Solution. Note that in the $n$-th group there are $n$ terms and the first one is $\frac{n(n-1)}{2}+$ 1. The last term of the $n$-th group is $\frac{n(n-1)}{2}+1+(n-1)=\frac{n^{2}+n}{2}$. Therefore, $S_{n}=\left(\frac{n(n-1)}{2}+1+\frac{n(n+1)}{2}\right) \cdot \frac{n}{2}=\frac{n\left(n^{2}+1\right)}{2}$. Hence, $S_{16}-S_{4}-S_{1}=$ $2056-34-1=2021$
1. Let's divide the sequence of natural numbers into groups:

$(1),(2,3),(4,5,6)(7,8,9,10), \ldots$

Denote $S_{n}$ as the sum of the $n$-th group of numbers. Find $S_{16}-S_{4}-S_{1}$.