3. From the set of natural numbers, the following sequence was formed: $1, 2+3, 4+5+6, 7+8+9+10$, and so on, where each subsequent sum has one more addend. What is the 2023rd term of the sequence?
Answer: $\frac{2023 *\left(2023^{2}+1\right)}{2}=4139594095$.

Solution. Let the $n$-th group start with the number $k$. Then these numbers are:

$k, k+1, \ldots k+(n-1)$ - a total of $n$ numbers. Their sum is the sum of an arithmetic progression with the first term equal to $k$, a common difference of 1, and can be calculated using the formula:

$$
S_{n}=\frac{k+(k+n-1)}{2} \cdot n=\frac{2 k+n-1}{2} \cdot n
$$

Notice that the number $k-1$, which precedes $k$, can be counted as the total number of all numbers in the groups from the 1st to the $(n-1)$-th. Since, by condition, the first group has 1 number, the second group has 2 numbers, and so on, the total number of numbers in the first $(n-1)$ groups will be: $\frac{1+(n-1)}{2} *(n-1)=\frac{n(n-1)}{2}=k-1$ Therefore, the desired formula is:

$$
\begin{gathered}
S_{n}=\frac{2 k+n-1}{2} * n=\frac{2\left(1+\frac{n(n-1)}{2}\right)+n-1}{2} * n=\frac{n\left(n^{2}+1\right)}{2}, S_{2023}=\frac{2023 \cdot\left(2023^{2}+1\right)}{2} \text { (no need to } \\
\text { compute further!) }
\end{gathered}
$$

Criteria. A correctly derived and justified answer - 7 points.

1 computational error - -1 point. The formulas for the arithmetic progression are used without mentioning the arithmetic progression, the score is not reduced. The formula for $S_{n}$ is derived or the number $\mathrm{k}$ (or $k-1$) is calculated - 2 points. Case analysis - 0 points.