We choose $n$ points on a circle and draw all the associated chords (we make sure that 3 chords are never concurrent). Into how many parts is the circle divided?
As in the previous exercise, we will count how many sectors are added when placing the $(n+1)$-th point. Number the vertices from 1 to $n$ and place an additional vertex between point 1 and point $n$. Consider the chord $C_{k}$ connecting the new point to point $k$. Another chord $\Gamma$ in the figure intersects $C_{k}$ if and only if one of its endpoints is to the right of $C_{k}$ and the other is to the left. To find a chord that intersects $C_{k}$, we need to choose one point from the $k-1$ points to the right and one from the $n-k$ points to the left, which gives $(k-1)(n-k)$. And as in the previous exercise, this means that the chord $C_{k}$ adds $(k-1)(n-k)+1$ sectors. Therefore,

$u_{n+1}-u_{n}=\sum_{k=1}^{n}(k-1)(n-k)+1=(n+1) \sum_{k=1}^{n} k-\sum_{k=1}^{n} k^{2}-n(n-1)=u_{n}+\frac{n^{3}-3 n^{2}+8 n}{6}$.

We initialize with $u_{0}=u_{1}=1$ and we find:

$$
u_{n}=\frac{n(n-1)\left(n^{2}-5 n+18\right)}{24}+1=\frac{n^{4}-6 n^{3}+23 n^{2}-18 n+24}{24}
$$