2020 AMC 10A Problem 8

Group the terms in blocks of four: $(1+2+3-4)$ $+(5+6+7-8)$ $+\cdots$ $+(197+198+199-200)$. The $j$th block is $(4j-3)+(4j-2)$ $+(4j-1)-4j=8j-6$. There are $50$ blocks, so the sum is $\sum_{j=1}^{50}(8j-6)=8\cdot\dfrac{50\cdot51}{2}$ $-6\cdot50=9900$. Thus, B is the correct answer.