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 \(a_n = \frac{1}{3}n^3 + \frac{1}{2}n^2 + \frac{1}{6}n = \frac{1}{6}n(n+1)(2n+1)\text{.}\)
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