2021 AMC 10A Problem 1

Parentheses always come first, so work out each group on its own and only then combine them.

1. First group: $2^2 - 2 = 4 - 2 = 2$.
2. Second group: $3^2 - 3 = 9 - 3 = 6$.
3. Third group: $4^2 - 4 = 16 - 4 = 12$.
4. Now put the three results back with the signs that sit between the groups. The second group is subtracted and the third is added:
$$2 - 6 + 12 = 8.$$

The answer is $\boxed{\textbf{(D)}\ 8}$.

A long expression becomes easy once you break it into small pieces and handle one piece at a time. The minus sign in front of the second group belongs to that group only; the third group has its own plus sign.

Notice also that each group has the form $n^2 - n = n \times (n - 1)$: the results $2 = 2 \times 1$, $6 = 3 \times 2$ and $12 = 4 \times 3$ are products of two neighbouring whole numbers. That pattern comes up often on the AMC.

Work inside each pair of parentheses first:
$$2^2 - 2 = 4 - 2 = 2,\qquad 3^2 - 3 = 9 - 3 = 6,\qquad 4^2 - 4 = 16 - 4 = 12.$$
Now combine them with the signs given in the expression:
$$2 - 6 + 12 = 8.$$

The answer is $\boxed{\textbf{(D)}\ 8}$.