A conjecture about perfect numbers

Perfect number

In number theory the sum of the divisors is denoted as $\sigma$: $$\sigma(n) = \Sigma_{d/n} d$$ and $s(n)=\sigma(n) - n$ is the sum of the proper divisors. A perfect number is equal to the sum of its proper divisors. All known perfect numbers are even, it is unknown if odd perfect numbers exist. The number $2^{p-1}(2^p-1)$ is perfect if and only if $(2^p-1)$ is prime.

Conjecture

Show that: if $p$ is odd then
$$ 2^{p-1}(2^p-1) = \sum_{k=1}^{\frac{p+1}{2}-1} (2k-1)^3$$
( Notice that $2^{p-1}(2^p-1)$ yields a perfect number if $(2^p-1)$ is prime. )

Example

$6$ is perfect, since $6 = 1 + 2 + 3.$
$28$ is perfect, since $28= 1 + 2 + 4 + 7 + 14.$

Any perfect number ( except 6 ) can be represented as a sum of cubes.
$\begin{array}{ccc}
\underline{p} & \underline{Pf} &\underline{s}\\
3 & 28 & 1^3 + 3^3 \\
5 & 496 & 1^3 + 3^3 + 5^3 + 7^3 \\
7 & 8128 & 1^3 + 3^3 + ... + 15^3 \\
13 & 33550336 & 1^3 + 3^3 + ... + 127^3
\end{array}$

Proof

My exercise for New Year's Day. ( You may have noticed that I like doing 'sums'. ) Later...

( Source:
- A primer of analytic number theory, From Pythagoras to Riemann by Jeffrey Stopple
)