6.2. Calculate: $1+2-3-4+5+6-7-8+\ldots+1982-1983-$ $-1984+1985+1986$.

$$
(4-6 \text { grades })
$$
6.2. Without touching the first and last terms of the given sum, we will group the other terms into sets of four consecutive terms, namely: $1+(2-3-4+5)+(6-7-8+9)+\ldots+$ ( $1982-1983-1984+1985)+1986$. It is easy to notice that the sum in each bracket is zero. Therefore, the total sum is $1+1986=1987$.