Determine the value of the expression

$$
1+2-3+4+5-6+7+8-9+10+11-12+\cdots+94+95-96+97+98-99
$$

(The expression consists of 99 terms. The operations alternate between two additions and one subtraction.)
Removing the initial 0 , the remaining 99 terms can be written in groups of the form $(3 k-2)+$ $(3 k-1)-3 k$ for each $k$ from 1 to 33 .

The expression $(3 k-2)+(3 k-1)-3 k$ simplifies to $3 k-3$.

Therefore, the given sum equals

$$
0+3+6+\cdots+93+96
$$

Since $k$ ran from 1 to 33 , then this sum includes 33 terms and so equals

$$
\frac{33}{2}(0+96)=33(48)=33(50)-33(2)=1650-66=1584
$$

ANSWER: 1584