What is the $21^{\varrho}$ term of the sequence

$$
1 ; 2+3 ; 4+5+6 ; 7+8+9+10 ; 11+12+13+14+15 ; \ldots ?
$$
Observe that the $21^{\text{st}}$ term is the sum of 21 consecutive numbers. Taking the first term of each term, that is, 1, 2, 4, 7, 11, $16, \ldots$, we have that

$$
\begin{aligned}
2 & =1+1 \\
4 & =2+1+1 \\
7 & =3+2+1+1 \\
11 & =4+3+2+1+1 \\
16 & =5+4+3+2+1+1
\end{aligned}
$$

Thus, the first term of the $21^{\text{st}}$ term is

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

and the $21^{\text{st}}$ term is

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