Working through an FP2 question on telescoping sums (one of my favourite topics - although FP2 is full of those), we determined that $r^2 = \frac{\br{2r+1}^3-\br{2r-1}^3-2}{24}$.

Adding these up for $r=1$ to $r=n$ gave the fairly neat result that $24\sum_{r=1}^{n} r^2 = \br{2n+1}^3 - 1 - 2n$.

Now, there are (at least) two ways to go here. The way I would habitually have done this, until fairly recently, was to expand the bracket and then factorise.

That’s not too awful (in fact, the expansion was part a) of the question), but it turns out to be more work than you need to do.

Instead, if you spot that you can write the right-hand side as $\br{2n+1}^3 - (2n+1)$, you can factorise without expanding! There’s clearly a $(2n+1)$ all the way through, so the sum is $\br{2n+1}\br{\br{2n+1}^2-1}$ - and that second bracket is the difference of two squares! The whole thing becomes $\br{2n+1}\br{2n+2}\br{2n}$; bringing out a factor of 4 makes it $4n\br{n+1}\br{2n+1}$, with very little work!

All that remains is to divide the whole thing by 24 to leave the familiar sum-of-squares formula. Sweet.