12. Find the 2020th term of the following sequence:
$$
1,1,3,1,3,5,1,3,5,7,1,3,5,7,9,1,3,5,7,9,11, \ldots
$$
Answer: 7
Solution: We have that for each $n \in \mathbb{N}$, the $(1+2+\cdots+n)$ th term is $2 n-1$. The first $n+1$ odd positive integers are then listed. Observe that the largest triangular number less than or equal to 2020 is $\frac{63(64)}{2}=2016$. Therefore, the 2020 th term is 7 .