We have,

n∑k=1k3=(n∑k=1k)2

2n∑k=1k5=−(n∑k=1k)2+3(n∑k=1k2)2

2n∑k=1k7=(n∑k=1k)2−3(n∑k=1k2)2+4(n∑k=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)=n∑k=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)B′m+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 m≥3

Therefore,

sm(n)has degree m+1 in nsm(0)=0s′m(0)=Bm(1)=0for odd m≥3(This makes n=0 a double zero of sm(n) for odd m≥3)sm(x−12){is even in xfor odd mis odd in xfor even m≥2

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(x−12)∣j=1,…,m}

For m>0, we find that s2m+1(x−12) has all the properties

required for membership in Vm.

Substituting x−12=n, we conclude that there exists a representation

s2m+1(n)=m∑j=1am,js2j(n)for m>0 with am,j∈Q

of s2m+1(n) as a linear combination of squares of sums.

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