# 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,

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.)

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

We have

where $\operatorname{B}_m(x)$ denotes the monic
Bernoulli polynomial
of degree $m$, which has the following useful properties:

Therefore,

Consider the vector space $V_m$ of univariate polynomials $\in\mathbb{Q}[x]$
with degree not exceeding $2m+2$, that are even in $x$ and have a double zero
at $x=\frac{1}{2}$.
Thus $V_m$ has dimension $m$ and is clearly spanned by

For $m>0$, we find that $s_{2m+1}(x-\frac{1}{2})$ has all the properties
required for membership in $V_m$.
Substituting $x-\frac{1}{2}=n$, we conclude that there exists a representation

of $s_{2m+1}(n)$ as a linear combination of squares of sums.