Generalizing the sum of consecutive cubes ∑nk=1k3=(∑nk=1k)2\sum_{k=1}^n k^3 = \Big(\sum_{k=1}^n k\Big)^2 to other odd powers

We have,

nk=1k3=(nk=1k)2

2nk=1k5=(nk=1k)2+3(nk=1k2)2

2nk=1k7=(nk=1k)23(nk=1k2)2+4(nk=1k3)2

and so on (apparently).

Is it true that the sum of consecutive odd m powers, for m>1, can be expressed as sums of squares of sums* in a manner similar to the above? What is the general formula?

*(Edited re Lord Soth’s and anon’s comment.)

Answer

This is a partial answer, it just establishes the existence.

We have
sm(n)=nk=1km=1m+1(Bm+1(n+1)Bm+1(1))
where Bm(x) denotes the monic
Bernoulli polynomial
of degree m, which has the following useful properties:
x+1xBm(t)dt=xm(from which everything else follows)Bm+1(x)=(m+1)Bm(x)Bm(x+12){is even in xfor even mis odd in xfor odd mBm(0)=Bm(1)=0for odd m3

Therefore,
sm(n)has degree m+1 in nsm(0)=0sm(0)=Bm(1)=0for odd m3(This makes n=0 a double zero of sm(n) for odd m3)sm(x12){is even in xfor odd mis odd in xfor even m2

Consider the vector space Vm of univariate polynomials Q[x]
with degree not exceeding 2m+2, that are even in x and have a double zero
at x=12.
Thus Vm has dimension m and is clearly spanned by
{s2j(x12)j=1,,m}
For m>0, we find that s2m+1(x12) has all the properties
required for membership in Vm.
Substituting x12=n, we conclude that there exists a representation
s2m+1(n)=mj=1am,js2j(n)for m>0 with am,jQ
of s2m+1(n) as a linear combination of squares of sums.

Attribution
Source : Link , Question Author : Tito Piezas III , Answer Author : ccorn

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