# Generalisation of the identity $\sum\limits_{k=1}^n {k^3} = \bigg(\sum\limits_{k=1}^n k\bigg)^2$

Are there any generalisations of the identity $\sum\limits_{k=1}^n {k^3} = \bigg(\sum\limits_{k=1}^n k\bigg)^2$ ?

For example can $\sum {k^m} = \left(\sum k\right)^n$ be valid for anything other than $m=3 , n=2$ ?

If not, is there a deeper reason for this identity to be true only for the case $m=3 , n=2$?

• Jun 27 '12 at 12:38
• Related: Proving the original identity without induction and proving it with induction. I am posting this here so that all of these questions are linked together, so that in the case someone finds only this one, they can also find the other two. Aug 30 '12 at 23:23

We can't have a relationship of the form $$\forall n\in\mathbb N^*, \sum_{k=1}^nk^a=\left(\sum_{k=1}^nk^b\right)^c$$ for $a,b,c\in\mathbb N$, except in the case $c=1$ and $a=b$ or $a=3$, $b=1$ and $c=2$. Indeed, we can write $$\sum_{k=1}^nk^a =n^{a+1}\frac 1n\sum_{k=1}^n\left(\dfrac kn\right)^a$$ hence $$\sum_{k=1}^nk^a\;\overset{\scriptsize +\infty}{\large \sim}\;n^{a+1}\int_0^1t^adt=\dfrac{n^{a+1}}{a+1}$$ and if we have the initial equality we should have $a+1 =(b+1)c$ and $a+1=(b+1)^c$. In particular, $(b+1)^{c-1}=c$. If $c>1$, then $c= (b+1)^{c-1}\geq 2^{c-1}\geq c$, and we should have $c=2$ and $b=1$, therefore $a=3$.

• Davide, I tweaked the spacing and sizes in the $\sim$ you had; the previous spacing looked too tight. Hope you don't mind. Sep 4 '11 at 12:49
• @J.M. Thanks, it's more readable thanks to that. Sep 4 '11 at 12:51

The Faulhaber polynomials are expressions of sums of odd powers as a polynomial of triangular numbers $T_n=\frac{n(n+1)}{2}$. Nicomachus's theorem, $\sum\limits_{k\leq n} k^3=T_n^2$, is a particular special case.

Other examples include

\begin{align*}\sum\limits_{k\leq n} k^5&=\frac{4T_n^3-T_n^2}{3}\\\sum\limits_{k\leq n} k^7&=\frac{6T_n^4-4T_n^3+T_n^2}{3}\end{align*}

Not a completely rigorous answer, but you should be able to turn it into one.

By comparing the sums to their corresponding integrals $\int_0^n \mathrm{d}x x^m$, you can see that $$\sum k^m = \frac{1}{m+1} n^{m+1} + \mathcal{O}(n^m).$$ Also, $$(\sum k)^q = \frac{1}{2^{q}} n^{2q} + \mathcal{O}(n^{2q-1}).$$ By comparing leading order terms, equality can only occur if $m+1 = 2q$ and if $m+1 = 2^q,$ which implies that $q = 2$ and $m = 3.$

Here is a curious (and related) identity which might be of interest to you. Let $D_{k} =${ $d$ } be the set of unitary divisors of a positive integer $k$, and let $\sigma_{0}^{*} \colon \mathbb{N} \to \mathbb{N}$ denote the number-of-unitary-divisors (arithmetic) function. Then it is relatively straightforward to prove \begin{eqnarray} \sum_{d \in D_k} \sigma_{0}^{*}(d)^{3} = \left( \sum_{d \in D_k} \sigma_{0}^{*}(d) \right)^{2} \qquad k \in \mathbb{N}. \end{eqnarray}

Note that $\sigma_{0}^{*}(k) = 2^{\omega(k)}$, where $\omega(k)$ is the number distinct prime divisors of $k$. For example, \begin{eqnarray} 1^{3} + 2^{3} + 2^{3} + 2^{3} + 4^{3} + 4^{3} + 4^{3} + 8^{3} = (1 + 2 + 2 + 2 + 4 + 4 + 4 + 8)^{2} \end{eqnarray}