We can find \(d\) if we know what \(a_0\) is. Working backwards from the third differences, we find \(a_0 = 0\) (unsurprisingly, since there are no squares on a \(0\times 0\) chessboard). Thus \(d = 0\text{.}\) Now plug in \(n = 1\text{,}\) \(n =2\text{,}\) and \(n =3\text{:}\)