[Ne[S]zA]

$$\dfrac{\dfrac{n(n+1)(2n+1)}{6}}{\dfrac{n(n+1)(2n+1)}{6}-\dfrac{n(n+1)}{2}}=\dfrac{231}{228}$$
$$\dfrac{2n+1}{n-1}=\dfrac{231}{114}$$
$$n=115$$