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Proof 

Let \(S = \sum_{r=1}^n r^2\).

Then \(S = \\
1+\\
2+2+\\
3+3+3+\\
\cdots \\
n+n+n+\cdots+n\)

Also, \(S = \\
n+\\
n+(n-1)+\\
n+(n-1)+(n-2)+\\
\cdots \\
n+(n-1)+(n-2)+\cdots+2+1\)

And \(S = \\
n+\\
(n-1)+n+\\
(n-2)+(n-1)+n+\\
\cdots \\
1+2+3+\cdots+(n-1)+n\)

So \(S+S+S = \\
(2n+1)+\\
(2n+1)+(2n+1)+\\
(2n+1)+(2n+1)+(2n+1)+\\
\cdots \\
(2n+1)+(2n+1)+(2n+1)+\cdots+(2n+1)+(2n+1)\)

i.e., \(3S = (2n+1) \times \frac{n(n+1)}{2}\) or \(S = \frac{1}{6} n(n+1)(2n+1)\) (via Jeremy Kun)

Proof 

@icecolbeveridge @ColinTheMathmo It's a rather lovely proof. 😀 I've learnt several new ways of proving it today which is rather exciting.

Proof 

@karenshancock The one that @icecolbeveridge has shown us is an example of writing the right thing lots of times and seeing how they connect. In contrast, the idea of writing several terms of the sequence and taking finite differences is perhaps a way of discovering the form of solution that we're looking for.

That technique also works for, say, the Fibonacci sequence - it a good tool for exploring.

Proof 

@icecolbeveridge I love that! It might actually stick in my head, too. I can never remember a quick way of deriving that formula

Proof 

@christianp I love that it has the same flavour as the usual way for proving the sum of integers -- I agree with @ColinTheMathmo that it's probably more of a 'remember' proof than a 'discover' one.

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