1A. Determine the 2015th term in the sequence $1,2,-2,3,-3,3,4,-4,4,-4,5,-5,5,-5,5, \ldots$
Solution. Let the 2015th term in the sequence be the number $k$. Note that the last occurrence in the sequence of a number with an absolute value of 2 has an order number of 3; the number with an absolute value of 3 has an order number of 6; the number with an absolute value of 4 has an order number of 10, and so on. It is easy to prove that the last occurrence of a number whose absolute value is $k$ has an order number of $\frac{k(k+1)}{2}$. Since $\frac{62 \cdot 63}{2}=1953$ and $\frac{63 \cdot 64}{2}=2016$, it follows that the number at the 2015th position in the sequence is

$$
(-1)^{2015-1953+1} 63=-63
$$