Sums of Integer Powers--The Bernoulli, Binomial,
Stirling and Faulhaber Expansions
Norman Brenner
George Washington University
DRAFT VERSION
1960; 2002 Sep 12, 20, Oct 8; 2003 Apr 15 (shift to 1..n); Apr 18
(add binom); May 21; May 22 (shift back to 0..n-1); 2009 Sep 8
Perhaps publish in College Math Monthly (edited by Lowell
Leinecke, known to jw@cc)
I. Statement of the problem
Consider computing the sum of consecutive integers to a given power; we define
for all values of p:
n 1
Sp(n) :=

ip , p>0
i 0
Every schoolchild has seen the formula for the sum of the integers themselves:
S1(n) = n(n-1)/2
But the formulas for higher powers are unfamiliar to most people. In this
paper, we give four explicit formulas for such sums. One formula may be
original, and all the proofs are. The purpose of this paper, then, is pedagogic
exposition.
A special definition for the case p=0 is necessary because of the indeterminacy
of 00; taking it equal to 1, we define:
S0(n) := n
This value is chosen to be consistent with the recursion and differential
relations below.
[Alternately, we could work with the sum from i=1 through n; but the range as
indicated tends to yield simpler formulas in the rest of this work.]
II.
Recursion relation
We can easily derive a recursion relation among these sums.
From the definition for p>0,
n
Sp(n+1) =

i 0
ip
Consider Sp(n+1).
Rewrite this sum in two different ways.
First, split off the largest term:
n 1

Sp(n+1) = np +
ip
i 0
= np + Sp(n)
Alternately, replace the summation index by defining i:=j+1 :
n 1
Sp(n+1) =

(j+1)p
j  1
n 1
=
0p

+
(j+1)p
j 0
The kernel of this latter summation can be expanded by the binomial theorem, and
then the order of the two summations exchanged:
n 1
Sp(n+1) =
j 0
=

k
The
the
the
all
 p k
  j
k 
 
k
 p
 
k 
n 1

jk
j 0
limits of the summation variable k here are defaulted to –  to +  ; since
binomial coefficient is zero outside the range k=0..p, the latter becomes
de facto range. However, it is simpler to write index k bare, meaning “over
possible values of k”.
The inner summation is exactly the definition of Sk(n), so:
Sp(n+1) =

k
 p
  Sk(n)
k 
Combining the alternate expansions,
np +
Sp(n) =

k
 p
  Sk(n)
k 
Rearranging,
np =

k
 p
  Sk(n) - Sp(n)
k 
A helpful notation is to define an abstract symbol "S", whose p'th power is
defined to be Sp(n); then we may write more concisely,
np = (S+1)p - Sp,
p>0
A similar formula was published by X. Bernoulli in 17??.
[Look it up!]
III. Table of explicit expressions
With this recursion relation, we can now build a table of explicit formulas for
the Sp(n). Start by substituting p=1 in it:
n1 = (S+1)1 – S1
i.e.
n = (S1+S0) – S1
or
n = S0
Working upward:
S0(n)
S1(n)
S2(n)
S3(n)
S4(n)
S5(n)
S6(n)
S7(n)
=
=
=
=
=
=
=
=
n
n2/2
n3/3
n4/4
n5/5
n6/6
n7/7
n8/8
-
n/2
n2/2
n3/2
n4/2
n5/2
n6/2
n7/2
+
+
+
+
+
+
n/6
n2/4
n3/3 - n/30
5 n4/12 - n2/12
n5/2 - n3/6 + n/42
7 n6/12 – 7 n4/24 + n2/12
It is a striking result (Nicomachus's theorem) that
S3(n) = [S1(n)]2
[Author’s aside. An elegant geometric proof of this formula exists by unpacking
each cube into slabs, which are then attached to a square to form a larger
square. An algebraic proof is, of course, elementary, by performing
mathematical induction on n.]
We note also, for future reference, that
3S5(n) +
S3(n) = 4[S1(n)]3
and
S7(n) + S5(n) = 2[S1(n)]4
which will fall out below from another recursion relation.
IV.
Numerical checks
Simple numerical checks on the correctness of these formulas are available by
noting that, from the definition,
Sp(1) = 0
Sp(2) = 1
Sp(3) = 1 + 2p
Check e.g.
S7(1) = 1/8 - 1/2 + 7/12 – 7/24 + 1/12 = 0.
This is correct.
V. Worpitzky’s Expansion in terms of Binomial Coefficients
Rather than express Sp(n) as a sum of powers of n, it may be useful to express
them as a sum over binomial coefficients of comparable magnitude.
In lieu of
proof (cf. Worpitzky, J., J. reine angew. Math. 94, 203-232, 1883), we simply
give the first few here.
n
S0(n) =  
1 
n
S1(n) =  
2
n
S2(n) =  
3 
n
S3(n) =  
4
n
S4(n) =  
5 
n
S5(n) =  
6
 n  1


3


n

 1
 n  2
 + 

+ 4 
4
4




n

1
n


  2
 n  3
 + 11 

 + 
+ 11 
5
5
5






n

1
n

2
n




  3
 n  4
 + 66 
 + 
 + 26 

+ 26 
6
6
6
6








+
By substituting n=2 and n=3, and using the numerical values of Sp(n) given
above, we solve the linear equations for different n and easily determine the
leading coefficients to be:
n

n 1
 + (2p – p – 1) 
 +
Sp(n) = 
 p  1
 p  1

3p - 2p(p+1) +
 p  1


2


 n  2

 + ...
 p  1
Also, the coefficients within one expression for Sp(n) are symmetric.
general expression is
p 1
Sp(n) =

j 0
j 1


k 0
 p  1

(-1)k (j+1-k)p 
k


The
n  j


 p  1
and the inner sum is the same if j be replaced by p-1-j (i.e. symmetric
coefficients).
This is an attractively simple and easily computed formula.
The coefficients are called Eulerian numbers (not to be confused with Euler
numbers). Their row sum for fixed p is (p+1)!
Notice that all binomial coefficients in the expansion for Sp(n) are terms of
order of magnitude np+1, and can be readily computed in sequence by a single
multiplication and division. Also, the Eulerian coefficients can be directly
calculated without recurrence, making these formulas easier to find, even though
longer than ones below. They are not listed in the standard handbook by
Abramowitz & Stegun, but are described in Concrete Mathematics, by Graham, Knuth
& Patashnik (1994).
VI.
Stirling’s Expansion in terms of Binomial Coefficients
An alternative expression of Sp(n) as a sum of binomial coefficients of
increasing magnitude, for which the coefficients are the Stirling coefficients
of the second kind, will be found in Abramowitz & Stegun, Handbook of
Mathematical Functions, formula 24.1.4 C(2):
[Stirling “numbers”?]
p
Sp(n) =

k 0
n 
k!  p(k) 

k
1


[We use a script T here because we do not have A&S’s symbol, a script S.]
E.g. here is S4(n):
n
n
n
n
S4(n) =   + 14   + 36   + 24  
2
3 
4
5 
It will be
therefore,
n5. Also,
unlike the
seen that the lower indices of these binomial coefficients differ;
they represent ascending powers of n, from order of n2 to order of
the Stirling coefficients are not symmetric within a single Sp(n),
coefficients in the previous expansion.
Both of these expansions, in terms of binomial coefficients, are rather more
bulky than the more compact expansions that we now turn to.
VII. Differentiation formula and Bernoulli’s Expansion
We may try differentiating
empirically that
Sp(n) as given in the table above, and discover
dSp(n)/dn = p Sp-1(n) + Bp
where Bp are a series of constants to be determined. This differentiation
formula is easily proved by mathematical induction using the recursion formula
above.
[Insert proof by induction]
The Bernoulli numbers may be defined by the recursion formula remaining after
Sp(n) is removed from the derivative of the basic recursion formula. Here is a
table of the first few Bernoulli numbers:
B0 = 1
B1 = -1/2
B2 = 1/6
B3 = 0
B4 = -1/30
B5 = 0
B6 = 1/42
B7 = 0
B8 = -1/30
B9 = 0
B10 = 5/66
B11 = 0
B12 = -691/2730
B13 = 0
B14 = 7/6
B15 = 0
B16 = -3617/510
Note that, except for B1, all odd-indexed B2q+1 are 0, while the even-indexed
ones alternate in sign. Further, the denominator of any Bernoulli number is
readily computable, whereas a formula for the numerator is not known to me.
[Insert Bernoulli denominator formula]
An explicit formula for the coefficients of the nk in any
dependent on the Bernoulli numbers:
p
Sp(n) = (p+1)-1

i 0
Sp(k) is derivable,
 p  1

 Bi np+1-i
i

Using the symbol “B”, whose j’th power is a notation for Bj, we may write this
more compactly as
Sp = [(n+B)p+1 – Bp+1]/(p+1)
In effect, this is the „matrix inverse” to the recursion equation shown above
for the Sp. Note that this can be written suggestively as a definite integral:
⌠n
Sp = │ (x+B)p dx
⌡0
Incidentally, it is not possible to compute S-1(n) (called the harmonic sums)
viaa the above differentiation formula, since substituting p=0 therein yields
dS0(n)/dn = 0
S-1(n) + B0
i.e.
1 = 0
S-1(n) + 1
which tells us nothing about
S-1(n), except that it is finite.
VIII.
Generator function
A generator function for the sums may be defined by
G(n,t) :=

Sp(n) tp/p!
p
Here, t is a dummy placeholder variable, whose sole purpose is to mark and
separate the coefficients. (As before, the bare index p is to be interpreted as
ranging over all possible values of p, in this case, for p=0..  .) Inserting
the definition of Sp(n), and exchanging summations,
G(n,t) = (ent - 1)/(et - 1)
By taking the partial derivative of G with respect to n, we may again prove the
formula for the derivative dSp(n)/dn.
The generator formula for the Bernoulli numbers is similar; first define the
Bernoulli polynomials Bp(x) by
text/(et - 1) :=

Bp(x) tp/p!
p
Then define
Bp := Bp(0)
That is,
t/(et - 1) =

Bp tp/p!
p
If we replace the dummy variable t in G(n,t) by it, where i is the imaginary
unit, and extract the pure imaginary part, we have:
Imag(G(n,it)) = [G(n,it) - G(n,-it)]/(2i)
= [(enit - 1)/(eit - 1) - (e-nit - 1)/(e-it - 1)]/(2i)
= [enit + e(1-n)it - eit - 1] / [2i(eit - 1)]
and, from the power series definition,
Imag(G(n,it)) =
 
Sp(n) (it)p/p! -

p
Only the odd-indexed terms do not cancel.
=

Sp(n) (-it)p/p!

/ (2i)
p
Substituting p:=2q+1,
S2q+1(n) (-1)q t2q+1/(2q+1)!
q
A recursion equation for the odd-indexed sums will appear below. Meantime,
notice that in the left-hand side, if n is replaced by 1–n, there is no change
in the expression. This suggests defining a variable which is also invariant
when n is replaced by 1–n, e.g.
M := n(n-1)
In this case,
n = (1 +
4M  1 )/2
and
1-n = (1 -
4M  1 )/2
Now, the exponential functions on the left-hand side are expansible in terms of
powers of n and 1-n.
The odd-numbered powers of the square root will cancel
out, leaving only integer powers of 4M+1.
Therefore, the left-hand side
becomes a (very complicated) polynomial in terms of M, and not in terms of n
directly.
We will see explicit formulas in terms of M, below.
[We have found no further need for the generator function.]
[Another possible approach, suggested by Balaji, is to apply a Fourier
transform.]
IX.
Explicit factoring and Faulhaber’s expansion
A possibly fruitful path to explore lies in factoring the explicit expressions
above:
S0(n)
S1(n)
S2(n)
S3(n)
S4(n)
S5(n)
S6(n)
S7(n)
=
=
=
=
=
=
=
=
n
n (n-1) / 2
n (n-1) (2n-1) / 6
n2 (n-1)2 / 4
n (n-1) (2n-1) (3n2 - 3n - 1) / 30
n2 (n-1)2 (2n2 - 2n - 1) / 12
n (n-1) (2n-1) (3n4 - 6n3 + 3n + 1) / 42
n2 (n-1)2 (3n4 - 6n3 - n2 + 4n + 2) / 24
By direct substitution in this table, it appears that
Sp(1-n) = (-1)p+1 Sp(n),
p>0
As before, this suggests defining a variable which is also invariant when n is
replaced by 1–n. On further examination, we see that the parameter
M := n(n-1)
appears as a factor in every Sp(n) for p>0, and its derivative
M' := dM/dn = 2n - 1
appears in every S2q(n) for q>0. Therefore, rewriting the table in terms of M,
we have more compact expressions:
S0(n) = M'/2 + 1/2
S1(n) = M/2
S2(n)
S3(n)
S4(n)
S5(n)
S6(n)
S7(n)
=
=
=
=
=
=
(M'/2) M/3
M2/4
(M'/2)(M2 - M/3)/5
(M3 - M2/2)/6
(M'/2)(M3 - M2 + M/3)/7
(M4 - 4 M3/3 + 2 M2/3)/8
It will be helpful to give the odd-indexed sums a new name; define
Fq(M) := S2q+1(n)
and call these the Faulhaber polynomials.
obtainable by differentiation:
The even-indexed sums would thence be
S2q(n) = M' [ dFq(M)/dM ] /(2q+1) – B2q+1
Of course, B2q+1 = 0 except for B1 = -1/2.
These compact expansions in terms of M (with M’ written out as 2n-1) were
published by Johann Faulhaber in 1631, proven by Jacobi in 1834, and
rediscovered by D. E. Knuth in 1993: “Johann Faulhaber and Sums of Powers”,
Mathematics of Computation 61 (1993), 277-294. A short biography of Faulhaber
is on the web site
http://www-groups.dcs.stand.ac.uk/~history/Mathematicians/Faulhaber.html
[I have not actually read any of these last three papers.]
X. Recursion relation for the Faulhaber polynomials
It is elementary to derive a recursion formula for the Faulhaber polynomials.
[This proof is original with me, so far as I know.] From its definition,
Mq = nq (n-1)q
We may expand this immediately with the binomial theorem:
Mq =

i
q
  (-n)q+i
i 
(*)
For later use, substitute i:=j-q so that
Mq =

j
q


 (-n)j
j

q


(**)
We will come back to these expansions later, so we give them markers.
Alternately, we can first rewrite Mq as
Mq = [(n-1)+1]q (n-1)q
and expand this with respect to n-1 by the binomial theorem:
Mq =

i
q
  (n-1)q+i
i 
[Proof needed?]
Next, expand (n-1)q+I by the binomial theorem:
Mq =

q
  (-1)q+i
i 
i

q  i

 (-n)j
j 
j
Exchanging summations here and equating with coefficients of (-n)j in the
previous expansion (**), we prove a useful identity for binomial coefficients:

q


 =
 j  q
(-1)q+i
i
q q  i

  
i   j 
(***)
Return now to equation (*). If we now restrict q>0, then a fortiori also q+i>0,
and so we may substitute for nq+i therein the recursion identity for the Sp(n)
above:
Mq =

i
q
  (-1)q+i
i 


j
q  i

 Sj(n) - Sq+i(n)
j 

Apply the outer summation to each term in the bracketed sum and exchange
summations in the first term:
Mq =


i
=

q
  (-1)q+i
i 
Sj(n)
j



j
(-1)q+i
i
q  i

 Sj(n)
j



q q  i

  
j
i

 

-

i
-

i
q
  (-1)q+i Sq+i(n)
i 
q
  (-1)q+i Sq+i(n)
i 
The bracketed expression in the central term appears in the identity (***) we
derived above. Inserting it,
Mq =

j
q


 Sj(n)  j  q

i
q
  (-1)q+i Sq+i(n)
i 
Substitute i:=j-q in the last term:
Mq =

j
q


 Sj(n)  j  q

j
q


 (-1)j Sj(n)
j

q


Notice that terms with even-valued j cancel out, leaving only the odd j:=2k+1
terms:
Mq = 2

k
q


 S2k+1(n)
 2k  1  q 
Substitute k:=q–p, and the definition of Faulhaber polynomials, and this
becomes:
Mq = 2

p
q


 Fq-p(M),
 2 p 1
q>0
This is the desired recursion relation. Again, use an abstract symbol "F",
whose p'th power is Fp(M), and we may write more concisely,
Mq = F(q-1)/2 [ (F1/2+1)q – (F1/2-1)q ],
q>0
This formula can be rewritten as a matrix multiplication:
M = (P + P-1)·F
Where the matrix Pjk is the Pascal triangle extended with 0’s (i.e. the entries
are the binomial coefficients), and its inverse matrix has the same entries but
with alternating signs. M and F are of course the vectors of the powers of M
and the symbol F.
To express F in terms of M would be equivalent to inverting
the matrix multiplier shown.
XI.
Numerical checks for the Faulhaber recursion formula
When q=1, this is
M1 = F0/2 [ (F1/2+1)1 – (F1/2-1)1 ]
or
M1 = 2F0
i.e.
M = 2F0(M)
When q=2, this is
M2 = F1/2 [ (F1/2+1)2 – (F1/2-1)2 ]
or
M2 = 4F1
i.e.
M2 = 4F1(M)
which we saw earlier in the form
S3(n) = M2/4
When q=3, this is
M3 = F [ (F1/2+1)3 – (F1/2-1)3 ]
or
M3 = 6F2 + 2F
i.e.
M3 = 6F2(M) + 2F1(M)
which we saw earlier in the form
3S5(n) +
S3(n) = 4[S1(n)]3
When q=4, this is
M4 = F3/2 [ (F1/2+1)4 – (F1/2-1)4 ]
or
M4 = 8F3 + 8F2
i.e.
M4 = 8F3(M) + 8F2(M)
We saw this same equation earlier in the form
S7(n) + S5(n) = 2[S1(n)]4
When q=5, this is
M5 = F2 [ (F1/2+1)5 – (F1/2-1)5 ]
or
M5 = 10F4 + 20F3 + 2F2
i.e.
M5 = 10F4(M) + 20F3(M) + 2F2(M)
XII. Differential equation for the Faulhaber polynomials
Since we have a differential equation for Sp(n) with respect to n, we may
substitute the function Fq(M) and replace d/dn by (2n-1)d/dM. We find
2Fq’ + (4M+1)Fq” = (2q+1)(2q Fq-1 + B2q),
q>-1
[Check!]
Possibly useful is
Fq’(M) = [(2q+1)S2q(n) + B2q+1]/(2n-1) ,
q>-1
Recall that B2q+1 = 0 except for B1 = -1/2.
[A generating function for the Faulhaber polynomials would be useful.]
XIII.
Explicit expressions for the Faulhaber polynomials
Using the recursion formula, we may compute explicit expressions for the Fq(M),
for larger values of q; the following table extends up to F8(M) (=S17(n)):
F0(M)
F1(M)
F2(M)
F3(M)
F4(M)
F5(M)
F6(M)
F7(M)
F8(M)
- 10851
= M/2
= M2/4
= (M3 - M2/2)/6
= (M4 - 4 M3/3 + 2 M2/3)/8
= (M5 - 5 M4/2 + 3 M3 - 3 M2/2)/10
= (M6 - 4 M5 + 17 M4/2 - 10 M3 + 5 M2)/12
= (M7 - 35 M6/6 + 287 M5/15 - 118 M4/3 + 691 M3/15 - 691 M2/30)/14
= (M8 - 8 M7 + 112 M6/3 - 352 M5/3 + 718 M4/3 - 280 M3 + 140 M2)/16
= (M9 - 21 M8/2 + 66 M7/3 - 293 M6/3 + 9114 M5/10 - 3711 M4/2 + 10851 M3/5
M2/10)/18
[Hmm. F2(M) is evenly divisible by M-1/2 and F4(M) is evenly divisible by M-1.
But F6(M) is divisible only by M-1.444844178. Bah!]
[The coefficients multiplying the powers of M form a triangular matrix which is
the inverse of the matrix of binomial coeffients multiplying the F’s in the
recursion relation.
Is there a known way of inverting it?]
Note that the last two terms of any expansion (for q>1) may be written in terms
of the Bernoulli numbers:
Fq(M) = .. + (-1)q (2q+1) B2q (M3 - M2/2)
This is easily shown by differentiating Fq(M) twice.
[Check!]
XIV. Explicit expressions for the lower Faulhaber coefficients
Write the summation expansions in the format
Fq(M) = aq Mq+1/(q+1) - bq Mq/q + cq Mq-1/(q-1) - dq Mq-2/(q-2) + ..
[What about the denominator of 2(q+1) in all coefficients?]
[Wha??]
For reference, here are the numerical values of some of the coefficients, taken
from the explicit expressions above:
q
0
1
2
3
4
5
6
7
8
aq/(q+1) -bq/q
1/2
0
1/4
0
1/6
1/12
1/8
1/6
1/10
1/4
1/12
1/3
1/14
5/12
1/16
1/2
1/18
7/12
We find
aq = 1 / 2
cq/(q-1)
0
0
0
1/12
3/10
17/24
41/30
7/3
11/3
-dq/(q-2) eq/(q-3) -fq/(q-4) gq/(q-5) -hq/(q-6)
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
3/20
0
0
0
0
5/6
5/12
0
0
0
59/21
691/210
691/420
0
0
22/3
359/24
35/2
35/4
0
293/18 1519/30
1237/12 3617/30
3617/60
bq
cq
dq
eq
=
=
=
=
q(q-1) / 12
q(q-1)(q-2) (7q-1) / 6!
q(q-1)(q-2)(q-3) (31q2 - 27q - 10) / 6x7!
q(q-1)(q-2)(q-3)(q-4) (127q3 - 310q2 + 37q + 90) / 30x8!
More suggestively, we may write in terms of the Bernoulli numbers:
aq =
bq =
cq =
dq =
eq =
fq =
gq =
q
 
1 
q
 
2
q
 
3 
q
 
4
q
 
5 
q
 
6
q
 
7
1! (-B0/0!) [(2-1-1)q-1]
2!
(B2/2!) [(21-1)q0]
3! (-B4/4!) [(23-1)q - 1]
4!
(B6/6!) [(25-1)q2 - 27q – 10]
5! (-B8/8!) [(27-1)q3 - 310q2 + 37q + 90]
6!
(B10/10!) [(29-1)q4 + (- 12674q3 + 14161q2 + 2486q - 3864)/5]
7! (-B12/12!) [(211-1)q5 + (- 11974437q4 + 31092673q3 - 22432587q2 -
7706534q + 6399960)/691]
[That |B2k| = (k-1)!(k-1)!/((k+1)!(2k+1)) for k=1,2,3,4 seems to just be a
coincidence here. Because for k=5,6,7,8, B2k = 5/66, -691/2730, 7/6, -3617/510,
while the formula yields 4/55, 20/91, 6/7, 70/17.]
A simple way to generate these coefficient expressions is to insert them into
the recursion relation above, and solve for successively smaller powers of M.
aq is determined first, then bq in terms of it, &c. Or the differential
equation for Fq can be used; specifically, equating the coefficients of M q-6:
-(q-5)fq + [2 + 4(q-6)]gq = 2q(2q+1)gq-1/(q-6)
Define the polynomials
FF(q) := 511q4 - ..
GG(q) := 2047q5 - ..
so that
q
  (6! B10/10!) FF(q)
6
q
gq =   (7! (-B12)/12!) GG(q)
7
fq =
Then, given FF(q), the equation to be solved for GG(q) is
0 = (B1012!)/(B1210!)(q-5)FF(q)/2 + (2q-11)(q-6)GG(q) – (2q+1)(q-7)GG(q-1)
Set q to 6 different numerical values, and solve the resulting 6 simultaneous linear
equations for the numerical coefficients of polynomial GG(q).
[Build a generator function for these expansions??]
[How confirm from these models that the diagonal coefficients are Bernoulli
numbers?]
XV.
A numerical check
When n=1, then M=2, and F6(1)=1, so
F6(1) ?= (2^7 - 35 2^6/6 + 287 2^5/15 - 118 2^4/3 + 691 2^3/15 691 2^2/30)/14
= 2^7/14 - 35 2^6/6x14 + 287 2^5/15x14 - 118 2^4/3x14 + 691 2^3/15x14 691 2^2/30x14
= 2^6/7 - 5 2^4/3 + 41 2^4/15 - 59 2^4/3x7 + 691 2^2/15x7 - 691/15x7
= 16 (4/7 - 5/3 + 41/15 - 59/3x7) + 691 3/15x7
= 16 (4x15 - 5x35 + 41x7 - 59x5)/3x5x7 + 691/5x7
= 16 (60 - 175 + 287 - 295)/3x5x7 + 691/5x7
= 16 (- 123)/3x5x7 + 691/5x7
= 16 (- 41)/5x7 + 691/5x7
= -656/5x7 + 691/5x7
= 35/5x7
= 1
Check.
XVI. Closed form for the Faulhaber coefficients
The Bernoulli expansion for S2q+1(n) is
Fq(M)
 S2q+1(n) = (n+B)2q+2/(2q+2) – B2q+2/(2q+2)
We may substitute for M herein. Expanding must produce terms which agree with
the Faulhaber expansion in terms of M. Substitute
n =
M  1 / 4 + 1/2
giving
Fq(M) = (
M  1 / 4 + 1/2 + B)2q+2/(2q+2) – B2q+2/(2q+2)
Expand binomially:
Fq(M) = (2q+2)-1

i
 2q  2 

 (M + 1/4)(2q+2-i)/2 (1/2 + B)i – B2q+2/(2q+2)
i


Since no square roots involving M appear in the Faulhaber expansion, we expect
to find that the expressions when i is odd will be 0 when we expand. In fact,
this is because the latter factor has a closed form value:
(1/2 + B)i = (21-i - 1) Bi
(cf. Abramowitz & Stegun, 23.?.?).
Since Bi is 0 for all odd i except i=1, and the multiplicative factor is 0 for
i=1, only even values of i survive in the summation. Therefore, substitute
i:=2j immediately:

Fq(M) = (2q+2)-1
j
 2q  2 

 (M + 1/4)q+1-j (21-2j - 1) B2j –
2
j


B2q+2/(2q+2)
Now expand the factor in M + 1/4 binomially:

Fq(M) = (2q+2)-1
j
 2q  2 


2 j


k
q 1

k
j
 Mq+1-j-k (21-2j-2k - 4-k)B2j – B2q+2/(2q+2)

Exchange the inner and outer summation, and define coefficients Cq(i) by
q
Fq(M) :=

Cq(i) Mq+1-i – B2q+2/(2q+2)
i 0
where we substitute k:=i-j and we find a closed form:
i
Cq(i)
:=
(2q+2)-1

j 0
 2q  2   q  1 

 
2 j
 i  j
j  1-2i
 (2
– 4j-i)B2j

Let us compute some of these coefficients.
The largest term in Fq(M) is where
i=0. In this case, the summation definition of Cq(i) is over only j=0, so
Cq(0) = (2q+2)-1 (1)(1)(21 – 40)B0 = (2q+2)-1
which agrees with the formula for aq/(q+1) above. In the next smaller term, for
i=1, the sum over j ranges over the indices 0 and 1, so
 2q  2   q  1 -1
 2q  2   q  -1

 
 (2 – 4-1)B0 + 
   (2 – 40)B2
0
 1 
2
 0

Cq(1) = (2q+2)-1
= (2q+2)-1 [(q+1)(1/4) + ((2q+2)(2q+1)/2)(–1/2)(1/6)]
= 1/8 + (2q+1)(–1/24)
= -(q-1)/12
which agrees exactly with -bq /q above.
Next,
2
Cq(2)
=
(2q+2)-1
= (2q+2)-1
 2q  2   q  1  j  1-4

 
 (2 – 4j-2)B2j
2
j
2

j
j 0 


2q

2
q

1

0


 -3

 
 (2 – 40-2)B0
0
2

0




[
+
 2q  2   q  1  1 -3

 
 (2 – 41-2)B2
2
2

1




+
= (2q+2)-1
[
 2q  2   q  1  2  -3

 
 (2 – 42-2)B4
4
2

2



 q  1 -3

 (2 – 4-2)
2


]
 2q  2   q  -3

   (2 – 4-1)(1/6)
2

 1 
 2q  2  -3
 (2 – 40)(-1/30)
+ 
4


+
]
= (2q+2)-1
[  q2  1 (1/16) +


 2q  2 
 2q  2 

 q(-1/8)(1/6) + 
 (-7/8)(-1/30)
2

4

]
= q(q-2)(7q-1)/720
which agrees exactly with cq /(q-1) above.
This closed form expression for Cq(i) is not completely satisfactory, since it
does not explicitly display the factors of binomial coefficients in q, which we
have deduced experimentally above. Nor does it suggest a form for the
polynomial in q which is a factor in each coefficient.
[Hint, hint.
Prove at least the presence of factor (q above i). Or, try to
compute the last coefficients in each Fq, especially that the last two have a
ratio of -1/2.]