Neerajan and Eilidh both noticed that the number of L-triominoes needed to tile a size $n$ L-triominoes was equal to $n^2$.

Philip from Wilson's School explained his thinking:

To tile a size 4 triomino you would simply put 4 size 2 triominoes together the same way you put four size 1 triominoes together to make a size 2.

Therefore you would put 4 size 4 triominoes together in the same way to form a size 8 triomino. Then 4 size 8 triominoes to make a size 16, and so on.

The triominoes could therefore tile size 8, 16, 32... $2^n$. These can therefore be made up of size 1 triominoes.

As it takes $4^1$ size 1 triominoes to tile a size 2 triomino, and $4^2$ size 1 triominoes to tile a size 4 triomino, it would therefore take $4^3=64$ size-1 triominoes to tile a size 8 triomino.
It would then take $4^4=256$ size 1 triominoes to tile a size 16 triomino and $4^5=1024$ size 1 triominoes to tile a size 32 triomino.
Finally, it would take $4^x$ triominoes to tile a size $2^x$ triomino.

To tile the size 3 triomino you simply use the 2x3 rectangular blocks at the top and on the right and then the rest you use a size 2 triomino and the size 1 in the middle.

To tile a size 5 triomino you simply add another size 2 triomino in the bottom left and for the rest you can use the 2x3 rectangular boxes. It's the same with a size 7 triomino, add a size 2 and then use 2x3 boxes to tile the rest. This works with all odd numbers.

Therefore all size triominoes can be tiled. If the size number is even you use the first method and if the size number is odd you use the second.

Isaac from Hampton School also sent us his solution, which you can read here.