For n=3 (8 by 8) use four sets of the 4 by 4 configuration. Use three of them
in such a way that the three extra squares form one extra triominoe, the fourth
one should have the extra square in the corner.:

For each additional n repeat the n-1 by n-1 configuration four times, rotating in such as way as to create one extra triominoe.

Here is a nice graphic from Chris Johnson that explains the solution more clearly.

Thanks to Guy de Kindler for this one.

Michael Shackleford, A.S.A.