Tiling a 2×N rectangle with 1×1 and 2×1 tiles

Question

I am trying to prove why this relation is true:

I think this is some exhaustive method, but I am not quite seeing how.

I believe you can start by assuming, there is no tile that is allowed to be placed in a vertical orientation, so you get the t(n)*t(n) solution by the multiplication principle, that takes care of the part outside of the summation. Then you have to consider the case where you have a 2x1 space that is allowed a vertical tile placement, then 2x2 2x3 and so on... This is what I've been trying to play with and I can't put it together,I don't see how that's exhaustive, please help.

note: That 1x2 tile can be placed as a 2x1 tile as well.

Answer 1

Each element of the summation corresponds to placing the first vertical $2\times 1$ tile at position $k$ along the grid. In each case considered, before that position, the options are all those for the two independent rows of length $k{-}1$; after that, it is for the shorter $2\times n $ grid with $d(n-k)$ options.