Find the sum of the series 1 - 2 + 3 - 4 + ... - 98 + 99.

Concept: For an alternating sum of the first n positive integers, S = 1 - 2 + 3 - 4 + ... +/- n, pairing consecutive terms (odd - even) gives -1 per pair; when n is odd, one term is left over, giving the identity S = (n + 1) / 2, and when n is even, S = -n / 2.

Application:

1. The series runs from 1 to 99, so n = 99, an odd number, leaving the last term unpaired.

2. Group the terms in consecutive pairs: (1 - 2), (3 - 4), (5 - 6), ..., (97 - 98); each pair equals -1.

3. There are 98 / 2 = 49 such pairs, so their combined subtotal is 49 x (-1) = -49.

4. The lone unpaired term is 99 (the last term of the series).

5. Add the subtotal and the unpaired term: -49 + 99 = 50.

Cross-check:

Using the identity for odd n directly, S = (n + 1) / 2 = (99 + 1) / 2 = 50, which matches the pairing method, confirming the sum is 50.