# 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?

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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.

There is a more beautiful Gauss-style proof that involves writing the numbers in triangles instead of in a line.

I leave the details to you.

HINT: $(k + 1)^3 – k^3 = 3k^2 + 3k + 1$. Telescope the left hand side, solve for $k^2$.

If you need more of a hint I’ll be glad to elaborate later. In case you’d like a reference, this is one of the first exercises in Spivak’s Calculus (I don’t have the latest edition, but it’s in the section “Numbers of Various Sorts.”)

EDIT

Since you’re only interested in the “Gaussian” method of summing this series, I suggest you take a look at this Wikipedia article on Arithmetic progression. It shows how you can use this specific trick for finding the sum of arbitrary arithmetic series. Unfortunately, your sum is not of this kind, so cannot be summed by that simple method.

I have no doubt that if you fumble around with the series for long enough, you’ll encounter some trick that will allow you to sum it (maybe the fact that the sum of the first $n$ odd numbers is a square?). No doubt a lot of research has been done on the so-called square pyramidal numbers (check out the list of references!) The Wikipedia entry on them has a picture of what you’re actually summing (finding the number of balls in a square bottomed pyramid), so maybe you can see why they aren’t as easy to sum as the triangular numbers, which can easily be arranged into squares. The MathOverflow link by aelguindy gives a “visual proof” of how the formula is derived.

Sorry I could not be of any more help.

Since I think the solution Tyler proposes is very useful and accesible, I’ll spell it out for you:

We know that

$$(k+1)^3-k^3=3k^2+3k+1$$

If we give the equation values from $1$ to $n$ we get the following:

$$(\color{red}{1}+1)^3-\color{red}{1}^3=3\cdot \color{red}{1}^2+3\cdot \color{red}{1}+1$$
$$(\color{red}{2}+1)^3-\color{red}{2}^3=3 \cdot \color{red}{2}^2+3 \cdot \color{red}{2}+1$$
$$(\color{red}{3}+1)^3-\color{red}{3}^3=3 \cdot \color{red}{3}^2+3 \cdot \color{red}{3}+1$$
$$\cdots=\cdots$$
$$(\color{red}{n-1}+1)^3-(\color{red}{n-1})^3=3(\color{red}{n-1})^2+3(\color{red}{n-1})+1$$
$$(\color{red}{n}+1)^3-\color{red}{n}^3=3\color{red}{n}^2+3\color{red}{n}+1$$

We sum this orderly in columns.

Note that in the LHS the numbers cancel out with each other, except for the $(n+1)^3$ and the starting $-1$ ($2^3-1^3+3^3-2^3+4^3-3^3+\cdots+n^3-(n-1)^3+(n+1)^3-n^3$). We get:

$$(n+1)^3-1 = 3(1+2^2+3^2+\cdots +(n-1)^2+n^2)+ 3(1+2+3+\cdots +(n-1)+n)+(\underbrace{1+1+\cdots+1}_{n})$$

We can write this in sigma notation as:

$$(n+1)^3-1=\sum\limits_{k=1}^n(3k^2+3k+1)$$

Naming our sum $S$ we have that:

$$(n+1)^3-1=3S+\sum\limits_{k=1}^n(3k+1)$$

We know how to compute the sum in the RHS, because

$$\sum\limits_{k=1}^n 3k =3\frac{n(n+1)}{2}$$

$$\sum\limits_{k=1}^n 1 =n$$

(We’re summing $n$ ones in the last sum.)

$$(n+1)^3-1=3S+3 \frac{n(n+1)}{2}+n$$

$$n^3+3n^2+3n=3S+\frac{3}{2}n^2+\frac{3}{2}n+n$$

$$n^3+\frac{3n^2}{2}+\frac{n}{2}=3S$$

$$\frac{n^3}{3}+\frac{n^2}{2}+\frac{n}{6}=S$$

This factors to

$$\frac{n(2n+1)(n+1)}{6}=S$$

which is what you wanted.