Find the 21st term of the sequence that starts like this:

$$
1 ; 2+3 ; 4+5+6 ; 7+8+9+10 ; 11+12+13+14+15 ; \ldots
$$
Observe that the 21st term is the sum of 21 consecutive numbers. To find out what these numbers are, we just need to find the first term of the 21st term, which is the sum of 21 consecutive numbers. We observe that, starting from the second term, the first term of

$$
\begin{aligned}
2^{\text{nd}} \text{ term is } 2 & =1+1, \text{ that of } \\
3^{\text{rd}} \text{ term is } 4 & =2+1+1, \text{ that of } \\
4^{\text{th}} \text{ term is } 7 & =3+2+1+1, \text{ that of } \\
5^{\text{th}} \text{ term is } 11 & =4+3+2+1+1, \text{ that of } \\
6^{\text{th}} \text{ term is } 16 & =5+4+3+2+1+1,
\end{aligned}
$$

and so on. Therefore, we can see that the first term of the $21^{\text{st}}$ term is

$$
20+19+\cdots+3+2+1+1=\frac{21 \times 20}{2}+1=211
$$

and that, therefore, the 21st term is

$$
211+212+\cdots+230+231=\frac{1}{2}(211+231) \times 21=221 \times 21=4641
$$