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!
 A: We have
$$
\sum_{k=1}^{n}k^3 = 1 + 8 + 27 + \ldots + n^3 = \\
\underbrace{1}_{1^3} + \underbrace{3+5}_{2^3} + \underbrace{7 + 9 + 11}_{3^3} + \underbrace{13 + 15 + 17 + 19}_{4^3} + \ldots = \\
\underbrace{\underbrace{\underbrace{1}_{1^2} + 3}_{2^2} + 5}_{3^2} + \ldots
$$
which is
$$
\left( \sum_{k=1}^{n}k \right)^2
$$
A: Maybe this will help you visualize it:

Source.
or this one which is clearer:
$\phantom{XXXXXXXX}$
A: Here are some 'Proof without Words' for this identity;
As Hakim showed (but this one might be slightly clearer);

Here's another very clean illustration by Anders Kaseorg;

Here, the total top volume is undoubtedly $1^2+2^2+3^3+\cdots +n^2,$ while on the volume of the bottom part is;
$$1\left(1+2+\cdots+n\right)+2\left(1+2+\cdots+n\right)+\cdots +n\left(1+2+\cdots+n\right)=\left(1+2+\cdots+n\right)^2.$$
A: 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
$.
