r2+...+rn−1)!r1!r2!...rn!∏i=1n(−an−i)ri{\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}
an−m=∑ri=0⌊mi⌋∏i=1m(−si)riiriri!{\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}!}}}