There is a famous proof of the Sum of integers, supposedly put forward by Gauss.
$S=\sum\limits_{i=1}^{n}i=1+2+3+\cdots+(n-2)+(n-1)+n$
$2S=(1+n)+(2+(n-2))+\cdots+(n+1)$
$S=\frac{n(1+n)}{2}$
I was looking for a similar proof for when $S=\sum\limits_{i=1}^{n}i^2$
I've tried the same approach of adding the summation to itself in reverse, and I've found this:
$2S=(1^2+n^2)+(2^2+n^2+1^2-2n)+(3^2+n^2+2^2-4n)+\cdots+(n^2+n^2+(n-1)^2-2(n-1)n$
From which I noted I could extract the original sum;
$2S-S=(1^2+n^2)+(2^2+n^2-2n)+(3^2+n^2-4n)+\cdots+(n^2+n^2-2(n-1)n-n^2$
Then if I collect all the $n$ terms;
$2S-S=n\cdot (n-1)^2 +(1^2)+(2^2-2n)+(3^2-4n)+\cdots+(n^2-2(n-1)n$
But then I realised I still had the original sum in there, and taking that out mean I no longer had a sum term to extract.
Have I made a mistake here? How can I arrive at the answer of $\dfrac{n (n + 1) (2 n + 1)}{6}$ using a method similar to the one I expound on above? I.e following Gauss' line of reasoning?