2020 AMC 10A Problem 8

The signs repeat every four terms (plus, plus, plus, minus), so cut the sum into blocks of four.

1. There are $200$ terms, so there are $200 \div 4 = 50$ blocks.
2. Work out the first few blocks:
- $1 + 2 + 3 - 4 = 2$
- $5 + 6 + 7 - 8 = 10$
- $9 + 10 + 11 - 12 = 18$
3. Each block goes up by $8$. That makes sense: every one of the four numbers is $4$ bigger than in the block before, so the block changes by $4 + 4 + 4 - 4 = 8$.
4. The last block is $197 + 198 + 199 - 200 = 394$. (Check: $2 + 49 \times 8 = 394$.)
5. Add the $50$ block values $2, 10, 18, \ldots, 394$. They go up evenly, so their sum is the number of terms times the average of the first and last:
$$50 \times \frac{2 + 394}{2} = 50 \times 198 = 9900.$$

The answer is $\boxed{\textbf{(B)}\ 9,900}$.

When a pattern of signs repeats, grouping one full pattern at a time turns a messy sum into a neat list. Here the list of block sums goes up by the same amount each time, and a list like that adds up to (how many) $\times$ (average of first and last).

Pretend every sign is plus: $1 + 2 + \cdots + 200 = \frac{200 \times 201}{2} = 20100$.

Now fix the multiples of $4$. Each one was added but should be subtracted, so take it away twice. The multiples of $4$ are $4, 8, \ldots, 200$, which is $4 \times (1 + 2 + \cdots + 50) = 4 \times 1275 = 5100$. Taking that away twice:
$$20100 - 2 \times 5100 = 9900.$$

Every fourth term is subtracted, so cut the $200$ terms into $50$ blocks of four. The $k$-th block is
$$(4k-3) + (4k-2) + (4k-1) - 4k = 8k - 6.$$
Check: $k = 1$ gives $1 + 2 + 3 - 4 = 2 = 8 - 6$.

Summing over $k = 1, 2, \dots, 50$:
$$\sum_{k=1}^{50} (8k - 6) = 8 \cdot \frac{50 \cdot 51}{2} - 6 \cdot 50 = 8 \cdot 1275 - 300 = 10200 - 300 = 9900.$$

The answer is $\boxed{\textbf{(B)}\ 9,900}$.