The maverick nature of the sum of squares proof If you do not know how to prove $\sum_{i=0}^n i^2 = \frac{(n^2+n)(2n+1)}{6}$, please try it before reading further.
I managed to prove that $\sum_{i=0}^n i = \frac{(n^2+n)}{2}$, using the delightful finding the constants from the first and latter halves of the sum cancel each other. However, when consequently attempting to prove the sum of squares, I found to my great consternation that the same trick does not work, as the squaring of the parts prevents the latter half being negative. When I finally gave up and looked at the derivation, its literally geometric nature surprised me. 


*

*Is using such an approach to proving an algebraic formula considered ‘rigorous’?

*If the three triangles were manifested as a grid of translucent material, with the opacity of each square given by the normalized values of the triangle elements in the proof, then the fact that there is a stacking configuration of the three plates where every sum equals $2n + 1$ would be visually intuitive, and could even be discovered by accident. But since the genius who discovered this method of proof most likely did not have such lovely artefacts to play with, I wonder how he/she did it. What could his/her thought process have been?
I would appreciate any other comments about the derivation as well.
 A: Call your sum $P_n$ so$$P_n=\sum_{i=1}^ni^2=\sum_{i=1}^n\sum_{j=1}^i(2j-1)=\sum_{j=1}^n\sum_{i=j}^n(2j-1)=\sum_{j=1}^n(2j-1)(n+1-j)\\=\sum_{j=1}^n(2j-1)(n+1)-\sum_{j=1}^n(2j-1)j=n^2(n+1)-2P_n+\frac12n(n+1).$$Therefore$$3P_n=\frac12n(n+1)(2n+1).$$
A: I like using
ascending factorials,
$p_n(x)
=\prod_{k=0}^{n-1}(x+k)
$.
These have the nice property that
$\begin{array}\\
p_n(x+1)-p_n(x)
&=\prod_{k=0}^{n-1}(x+1+k)-\prod_{k=0}^{n-1}(x+k)\\
&=\prod_{k=1}^{n}(x+k)-\prod_{k=0}^{n-1}(x+k)\\
&=(x+n)\prod_{k=1}^{n-1}(x+k)-x\prod_{k=1}^{n-1}(x+k)\\
&=((x+n)-x)\prod_{k=1}^{n-1}(x+k)\\
&=n\prod_{k=0}^{n-2}(x+1+k)\\
&=np_{n-1}(x+1)\\
\text{or}\\
\dfrac{p_n(x+1)-p_n(x)}{n}
&=p_{n-1}(x+1)\\
\end{array}
$
This gives the nice
telescoping sum
$\begin{array}\\
\sum_{x=1}^{m} p_n(x)
&=\sum_{x=0}^{m-1} p_n(x+1)\\
&=\sum_{x=0}^{m-1}\dfrac{p_{n+1}(x+1)-p_{n+1}(x)}{n+1}\\
&=\dfrac{p_{n+1}(m)-p_{n+1}(0)}{n+1}\\
&=\dfrac{p_{n+1}(m)}{n+1}\\
\end{array}
$
Since
$p_1(x) = x,
p_2(x) = x(x+1),
p_3(x) = x(x+1)(x+2)
$,
this gives,
for $n = 1$,
$\begin{array}\\
\sum_{x=1}^{m} x
&=\sum_{x=1}^{m} p_1(x)\\
&=\dfrac{p_{2}(m)}{2}\\
&=\dfrac{m(m+1)}{2}\\
\end{array}
$
and,
for $n = 2$,
$\begin{array}\\
\sum_{x=1}^{m} x(x+1)
&=\sum_{x=1}^{m} p_2(x)\\
&=\dfrac{p_{3}(m)}{3}\\
&=\dfrac{m(m+1)(m+2)}{3}\\
\text{so}\\
\sum_{x=1}^{m} x^2
&=\sum_{x=1}^{m}( x(x+1) -x)\\
&=\sum_{x=1}^{m} x(x+1)-\sum_{x=1}^{m} x\\
&=\dfrac{m(m+1)(m+2)}{3}-\dfrac{m(m+1)}{2}\\
&=\dfrac{2m(m+1)(m+2)-3m(m+1)}{6}\\
&=\dfrac{m(m+1)(2(m+2)-3)}{6}\\
&=\dfrac{m(m+1)(2m+1)}{6}\\
\end{array}
$
Continuing this further
leads to the
unsigned Stirling numbers.
