Intuitive ways to get formula of cubic sum Is there an intuitive way to get cubic sum? From this post: combination of quadratic and cubic series and Wikipedia: Faulhaber formula, I get $$1^3 + 2^3 + \dots + n^3 = \frac{n^2(n+1)^2}{4}$$
I think the cubic sum is squaring the arithmetic sum $$1^3 + 2^3 + \dots + n^3 = (1  + 2 + \dots + n)^2$$
But how to prove it? Please help me. Grazie!

Here's a proof by induction
I came up with
about two years ago.
Let
$s_k(n)
=\sum_{i=1}^n i^k
$.
Want to show that
$s_3(n) = (s_1(n))^2
$.
If you want to
prove by induction
that $a(n) = b(n)$,
there are two possibilities:
(1) Show
$a(0)-b(0) = 0$
and
$a(n)-b(n) = 0
\implies
a(n+1)-b(n+1) = 0$;
(2) show
$a(0) = b(0)$
and
$a(n) = b(n)
\implies
a(n+1)-a(n)
=b(n+1)-b(n)
$.
These are,
of course,
equivalent,
but I find that (2)
is often easier
since the math
is simpler
in the initial parts.
In this case,
$s_3(n+1)-s_3(n)
=(n+1)^3
=n^3+3n^2+3n+1
$
and
$(s_1(n+1))^2-(s_1(n))^2
=(s_1(n)+n+1)^2-(s_1(n))^2\\
=s_1^2(n)+2(n+1)s_1(n)+(n+1))^2-(s_1(n))^2\\
=2(n+1)s_1(n)+n^2+2n+1\\
$
so we want to show that
$n^3+3n^2+3n+1
=2(n+1)s_1(n)+n^2+2n+1
$
or
$n^3+2n^2+n
=2(n+1)s_1(n)
$
or
$n(n^2+2n+1)
=2(n+1)s_1(n)
$
or
$n(n+1)^2
=2(n+1)s_1(n)
$
or
$s_1(n)
=\dfrac{n(n+1)}{2}
$.
This can either
be assumed known
or, in turn,
proved by induction using
$\dfrac{(n+1)(n+2)}{2}
-\dfrac{n(n+1)}{2}
=\dfrac{n^2+3n+2-(n^2+n)}{2}
=\dfrac{2n+2}{2}
=n+1
$.
