Problem: What is the value of
(22−2)−(32−3)+(42−4)?\left(2^{2}-2\right)-\left(3^{2}-3\right)+\left(4^{2}-4\right)? (22−2)−(32−3)+(42−4)?
Answer Choices:
A. 111 B. 222 C. 555 D. 888 E. 121212
Solution:
(22−2)−(32−3)+(42−4)=2−6+12= (D) 8\left(2^{2}-2\right)-\left(3^{2}-3\right)+\left(4^{2}-4\right)=2-6+12=\text { (D) } \boxed{8}(22−2)−(32−3)+(42−4)=2−6+12= (D) 8.
The problems on this page are the property of the MAA's American Mathematics Competitions