Faulhaberova formula predstavlja sumu:
∑
k
=
1
n
k
p
=
1
p
+
2
p
+
3
p
+
⋯
+
n
p
{\displaystyle \sum _{k=1}^{n}k^{p}=1^{p}+2^{p}+3^{p}+\cdots +n^{p}}
Dobila je ime po nemačkom matematičaru Johanu Faulhaberu . Formula se može predstaviti preko Bernulijevih brojeva kao:
∑
k
=
1
n
k
m
=
1
m
+
1
∑
k
=
0
m
(
m
+
1
k
)
B
k
n
m
+
1
−
k
{\displaystyle \sum _{k=1}^{n}k^{m}={1 \over {m+1}}\sum _{k=0}^{m}{m+1 \choose {k}}B_{k}\;n^{m+1-k}}
1
+
2
+
3
+
⋯
+
n
=
n
(
n
+
1
)
2
=
n
2
+
n
2
{\displaystyle 1+2+3+\cdots +n={n(n+1) \over 2}={n^{2}+n \over 2}}
1
2
+
2
2
+
3
2
+
⋯
+
n
2
=
n
(
n
+
1
)
(
2
n
+
1
)
6
=
2
n
3
+
3
n
2
+
n
6
{\displaystyle 1^{2}+2^{2}+3^{2}+\cdots +n^{2}={n(n+1)(2n+1) \over 6}={2n^{3}+3n^{2}+n \over 6}}
1
3
+
2
3
+
3
3
+
⋯
+
n
3
=
(
n
2
+
n
2
)
2
=
n
4
+
2
n
3
+
n
2
4
{\displaystyle 1^{3}+2^{3}+3^{3}+\cdots +n^{3}=\left({n^{2}+n \over 2}\right)^{2}={n^{4}+2n^{3}+n^{2} \over 4}}
1
4
+
2
4
+
3
4
+
⋯
+
n
4
=
6
n
5
+
15
n
4
+
10
n
3
−
n
30
{\displaystyle 1^{4}+2^{4}+3^{4}+\cdots +n^{4}={6n^{5}+15n^{4}+10n^{3}-n \over 30}}
1
5
+
2
5
+
3
5
+
⋯
+
n
5
=
2
n
6
+
6
n
5
+
5
n
4
−
n
2
12
{\displaystyle 1^{5}+2^{5}+3^{5}+\cdots +n^{5}={2n^{6}+6n^{5}+5n^{4}-n^{2} \over 12}}
1
6
+
2
6
+
3
6
+
⋯
+
n
6
=
6
n
7
+
21
n
6
+
21
n
5
−
7
n
3
+
n
42
{\displaystyle 1^{6}+2^{6}+3^{6}+\cdots +n^{6}={6n^{7}+21n^{6}+21n^{5}-7n^{3}+n \over 42}}
Definišemo li sumu
S
p
(
n
)
=
∑
k
=
1
n
k
p
.
{\displaystyle S_{p}(n)=\sum _{k=1}^{n}k^{p}.}
Tada je:
S
p
(
1
)
=
1
{\displaystyle S_{p}(1)=1}
S
p
(
n
+
1
)
=
S
p
(
n
)
+
(
n
+
1
)
p
.
{\displaystyle S_{p}(n+1)=S_{p}(n)+(n+1)^{p}.}
Pokušajmo sada da
S
p
(
n
)
{\displaystyle S_{p}(n)}
∑
k
=
0
∞
a
p
,
k
n
k
+
1
{\displaystyle \sum _{k=0}^{\infty }a_{p,k}n^{k+1}}
Uvrstimo li to u drugi izraz u ovom poglavlju dobijamo:
∑
k
=
0
∞
a
p
,
k
(
n
+
1
)
k
+
1
=
∑
k
=
0
∞
a
p
,
k
n
k
+
1
+
(
n
+
1
)
p
{\displaystyle \sum _{k=0}^{\infty }a_{p,k}(n+1)^{k+1}=\sum _{k=0}^{\infty }a_{p,k}n^{k+1}+(n+1)^{p}}
Koristimo binomnu teoremu , pa sledi:
∑
k
=
0
∞
a
p
,
k
[
(
∑
j
=
0
k
+
1
(
k
+
1
j
)
n
j
)
−
n
k
+
1
]
=
(
n
+
1
)
p
{\displaystyle \sum _{k=0}^{\infty }a_{p,k}\left[\left(\sum _{j=0}^{k+1}{\binom {k+1}{j}}n^{j}\right)-n^{k+1}\right]=(n+1)^{p}}
∑
k
=
0
∞
a
p
,
k
∑
j
=
0
k
(
k
+
1
j
)
n
j
=
∑
j
=
0
∞
(
p
j
)
n
j
.
{\displaystyle \sum _{k=0}^{\infty }a_{p,k}\sum _{j=0}^{k}{\binom {k+1}{j}}n^{j}=\sum _{j=0}^{\infty }{\binom {p}{j}}n^{j}.}
Dvostruku sumu na levoj strani preuredimo uzimajući u obzir j ≤k
∑
j
=
0
∞
n
j
∑
k
=
j
∞
(
k
+
1
j
)
a
p
,
k
=
∑
j
=
0
∞
(
p
j
)
n
j
.
{\displaystyle \sum _{j=0}^{\infty }n^{j}\sum _{k=j}^{\infty }{\binom {k+1}{j}}a_{p,k}=\sum _{j=0}^{\infty }{\binom {p}{j}}n^{j}.}
i konačno se dobija:
∑
k
=
j
∞
(
k
+
1
j
)
a
p
,
k
=
(
p
j
)
.
{\displaystyle \sum _{k=j}^{\infty }{\binom {k+1}{j}}a_{p,k}={\binom {p}{j}}.}
Desna strana je jednaka nuli za j >p
a
p
,
k
=
0
{\displaystyle a_{p,k}=0}
k >p j !Pohhamerovoga simbola vredi:
∑
k
=
j
p
(
k
+
1
)
j
a
p
,
k
=
(
p
)
j
{\displaystyle \sum _{k=j}^{p}(k+1)_{j}a_{p,k}=(p)_{j}}
∑
k
=
j
p
(
k
+
1
)
t
(
k
+
1
−
t
)
j
−
t
a
p
,
k
=
(
p
)
t
(
p
−
t
)
j
−
t
{\displaystyle \sum _{k=j}^{p}(k+1)_{t}(k+1-t)_{j-t}a_{p,k}=(p)_{t}(p-t)_{j-t}}
Supstitucijom k =k' +t
∑
k
′
=
j
−
t
p
−
t
(
k
′
+
1
)
j
−
t
[
(
k
′
+
t
+
1
)
t
(
p
)
t
a
p
,
k
′
+
t
]
=
(
p
−
t
)
j
−
t
.
{\displaystyle \sum _{k'=j-t}^{p-t}(k'+1)_{j-t}\left[{\frac {(k'+t+1)_{t}}{(p)_{t}}}a_{p,k'+t}\right]=(p-t)_{j-t}.}
odnosno:
(
k
′
+
t
+
1
)
t
(
p
)
t
a
p
,
k
′
+
t
=
a
p
−
t
,
k
′
.
{\displaystyle {\frac {(k'+t+1)_{t}}{(p)_{t}}}a_{p,k'+t}=a_{p-t,k'}.}
Za k' =0
a
p
,
t
=
(
p
)
t
(
t
+
1
)
t
a
p
−
t
,
0
=
(
p
t
)
a
p
−
t
,
0
t
+
1
.
{\displaystyle a_{p,t}={\frac {(p)_{t}}{(t+1)_{t}}}a_{p-t,0}={\binom {p}{t}}{\frac {a_{p-t,0}}{t+1}}.}
a to upravo odgovara Bernulijevim brojevima , tako da konačno dobijamo:
∑
k
=
1
n
k
p
=
∑
j
=
0
p
(
p
j
)
B
p
−
j
j
+
1
n
j
+
1
{\displaystyle \sum _{k=1}^{n}k^{p}=\sum _{j=0}^{p}{\binom {p}{j}}{\frac {B_{p-j}}{j+1}}n^{j+1}}
uredi
∑
k
=
1
n
k
p
=
B
p
+
1
(
n
+
1
)
−
B
p
+
1
(
0
)
p
+
1
,
{\displaystyle \sum _{k=1}^{n}k^{p}={\frac {B_{p+1}(n+1)-B_{p+1}(0)}{p+1}},}
a tu su
B
p
(
x
)
{\displaystyle B_{p}(x)}
Bernulijevi polinomi .
uredi
Faulhaberova formula Donald E. Knuth: Johann Faulhaber and Sums of Powers. Math. Comp. 61 (1993), no. 203, S. 277-294
Abramowitz, Milton; Stegun, Irene A., eds. (1965), Handbook of Mathematical Functions with Formulas, Graphs, and Mathematical Tables, New York: Dover, ISBN 978-0486612720
Bernulijevi brojevi 
![{\displaystyle \sum _{k=0}^{\infty }a_{p,k}\left[\left(\sum _{j=0}^{k+1}{\binom {k+1}{j}}n^{j}\right)-n^{k+1}\right]=(n+1)^{p}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/52ae5d9e150bd8bceb33c20ef3adb615896a0520)
![{\displaystyle \sum _{k'=j-t}^{p-t}(k'+1)_{j-t}\left[{\frac {(k'+t+1)_{t}}{(p)_{t}}}a_{p,k'+t}\right]=(p-t)_{j-t}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f07d4eaa2d40947109ceb0f88f644fd4814d2d85)
