We have seen that the sequence \(1, 5, 14, 30, 55, \ldots\) is \(\Delta^3\)-constant, so we are looking for a degree 3 polynomial. That is,

\begin{equation*} a_n = an^3 + bn^2 + cn + d\text{.} \end{equation*}

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 \(n =3\text{:}\)

\begin{align*} 1 = \amp a + b + c\\ 5 = \amp 8a + 4b + 2c\\ 14 = \amp 27a + 9b + 3c\text{.} \end{align*}

If we solve this system of equations we get \(a = \frac{1}{3}\text{,}\) \(b = \frac{1}{2}\) and \(c = \frac{1}{6}\text{.}\) Therefore the number of squares on an \(n \times n\) chessboard is \(a_n = \frac{1}{3}n^3 + \frac{1}{2}n^2 + \frac{1}{6}n = \frac{1}{6}n(n+1)(2n+1)\text{.}\)