2020 AMC 10A Problem 8

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

A repeating sign pattern of period $4$ turns the sum into $50$ copies of one simple formula. Writing the block in terms of $k$ converts an irregular-looking series into an arithmetic series you can sum with standard tools. Verifying the block formula on $k = 1$ catches off-by-one errors.

Add everything, then correct the subtracted terms: $1 + 2 + \cdots + 200 = 20100$, and each multiple of $4$ was added when it should have been subtracted, so remove $2(4 + 8 + \cdots + 200) = 2 \cdot 4 \cdot 1275 = 10200$. Result $20100 - 10200 = 9900$.

• Treating the pattern as plus-plus-plus-minus over 200 terms but mis-tracking the sign of every fourth term, or summing 1 to 200 without subtracting 2 times each multiple of 4.
• Subtracting the multiples of $4$ only once (giving $20100 - 5100 = 15000$) instead of twice.