A New Formula for the Sum of Integer Powers Based on Elementary Matrix Theory

#A98 INTEGERS 25 (2025)
A NEW FORMULA FOR THE SUM OF INTEGER POWERS
BASED ON ELEMENTARY MATRIX THEORY
Pei an Ding
Wuxi Dipon School o A s and Science, Wuxi, Jiangsu, China
[email p o ec ed]
Yanjun Zhu
Wuxi Dipon School o A s and Science, Wuxi, Jiangsu, China
[email p o ec ed]
Lei Wang1
Scholas ic Excellence Resea ch Cen e , Wuxi Dipon School o A s and Science,
Wuxi, Jiangsu, China
[email p o ec ed]
Recei ed: 7/16/24, Re ised: 7/16/25, Accep ed: 10/7/25, Published: 11/5/25
Abs ac
In his s udy, we p esen a no el o mula o he sum o he k- h powe s o he i s
nposi i e in ege s, deno ed as S(n, k). Unlike exis ing o mula ions ha employ
binomial expansions, gene a ing unc ions, o make use o special numbe s such as
Be noulli o S i ling numbe s, ou app oach le e ages undamen al p inciples om
elemen a y ma ix heo y. Addi ionally, we demons a e ha o e y la ge alues
o n, he de i ed o mula can be pa ially simpli ied. By con as ing ou indings
wi h a p e iously es ablished Be noulli- ype o mula, we ul ima ely de i e a new
combina o ial iden i y.
1. In oduc ion
Ou in es iga ion begins wi h he well-es ablished exp ession o S(n, k), de ined as
he sum o he k- h powe s o he i s nposi i e in ege s:
S(n, k) = 1k+ 2k+· · · +nk.
The o mulas o S(n, k) ha e been he subjec o ex ensi e s udy o e se e al
cen u ies [3, 6, 9, 11]. In he pas decade, nume ous esea che s ha e con inued o
DOI: 10.5281/zenodo.17535263
1co esponding au ho
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explo e he de i a ion o new o mulas as well as he p oo o exis ing ones om
a ious me hodologies [2, 4, 5, 7, 8, 10]. A p ominen o mula in his domain is
de i ed om Pascal’s iden i y in conjunc ion wi h he binomial heo em, gi en by
S(n, k) = nk+1
k+ 1 +
k−1
X
=0 k
(−1)k− +1
k− + 1 S(n, ).(1)
This exp ession e eals a clea ecu ence ela ion be ween S(n, k) and S(n, ) (see
Equa ion (1)), hus, he compu a ion o S(n, k) necessi a es he p io de e mina ion
o S(n, ). Comp ehensi e analyses and p oo s conce ning his o mula can be ound
in e e ences [7, 8].
Subsequen ly, se e al esea che s ha e sough o de elop new o mulas ha elim-
ina e hese ecu ence ela ions, in oducing no el elemen s such as Be noulli num-
be s [3, 9], S i ling numbe s [5, 10], and hype ha monic numbe s [4]. Some o hese
o mula ions a e exp essed in he ollowing equa ions:
S(n, k) = nk+1
k+ 1 +nk
2+
k
X
i=2
Bi
ik
i−1nk−i+1,(2)
S(n, k) =
k
X
=0
!k
n+ 1
+ 1,(3)
S(n, k) = (−1)k+1
k+ 1
k+1
X
j=1
(−1)jj!k+ 1
jH(n)
j+1 −1
j+ 1.(4)
In hese equa ions, Bi ep esen s he Be noulli numbe s, k
deno es he S i ling
numbe s o he second kind, and H(n)
j+1 signi ies he (j+1)- h hype ha monic numbe .
De ailed discussions and p oo s o hese o mulas a e a ailable in he ci ed e e ences.
A close examina ion o he de i a ion p ocesses o o mulas (1)-(4) e eals ha
hey p edominan ly ely on binomial expansions, gene a ing unc ions, o he u i-
liza ion o special numbe s. In con as , ou s udy de i es a new o mula o S(n, k)
h ough he lens o ma ix heo y, employing se e al key esul s ela ed o ma ix
aces. Fu he mo e, we es ablish a new combina o ial iden i y by jux aposing ou
simpli ied o mula wi h a p e iously es ablished Be noulli- ype exp ession.
2. P elimina ies
2.1. Rela ionship Be ween S(n, k) and he T ace o Ma ices
I is well-es ablished ha eal- alued n×ndiagonal ma ices con ain nonze o en-
ies exclusi ely on hei main diagonal, wi h all o -diagonal elemen s equal o
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ze o. Conside a diagonal ma ix Awhose main diagonal consis s o he elemen s
1,2, . . . , n:
A=




1 0 · · · 0
0 2 · · · 0
.
.
..
.
.....
.
.
0 0 · · · n





.
U ilizing he undamen al ules o ma ix mul iplica ion, we ind ha he k- h powe
o ma ix Ais gi en by
Ak=




1k0· · · 0
0 2k· · · 0
.
.
..
.
.....
.
.
0 0 · · · nk





.
F om he de ini ion o he ace o a ma ix, we can de i e he ollowing ela ionship:
T Ak= 1k+ 2k+· · · +nk=S(n, k).
This ela ion indica es ha he analysis o he o mula o S(n, k) can be ans-
o med in o an in es iga ion o he ace T Ak. Fo small alues o k, such as
k= 1,2,3, he aces a e well-known:
T A1= 11+ 21+· · · +n1=n(n+ 1)
2,
T A2= 12+ 22+· · · +n2=n(n+ 1)(2n+ 1)
6,
T A3= 13+ 23+· · · +n3=n2(n+ 1)2
4.
Howe e , o la ge alues o k(k > 3), ob aining a gene al o mula becomes in-
c easingly complex. Insigh s om a p e iously es ablished heo em discussed in
Sec ion 2.2, which p esen s wo o mulas cha ac e izing he aces o 2 ×2 eal-
alued ma ices aised o he k- h powe , will be ins umen al in ad ancing ou
s udy.
2.2. A P e iously Es ablished Theo em
Theo em 1. Fo any posi i e in ege mand o 2×2 eal- alued ma ices M, he
ollowing ela ions hold:
T M2m= (2m)
m
X
=0
(−1)
!
(2m− −1)!
(2m−2 )! [De (M)] [T (M)]2m−2 ,(5)
and
T M2m+1= (2m+ 1)
m
X
=0
(−1)
!
(2m− )!
(2m+ 1 −2 )! [De (M)] [T (M)]2m+1−2 ,(6)
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whe e T (·)and De (·)deno e he ace and de e minan o he co esponding ma-
ices, espec i ely.
P oo . The p oo o he o mulas p esen ed in Theo em 1 can be es ablished using
he me hod o ma hema ical induc ion. A comp ehensi e explana ion o he p oo
can be ound in e e ence [1].
2.3. Decomposi ion o S(n, k)
To u ilize he conclusions de i ed om Theo em 1 in o mula ing a new exp es-
sion o T Ak, i is necessa y o es ic he dimension o ma ix A o 2 ×2
o o decompose he n×nma ix Ain o smalle 2 ×2 ma ices. Consequen ly,
since T Ak=S(n, k), we can exp ess S(n, k) as he sum o he aces o hese
decomposed 2 ×2 ma ices.
Lemma 1. Fo any posi i e in ege kand e en posi i e in ege n,S(n, k)can be
exp essed as he sum o ⌊n/2⌋ aces o 2×2 eal- alued ma ices,
S(n, k) = T Mk
1+ T Mk
2+· · · + T Mk
s,
whe e s=⌊n/2⌋, and Mi=i0
0 2s+ 1 −ia e 2×2 eal- alued ma ices o
1⩽i⩽s.
P oo . Le n= 2s. Thus, we can ew i e S(n, k) as
S(n, k) = 1k+ 2k+· · · + (2s)k.
The 2s e ms in S(n, k) can be g ouped in o s=⌊n/2⌋pai s. Fo each pai , such
as ik+ (2s+ 1 −i)k, we ha e Mi=i0
0 2s+ 1 −i, leading o he ela ion
ik+ (2s+ 1 −i)k= T Mk
i.
Since he e a e di e en Mis, we conclude ha
S(n, k) = T Mk
1+ T Mk
2+· · · + T Mk
s,
whe e M1, M2, . . . , Msa e 2 ×2 eal- alued ma ices o all 1 ⩽i⩽s. So, Lemma
1 holds.
Lemma 2. Fo any posi i e in ege kand odd posi i e in ege n,S(n, k)can be
exp essed as he sum o ⌊n/2⌋ aces o 2×2 eal- alued ma ices, plus nk,
S(n, k) = T Mk
1+ T Mk
2+· · · + T Mk
s+nk,
whe e s=⌊n/2⌋, and Mia e 2×2 eal- alued ma ices o 1⩽i⩽s.
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P oo . Since nis odd, we can se n= 2s+ 1, whe e sis a posi i e in ege . Thus,
we can ew i e S(n, k) as
S(n, k)=1k+ 2k+· · · + (2s)k+ (2s+ 1)k.
By ocusing on he i s 2s e ms, we can apply a p ocedu e analogous o ha used
in he p oo o Lemma 1. The e o e, we es ablish ha
S(n, k) = T Mk
1+ T Mk
2+· · · + T Mk
s+nk,
whe e M1, M2, . . . , Msa e 2×2 eal- alued ma ices o all 1 ⩽i⩽s. Thus, Lemma
2 holds.
2.4. De ini ion o he Func ion W(n, )
In his sec ion, we in oduce he unc ion W(n, ), which is a c ucial componen in
ou de i ed o mula. The unc ion is de ined as
W(n, ) =
n
X
i=1  i
n+ 1 1−i
n+ 1 .
Le ing xi=i
n+1 and (xi) = x
i(1 −xi) , we no e ha since 1 ⩽i⩽n, i ollows
ha 0 < xi<1. The unc ion (x) = x (1 −x) is cha ac e ized by a symme ical
shape a ound x= 0.5. Consequen ly, when nis e en, W(n, ) can be ew i en as
W(n, ) = (x1) + (x2) + · · · + (xn/2) + (xn/2+1) + · · · + (xn−1) + (xn).
F om he de ini ion o xi, we can deduce he ollowing ela ionships:
x1+xn=1
n+ 1 +n
n+ 1 =1 + n
n+ 1 = 1 = 2 ×0.5
implies (x1) = (xn),
x2+xn−1=2
n+ 1 +n−1
n+ 1 =1 + n
n+ 1 = 1 = 2 ×0.5
implies (x2) = (xn−1),
.
.
.
xn/2+xn/2+1 =n/2
n+ 1 +n/2+1
n+ 1 =n+ 1
n+ 1 = 1 = 2 ×0.5
implies (xn/2) = (xn/2+1).
Thus, we conclude ha
n/2
X
i=1  i
n+ 1 1−i
n+ 1 =W(n, )
2.

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3. Main esul s
3.1. A New Fo mula o S(n, k)
In his sec ion, we de i e ou p ima y o mula o S(n, k).
Theo em 2. Fo any posi i e in ege s kand n, he quan i y S(n, k)can be exp essed
as
S(n, k) = k(n+ 1)k
2
⌊k/2⌋
X
=0
(−1)
k− k−
W(n, ),(7)
whe e W(n, )is he unc ion de ined in Sec ion 2.4.
P oo . We conside wo cases.
Case I: nis e en. In his case, based on Lemma 1, we ha e
S(n, k) = T Mk
1+ T Mk
2+· · · + T Mk
s,
whe e Mia e he 2 ×2 ma ices de ined p e iously.
I kis e en, le k= 2m. U ilizing he o mula om Equa ion (5) alongside he
alues o De (Mi) and T (Mi), we ob ain
T M2m
i= (2m)
m
X
=0
(−1)
!
(2m− −1)!
(2m−2 )! [De (Mi)] [T (Mi)]2m−2
= (2m)
m
X
=0
(−1)
!
(2m− −1)!
(2m−2 )! [i(2s+ 1 −i)] [2s+ 1]2m−2
= (2m)(2s+ 1)2m
m
X
=0
(−1)
!
(2m− −1)!
(2m−2 )!
[i(2s+ 1 −i)]
[2s+ 1]2
= (2m)(2s+ 1)2m
m
X
=0
(−1)
2m− 2m−
 i
2s+ 1 1−i
2s+ 1 .
Thus, we can exp ess S(n, k) as
S(n, k) = (2m)(2s+ 1)2m
m
X
=0
(−1)
2m− 2m−
s
X
i=1  i
2s+ 1 1−i
2s+ 1 .
Subs i u ing n= 2sand k= 2m, we de i e
S(n, k) = k(n+ 1)k
⌊k/2⌋
X
=0
(−1)
k− k−
n/2
X
i=1  i
n+ 1 1−i
n+ 1 .
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Thus, we conclude ha
S(n, k) = k(n+ 1)k
2
⌊k/2⌋
X
=0
(−1)
k− k−
W(n, ).
I kis odd, le k= 2m+1. Using he o mula om Equa ion (6) wi h he ele an
alues o De (Mi) and T (Mi), we ob ain
T M2m+1
i= (2m+ 1)
m
X
=0
(−1)
!
(2m− )!
(2m+ 1 −2 )! [De (Mi)] [T (Mi)]2m+1−2
= (2m+ 1)
m
X
=0
(−1)
!
(2m− )!
(2m+ 1 −2 )![i(2s+ 1 −i)] [2s+ 1]2m+1−2
= (2m+ 1)(2s+ 1)2m+1
m
X
=0
(−1)
!
(2m− )!
(2m+ 1 −2 )!
[i(2s+ 1 −i)]
[2s+ 1]2
= (2m+ 1)(2s+ 1)2m+1
m
X
=0
(−1)
2m+ 1 − 2m+ 1 −

· i
2s+ 1 1−i
2s+ 1 .
Thus, we exp ess S(n, k) as
S(n, k) = (2m+ 1)(2s+ 1)2m+1
m
X
=0
(−1)
2m+ 1 − 2m+ 1 −

·
s
X
i=1  i
2s+ 1 1−i
2s+ 1 .
Subs i u ing n= 2s,k= 2m+ 1, we ind
S(n, k) = k(n+ 1)k
⌊k/2⌋
X
=0
(−1)
k− k−
n/2
X
i=1  i
n+ 1 1−i
n+ 1 .
Thus, we conclude ha
S(n, k) = k(n+ 1)k
2
⌊k/2⌋
X
=0
(−1)
k− k−
W(n, ).
Case II: nis odd. In his case, based on Lemma 2, we ha e
S(n, k) = T Mk
1+ T Mk
2+· · · + T Mk
s+nk.
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U ilizing simila a gumen s and analy ical s a egies as in Case I, we can show ha
when nis odd, S(n, k) can be exp essed as
S(n, k) = knk
2
⌊k/2⌋
X
=0
(−1)
k− k−
W(n−1, ) + nk.(8)
The de ailed p oo s o his po ion a e p o ided in he Appendix. Nex , we demon-
s a e ha Equa ions (7) and (8) a e equi alen . F om he de ini ion o S(n, k),
we ha e S(n, k) = S(n−1, k) + nk o all posi i e in ege s n. In pa icula , i nis
odd, hen n−1 is e en, allowing us o apply Equa ion (7) o a i e a Equa ion
(8). Consequen ly, Equa ion (7) is alid o odd alues o n. Thus, o any posi i e
in ege s kand n, we conclude ha
S(n, k) = k(n+ 1)k
2
⌊k/2⌋
X
=0
(−1)
k− k−
W(n, ).
The e o e, Theo em 2 is es ablished.
3.2. Asymp o ic Fo mula o S(n, k) as nApp oaches In ini y
In Sec ion 2.4, we de ined he unc ion W(n, ). He e, we ex end ou discussion o
his unc ion by examining he asymp o ic beha io as napp oaches in ini y. Recall
he exp ession o W(n, ), gi en by
W(n, ) =
n
X
i=1  i
n+ 1 1−i
n+ 1 .
Le ing xi=i
n+1 , we can exp ess his as
W(n, ) =
n
X
i=1
(xi),
whe e (xi) = x
i(1−xi) . No ably, since x0=0
n+1 = 0 and 1−xn+1 = 1−n+1
n+1 = 0,
we ind ha (x0) = 0 and (xn+1) = 0. Thus, we can ew i e W(n, ) as
W(n, ) =
n+1
X
i=0
(xi) = (n+ 1) ·
n+1
X
i=0
(xi)1
n+ 1,
he e, 1
n+1 se es as he wid h o each subin e al, ∆xi, and as nbecomes e y la ge,
∆xi≈dx. By he de ini ion o he de ini e in eg al, we can hus app oxima e
W(n, )≈(n+ 1) Z1
0
(x) dx= (n+ 1) Z1
0
x (1 −x) dx.
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De ini ion 1. (Be a unc ion). The ollowing unc ion B(a, b) is de ined as he
Be a unc ion, i i sa is ies
B(a, b) = Z1
0
xa−1(1 −x)b−1dx,
whe e a,ba e eal numbe s.
De ini ion 2. (Gamma unc ion). The ollowing unc ion Γ(n) is de ined as he
Gamma unc ion, i i sa is ies
Γ(n) = Z∞
0
xn−1e−xdx= (n−1)!,
whe e nis a posi i e in ege .
Lemma 3. The ollowing holds o all posi i e in ege s :
Z1
0
x (1 −x) dx=1
(2 + 1)2
.
P oo . By De ini ion 1, we ha e
B(a, b) = Z1
0
xa−1(1 −x)b−1dx.
Thus, i ollows ha
Z1
0
x (1 −x) dx=B( + 1, + 1).
By De ini ion 2, we ha e
Γ(n)=(n−1)!.
Employing he ela ionship be ween he Be a unc ion and he Gamma unc ion
[12], we ob ain
B(a, b) = Γ(a)Γ(b)
Γ(a+b).
This leads o
B( + 1, + 1) = Γ( + 1)Γ( + 1)
Γ(2 + 2) .
Finally, we de i e
B( + 1, + 1) = !· !
(2 + 1)! = !· !
(2 )!
1
(2 + 1) =1
(2 + 1)2
.
Thus, Lemma 3 is es ablished.