Consecutive natural numbers are added and subtracted according to the following guide:

$$
1+2-3-4+5+6-7-8+9+10-11-12+\ldots,
$$

that is, two positive and two negative addends always repeat.

Determine the value of such an expression whose last term is 2015.

(L. Hozová)
The sums of pairs of adjacent numbers with opposite signs are either -1 or 1. These values also alternate regularly. In the considered expression, several adjacent numbers thus cancel each other out:

$$
\begin{aligned}
& 1+2-3-4+5+6-7-8+9+10-11-12+13+\ldots \\
= & {[1+(2-3)]+[(-4+5)+(6-7)]+[(-8+9)+(10-11)]+[(-12+13)+\ldots} \\
= & {[1-1]+[1-1]+[1-1]+[1-\ldots} \\
= & 0+0+0+\ldots
\end{aligned}
$$

The first zero is obtained as the sum of the first triplet of numbers, and the other zeros are obtained as the sums of consecutive quadruplets of numbers. The sum of the given expression would thus be zero if it ended with the number 3, 7, 11, 15, etc. More generally, the sum of such an expression is zero precisely when it ends with a number that has a remainder of 3 when divided by four. Since the number 2015 has this property, the value of the given expression is zero.

Suggested scoring: 1 point for noticing that the sums of pairs of adjacent numbers with opposite signs are alternately -1 and 1; 2 points for noticing that the sums of consecutive quadruplets, or the first triplet of numbers, are equal to 0; 3 points for correctly determining and justifying the overall sum.

Note. The initial observations can be utilized in various ways, leading to a more detailed discussion of the signs of the numbers at the end of the given expression. From the given solution, it follows that this expression ends as follows:

$$
\ldots+[(-2012+2013)+(2014-2015)].
$$

The addends of the given expression can also be grouped, for example, as follows:

$$
[1+2-3-4]+[5+6-7-8]+[9+10-11-12]+\ldots=-4-4-4-\ldots
$$

Such quadruplets can be formed at most 503 times (2015 = 503 * 4 + 3). The overall sum is expressed according to this method as follows:

$$
503 \cdot(-4)+2013+2014-2015=-2012+2013+2014-2015=0.
$$