algebra precalculus - Gaussian proof for the sum of squares?

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?

Answer

You can use something similar, though it requires work at the end.

If $S_n = 1^2 +2^2 + \cdots + n^2$ then
$$S_{2n}-2S_n = ((2n)^2 - 1^2) + ((2n-1)^2-2^2) +\cdots +((n+1)^2-n^2)$$

$$=(2n+1)(2n-1 + 2n-3 + \cdots +1) = (2n+1)n^2$$ using the Gaussian trick in the middle.

Similarly $$S_{2n+1}-2S_n = (2n+1)(n+1)^2$$

So for example to work out $S_9$, you start

$$S_0=0^2=0$$

$$S_1=1 + 2S_0 = 1$$

$$S_2=3+2S_1=5$$

$$S_4=25+2S_2=30$$

$$S_9 = 225+2S_4 = 285$$

but clearly there are easier ways.